RMT 2014
Calculus Test Solutions
February 15, 2014
1. Let f ( f (x) = x 4 and let g(x) = x −4 . Compute f Compute f (2)g (2)g (2). Answer: Answer: 15
note that f (x) = 12x 12x2 and g (x) = 20x 20x−6 . Then Then f (x)g (x) = 20 12 x −4 . Plugging in x = 2 we get f get f (2)g (2)g (2) = 1216·20 = 3 5 = 15 .
Solution: We
· ·
·
2. There is a unique positive positive real number number a a such that the tangent line to y to y = x 2 + 1 at x at x = = a a goes through the origin. Compute a. Answer: Answer: 1
slope of the tangent line is 2a 2a. The equation equation for the tangent tangent line is (y (y (a2 +1)) = = y = = 0 gives us a2 1 = 2a2 , which has solution a = 1 . a). Setting x = y
Solution: The
2a(x
−
− −
−
−
3. Moor has $1000, $1000, and he is playing playing a gambling gambling game. game. He gets to pick a number number k between 0 and 1 (inclu (inclusiv sive). e). A fair coin is then then flipped. flipped. If the coin comes up heads, heads, Moor is given given 500 50000k additional additional dollars. dollars. Otherwise, Otherwise, Moor loses 1000k 1000k dollars dollars.. Moor’s Moor’s happines happinesss is equal to the log of the amount amount of money money that he has after after this this gam gambli bling ng game. Find Find the value value of k of k that Moor should select to maximize his expected happiness. Answer: Solution:
2 5
Suppose that Moor chooses a value value of k. k . We write down the expected value of Moor’s
happiness. If the coin comes up heads, Moor now has 1000 + 1000(5k 1000(5k) = 1000(5k 1000(5k + 11)) dollars dollars.. If the coin coin comes up tails, Moor now has 1000 1000 1000k k = 1000(1 k) dollars. Therefore, Therefore, the expected value value of Moor’s happiness is
−
−
1 1 H (k) = log(1000(5k log(1000(5k + 1)) + log(1000(1 k )). )). 2 2 We want to maximize this. To do this, we differentiate, differentiate, set the derivative derivative equal equal to zero, and look for critical values. Here,
−
1 H (k) = 2
5000 1000(5k 1000(5k + 1)
−
1000 1 = 1000(1 k ) 2
−
5 5k + 1
1
− 1 − k
= 0
10k = 4, and hence k hence k = 25 is the only critical value. − k), so 10k The maximal value of H (k) for k ∈ [0, [0, 1] must occur either at a critical value or an endpoint. 2 2 when 5k 5k + 1 = 5(1
Observe that among the three values H (0), (0), H (1), (1), and H ( 5 ), the largest is H ( 5 ). Theref Therefore ore,, 2 Moor maximizes his happiness by selecting k selecting k = . 5
4. The set of points (x, (x, y) in the plane satisfying x2/5 + y = 1 form a curve enclosing a region. Compute the area of this region.
||
Answer:
8 7
Solution: The
set of points satisfying the equation form a closed curve that encloses a region. Observe that this curve is preserved if we transform x x or y y , so it is symmetric in all 4 quadrants. In particular, we can find the area in the first quadrant, where x where x,, y > 0. > 0. In the 2 / 5 quadrant, we can rewrite our equation as y as y = 1 x . This curve curve intersects intersects the coordinate axes at (0, (0, 1) and (1, (1, 0), and it is continuous, so the area is
→ −
→ −
−
1
A =
0
RMT 2014
1
2 dx = . − x2/5 dx = 7
Calculus Test Solutions
February 15, 2014
RMT 2014
Calculus Test Solutions
February 15, 2014
The total area is therefore 4A 4A = 8/7 . 5. Compute Compute the improper integral integral
− − 2
4
x
x
x
0
4
−x
dx.
Answer: Answer: 4 Solution 1: First
have
4−x x
of all, we note the many symmetries of the given expression. Specifically, we
and we subtract subtract its reciprocal. We also recall that square roots, when we take their
√ √ − −
derivative, give us their reciprocal. This inspires the guess that the function f ( f (x) = is somehow important to our integral. Indeed, we find that f that f (x) =
− 2 0
4−x x
√ √ − x 2 = 2(2 − 0) =
x
dx = 2 x 4 4−x dx =
Solution Solution 2: Although
0
4.
1 2
4−x x
x 4−x
x 4
x
so that
solution 1 is, perhaps, the prettiest way of solving this problem, it is not necessar necessarily ily easy to notice notice.. A more more direct direct approac approach h uses uses a trig trig substit substituti ution. on. Specific Specifically ally,, 4−x 2 2 noting the importance of x and remembering remembering the Pythagorean Pythagorean identity identity sin x + cos x = 1, 2
− 4−x x
it makes sense to try the substitution x = 4sin θ. Then 2
2
=
dx = dx = 8 sin sin θ cos θ dθ, dθ, 4 sin sin θ = 0 when θ when θ = 0 and 4sin θ = 2 when θ when θ = π
4
π
(cot θ
0
dθ = 8 − tan θ) · 8sin θ cos θ dθ =
4
π 4.
cos2 x = sin2 x
cot x. Also, lso,
The integral becomes π
2
cos θ
0
4
2
sin θ dθ = dθ = 8
cos2θdθ. cos2θdθ.
0
This last integral may be easily computed by the substitution 2θ 2θ
θ : → θ:
π
8
π
4
cos2θ cos2θ dθ = dθ = 4
0
2
0
π
cos θ dθ = dθ = 4(sin θ) 0 = 4(1 2
4 . − 0) = 4 .
Solution Solution 3: The
simplest way to solve this problem is perhaps to write the integrand with a common denominator. This gives
√ √ −
√ x 2 2 (4 − x) − x 4 − 2x √ √ √ dx = dx = dx = dx = dx. − √ x 4−x x 4−x 4x − x2 0 0 0 Substitute u = 4x − x2 , du = du = (4 − 2x) dx. dx. Then our integral becomes 2 4 √ 4 4 − 2x √ 4x − x2 dx = √ 1u du = dx = du = 2 u 0 = 4 . 0 0 2
4
x
6. Comput Computee
−
lim x
x→∞
Answer:
RMT 2014
x2 ln
1 + x + x x
.
1 2
Calculus Test Solutions
February 15, 2014
RMT 2014
Calculus Test Solutions
February 15, 2014
The total area is therefore 4A 4A = 8/7 . 5. Compute Compute the improper integral integral
− − 2
4
x
x
x
0
4
−x
dx.
Answer: Answer: 4 Solution 1: First
have
4−x x
of all, we note the many symmetries of the given expression. Specifically, we
and we subtract subtract its reciprocal. We also recall that square roots, when we take their
√ √ − −
derivative, give us their reciprocal. This inspires the guess that the function f ( f (x) = is somehow important to our integral. Indeed, we find that f that f (x) =
− 2 0
4−x x
√ √ − x 2 = 2(2 − 0) =
x
dx = 2 x 4 4−x dx =
Solution Solution 2: Although
0
4.
1 2
4−x x
x 4−x
x 4
x
so that
solution 1 is, perhaps, the prettiest way of solving this problem, it is not necessar necessarily ily easy to notice notice.. A more more direct direct approac approach h uses uses a trig trig substit substituti ution. on. Specific Specifically ally,, 4−x 2 2 noting the importance of x and remembering remembering the Pythagorean Pythagorean identity identity sin x + cos x = 1, 2
− 4−x x
it makes sense to try the substitution x = 4sin θ. Then 2
2
=
dx = dx = 8 sin sin θ cos θ dθ, dθ, 4 sin sin θ = 0 when θ when θ = 0 and 4sin θ = 2 when θ when θ = π
4
π
(cot θ
0
dθ = 8 − tan θ) · 8sin θ cos θ dθ =
4
π 4.
cos2 x = sin2 x
cot x. Also, lso,
The integral becomes π
2
cos θ
0
4
2
sin θ dθ = dθ = 8
cos2θdθ. cos2θdθ.
0
This last integral may be easily computed by the substitution 2θ 2θ
θ : → θ:
π
8
π
4
cos2θ cos2θ dθ = dθ = 4
0
2
0
π
cos θ dθ = dθ = 4(sin θ) 0 = 4(1 2
4 . − 0) = 4 .
Solution Solution 3: The
simplest way to solve this problem is perhaps to write the integrand with a common denominator. This gives
√ √ −
√ x 2 2 (4 − x) − x 4 − 2x √ √ √ dx = dx = dx = dx = dx. − √ x 4−x x 4−x 4x − x2 0 0 0 Substitute u = 4x − x2 , du = du = (4 − 2x) dx. dx. Then our integral becomes 2 4 √ 4 4 − 2x √ 4x − x2 dx = √ 1u du = dx = du = 2 u 0 = 4 . 0 0 2
4
x
6. Comput Computee
−
lim x
x→∞
Answer:
RMT 2014
x2 ln
1 + x + x x
.
1 2
Calculus Test Solutions
February 15, 2014
RMT 2014
Calculus Test Solutions
Solution: We
rewrite this limit in a form that allows us to apply L’Hˆ opital’s opital’s Rule. That is,
−
= lim
x→∞
1 + x + x x
2
lim x
x→∞
x ln 2
x→∞
2
− ln
1+x 1+x x
1 x2
by L’Hˆopital’s opital’s Rule
2 x3
−
1 x2 = lim x→∞ 2 1 + x + x
− x
1 x
= lim
x 1 − x1 − 1+x 1+x (− x )
1 = lim x→∞ 2
February 15, 2014
1 x = lim x→∞ 2 1 + x + x
−
1 1 = (1 1 + x + x 2
1
− 0) =
1 . 2
7. For a given given x x > 0, let a let a n be the sequence defined by a 1 = x = x for for n = n = 1 and a and a n = x an− for n for n Find the largest x for for which which the limit limit lim an converges. 1
≥ 2.
n→∞
Answer:
1/e
e
order order for for lim an to have a limit L, it must be that xL = L, so that x = L1/L . n→∞ Otherwise, we would be able to extend the recurrence and converge to a different limiting value. df Thus, we seek the maximum of the function f function f ((L) = L 1/L . To do this, we solve dL = 0. Since
Solution: In
df d = e dL dL
ln L
= L 1/L
L
1 L2
−
ln l n L , L2
we see that L = e. Thus, Thus, the maximum value value for x is f ( f (e) = e1/e . To be sure sure that that this this is a maximum, we check as follows: d2 f dL2
=L
1
L
4
−
e
−
2
3L + ln (L) + 2(L 2(L
8. Evaluate Evaluate
2
2I =
2 2
−
=
2
−
e
3
−
< 0. 0 .
2
substitute the variable x by
−x and add the resulting integral to the original
1 + x + x2 dx + dx + 1 + 2x 1 + x + x2 1 + 2x
+
2
−
2
−
1 + x + x2 dx = dx = 1 + 2−x
(1 + x + x2 )2x 1 + 2x
So the given integral is I is I =
2
dx = dx =
2
−
2
1 + x + x2 1 + x + x2 + dx 1 + 2 x 1 + 2 −x
2 (1 + x + x2 ) −
· (1 + 2x) dx = dx =
1 + 2x
2
1 + x + x2 dx = dx = 4 +
2
−
16 28 = . 3 3
14 . 3
Note that more generally, for even functions f , f ,
RMT 2014
e
1
3
integral to get 2
−e
14
Solution: We
=
1 + x + x2 dx. 1 + 2x
−
Answer:
1)ln(L) + 1 − 1)ln(L
a f ( f (x) dx = 12 1+bx dx = −a 1+b
Calculus Test Solutions
a f (x) dx. dx. a f (
−
February 15, 2014
RMT 2014
Calculus Test Solutions
f (x) . Calculate 9. Let f f satisfy x = f = f ((x)ef ( Answer:
e
−1
February 15, 2014
e f (x) dx. dx. 0 f (
First, we compute the antideriv antiderivativ ative. e. Mak Makee the substitution substitution u = f ( f (x), so hence f d 1 −f −f −f du = du = f f (x) dx. dx. Note that f (x) = dx xe = e f (x)xe , so f so f (x) = ef +x = x(1+f (1+f )) . Thus,
Solution Solution 1:
−
f ( f (x) dx = dx = u u
f ( f (x)dx = dx =
du
u x(1+u (1+u)
= x = x(1 (1 + u + u)) du = du = ue ue u (1 + u + u)) du
ueu (1 + u + u))du = du = e eu u2
f (x) = e f ( − u + 1 = e
f ( f (x)2
To conclude, when x = 0, f ( f (x) = 0 and when x = e, e , f ( f (x) = 1. Thus Thus,, 0 1) e (0 0 + 1) = e 1 .
−
−
−
f (x) + 1 − f (
e f (x) dx = dx = e 1 (1 0 f (
−1+
that f that f is is monotonically increasing and is the inverse of the function g function g((y ) = ye y ey . Since f ( f (e) = 1, the area under f ( f (x) from 0 to e is the area of the rectangle with vertices (0, (0, 0), 0), (e, 0), 0), (0, (0, 1), 1), (e, 1) minus the area to the left of f ( f (x) from 0 to 1, and the latter is just the integral of g( g (y) from 0 to 1. So we have
Solution 2: Note
e
1
f ( f (x)dx = dx = e e
0
= e
10. Given Given that
∞ 1 n=1 n2
=
π2 6 ,
−
− 0
[ye y ]10 +
1
e dy = dy =
0
π2 12
2
ey dy = dy = [ey ]10 = e = e
− 1.
compute the sum ∞
ln
0
−
yey dy
0
1
y
n=1 Answer:
1
g (y )dy = dy = e e
−
1 2n n2
.
2
2
Solution: First
of all, for the sake of clarity, I omit details about certain calculations which are justifia justifiable ble so there there is a more more clear clear focus on the actual actual com comput putatio ation. n. Specific Specifically ally,, I take take derivatives and integrals of series without explaining, and I integrate functions with removable singularities, but the ordinary student would not pay attention to these technical issues anyway. I proceed now: The first step to obtaining any insight on this problem is to replace 21n with x with xn . This allows us to n take derivatives, getting rid of powers of n n in the denominator. Thus, we write f write f ((x) = n∞=1 xn and what we want to find is f ( f ( 12 ) given that f that f (0) (0) = 0 and f (1) f (1) = π6 . As mentioned before, we n− first take f take f (x) = n∞=1 x n and then then we we take take (xf (xf (x)) = n∞=1 xn−1 = 1−1 x . By reintegrating, xf (x) = ln(1 x) + C C but by plugging in x = 0 it is easy to check that C C = 00.. Then x ln(1 x ) ln(1 t ) − − f (x) = . Because Because f f (0) (0) = 0, f 0, f ((x) = 0 dt. dt. Thus, the answer we are looking for x t 2
− −
−
1 2
1
−
2
is equal to 0 ln(1t−t) dt. dt. This completes the first part of the solution. The second part consists of computing this integral.
−
1 2
We denote I = 0 ln 1t−t dt. dt. There There are two things things we know about this integr integral: al: that that finding finding the antiderivative, if it even exists, would be extremely challenging, and also a related formula
−
RMT 2014
Calculus Test Solutions
February 15, 2014
RMT 2014
1 0
Calculus Test Solutions
February 15, 2014
ln(1−t) π2 dt = t 6 which
is given. Noting that 12 is the midpoint of the interval [0, 1] in which the integral formula is relevant, we note that there are several transformations which give integrals 1 ln t on the interval [ 12 , 1]. Specifically, the substitution of x 1 x yields I = 1−t dt. In
−
addition, I = 2
π 6
1
→ − − 1 − ln(1t t) dt.
2 1 1 ln(1−t) ln(1−t) dt dt = π6 1 t t 0 2 ln t dt. The apparent symmetry of the 1−t
−
− −
−
1 2
− 1 2
Thus, we may write 2I =
ln 1−t + integrand immediately brings to mind the t function g(x) = ln x ln(1 x) as a potential antiderivative: indeed, when we apply the product 1 rule, we easily get g (x) = ln 1x−x 1ln−xx . Thus, 2I = π6 + ln t ln(1 t) . Because plugging in
1 2
−
−
2
−
−
1 2
t = 1 is undefined, we resort to using limits and easily obtain 0. Thus, 2I = π2 and I = 12
−
ln2 12 π2 = 2 12
−
π2 6
2
− ln2 12 = π6 − ln2 2
ln2 2 are both correct and equally valid answers. 2
Note, with only a little more work (and some formalizing), we can obtain the more general result (1−x)n ∞ xn that ∞ + = π6 ln x ln(1 x) when x (0, 1). n=1 n n=1 n
RMT 2013
2
2
2
−
−
Calculus Test
∈
February 2, 2013
RMT 2013
Calculus Test
February 2, 2013
x2 + 2x 15 . x→3 x2 4x + 3
1. Compute lim
−
−
2. Compute all real values of b such that, for f (x) = x 2 + bx
− 17, f (4) = f (4).
3. Suppose a and b are real numbers such that sin2 x
lim
x→0 eax
− bx −
1 = . 1 2
Determine all possible ordered pairs (a, b). 4
4. Evaluate
e
√ x
dx.
0
sin2 (5x)tan3 (4x) . 5. Evaluate lim x→0 (log(2x + 1))5 6. Compute
∞
π
3
sin2k xdx.
0
k=0
7. The function f (x) has the property that, for some real positive constant C , the expression f (n) (x) n + x + C
is independent of n for all nonnegative integers n, provided that n + x + C = 0. Given that
1
f (0) = 1 and
f (x) dx = C + (e
0
− 2), determine the value of C .
Note: f (n) (x) is the n-th derivative of f (x), and f (0) (x) is defined to be f (x). 8. The function f (x) is defined for all x 0 and is always nonnegative. It has the additional property that if any line is drawn from the origin with any positive slope m, it intersects the graph y = f (x) at precisely one point, which is √ 1m units from the origin. Suppose further that
≥
a
f has a unique maximum value at some real number x = a . Find
f (x) dx.
0
π/2
9. Evaluate
√ 0
√ sin x + cos x n
10. Evaluate lim
n→∞
RMT 2013
dx
k=1
2k 2k
−1
4.
∞ (cos x)2n
−1
2x
dx .
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
February 2, 2013
1. Answer: 4 x2 + 2x 15 (x = 2 x 4x + 3 (x
Solution: Note that
−
−
− 3)(x + 5) = x + 5 . lim x + 5 = 3 + 5 = − 3)(x − 1) x − 1 x→3 x − 1 3 − 1
4.
2. Answer: 3 Solution: We have that f (4) = 4b other, we see that b = 3 .
− 1 and f (4) = 2(4) + b = b + 8. Setting these equal to each
3. Answer: (2, 2) and (−2, −2)
Solution: Since this is in an indeterminate form, we can use L’Hospital’s Rule to obtain sin(2x) 1 = . ax x→0 ae b 2 lim
−
However, the numerator goes to zero, so the denominator must also go to zero to give us another indeterminate form. This implies that a = b. Using L’Hospital’s Rule again, we have that 2cos(2x) 1 = . x→0 a2 eax 2 lim
The numerator goes to 2, so the denominator must go to 4. Therefore, a = b = (a, b) = (2, 2) and ( 2, 2) .
±2, giving us
− −
4. Answer: 2e2 + 2
√ Let w = x so that w 2 = x and dx = 2w dw. Then the integral becomes 2
2
wew dw.
0
To find this integral, use integration by parts: u = w
→ du = dw;
dv = e w dw
wew dw = uv
− −
v du
= wew = (w Evaluating 2(w
→ v = ew
ew dw
− 1)ew .
− 1)ew at our limits of integration yields
2e2 + 2 .
5. Answer: 50 Solution 1: For any function f with f (0) = 0, we know that f (x) f (x) f (0) = lim = f (0). x→0 x x→0 x
−
lim
sin(5x), tan(4x), and log(2x + 1) are all 0 at x = 0, and their derivatives at 0 are 5, 4, and 2, respectively. So, divide numerator and denominator by x 5 and re-arrange to get 2
(5x)tan3 (4x)
sin lim x→0 (log(2x + 1))5
RMT 2013
= lim
x→0
2
· sin(5x) x
tan(4x) x log(2x+1) 5
3
52 43 = = 50 . 25
·
x
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
February 2, 2013
Solution 2: Recall from Taylor series that if f (0) = 0, then f (x) This allows us to write
≈ f (0)x when x is small.
sin2 (5x)tan3 (4x) (5x)2 (4x)3 lim = lim = 50 . x→0 (log(2x + 1))5 x→0 (2x)5 6. Answer:
√
3
Bring the sum into the integral, so we have
∞
π
3
0
sin2k xdx.
k=0
The integrand is a geometric series, so the answer is π
3
0
7. Answer:
1
−
1 dx = sin2 x
π
3
2
sec x dx = tan
0
π 3
−
tan(0) =
√
3 .
√
3−e
Solution: Since f (n) (x)/(n + x + C ) is independent of n, we can say that it is equal to g(x). Multiplying by (n + x + C ), we have that f (n) (x) = (n + x + C )g(x). Taking a derivative with respect to x, we obtain f (n+1) (x) = (n + x + C )g (x) + g(x). However, this is equal to (n + 1 + x + C )g(x) by the problem statement. Canceling terms, we obtain that g(x) = g (x). The only class of functions that is its own derivative is ae x , so we have that g(x) = ae x (for some constant a). Now, f (x) = (x + C + 1)aex , so f (0) = 1 gives us that a = 1/(C + 1). We also have that 1
1
f (x) dx =
0
0
x + C x e dx = C + (e C + 1
·
− 2).
Integration by parts gives us (e
− 1)C + 1 = C + (e − 2), C + 1
which simplifies to C 2 = 3 from which it follows that the answer is
− e,
√ 3 − e .
8. Answer: 1+log(2) 4 Solution 1: First, express x and y as functions parametrized by m. We have the system y = mx 1 x2 + y 2 = . m
RMT 2013
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
Solving for y, we get y =
m . 1+m2
February 2, 2013
Hence, maximizing y is equivalent to maximizing
m . 1+m2
By
differentiating with respect to m, we see that the maximum occurs when m = 1, at the point 1 √ ( √ , 12 ). 2 Now we just need to compute the integral. However, this parametric form is not convenient. Instead, by drawing the line y = x, we notice that the integral splits up into a right isosceles triangle and a region between the line y = x and the y-axis. This suggests that we should 1 convert to polar coordinates. In fact, f (x) is equivalent to the graph r(θ) = , since a line
√ tan(θ)
at angle θ to the x-axis has slope tan(θ). The area we wish to compute is π/2
π/4
1 1 π/2 2 r(θ) dθ = cot(θ) dθ 2 2 π/4 1 π/2 = [log(sin(θ))]π/4 2 1 = (0 log(1/ 2)) 2 1 = log(2). 4
√
−
We add this area to the area of the triangle, which is
1 2
√ 1
2
2
=
1 4,
so our final answer is
1 + log(2) . 4 Solution 2: We begin as before to find a, but present a different method of computing the integral. Solving for x in terms of y, we get that x2 + y 2 = x/y =
⇒
yx2
We only care about the region where 1 x =
√ 1−4y
1−
2y
4
.
− x + y 3 = 0 =⇒
±
− 1
x =
1
±
− 1 2y
4y 4
.
≤ 1, since x, y ≤ √ 12 . Hence, we take
4y4 = 2xy
Notice that we can compute the desired quantity as 1 √
2
√ − − − 1 2
2
0
1
1 2y
4y 4
dy,
√ 1 , the area between the since within the square bounded by the coordinate axes and x, y 2 curve and the x-axis plus the area between the curve and the y-axis sum to the area of the whole square.
≤
RMT 2013
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
− − −
Now, using the substitution u = 1 √
2
1
0
1
4y4 , we get
1 2y
4y 4
0
(1 u)u du 4u2 4 1 (1 u)u du u)(1 + u) 0 (1 1 u du 0 1 + u 1 1 1 du 1 + u 0
− − − − −
dy =
1
1 4 1 4 1 4 1 [u log(1 + u)]10 4 1 log(2) . 4
= = = =
−
−
= The answer is 12 minus this quantity, so report
February 2, 2013
1 + log(2) . 4
9. Answer: 1/3 Solution 1: Observe that by pulling a factor of cos2 x out of the denominator, we can write the given integral as π/2 π/2 dx sec2 x dx = . (1 + tan x)4 cos2 x (1 + tan x)4 0 0
We now substitute u =
√
√ tan x + 1:
2
du = Thus, our integral is equal to
u4
1
which simplifies to
2
sec x sec x √ dx = dx. 2(u − 1) 2 tan x
∞ 2u − 2
√
du =
∞
2u−3
1
−
2 u−2 + u−3 3
∞
− 2u−4 du, 1 . 3
=
1
Solution 2: Let I be the value of the given integral. Note that 1 1 I = 2 2
π/2
sin(x) +
0
2
cos(x)
−2
which is the polar area bounded by the curve r(θ) = sin(θ) + axes for θ [0, π/2]. Converting to Cartesian coordinates, we get
∈
1 = r =
sin(θ) +
r sin(θ) +
cos(θ)
dx,
cos(θ)
−2
and the x and y
2
r cos(θ)
2
⇒ √ x + √ y = 1 √ √ =⇒ y = (1 − x)2 = 1 + x − 2 x. =
RMT 2013
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
February 2, 2013
Therefore, 1
1 I = 2
1 + x
0
− 2√ x dx
x2 4 3/2 = x + x 2 3 1 4 1 =1 + = 2 3 6 1 = I = 3
−
1
0
−
⇒
10. Answer: π 2 2−1 π
π
Solution 1: Observe that (cos x)2n looks like a bunch of spikes, centered at 0, π, 2π , . . . , each π/2 with mass I n = −π/2 (cos x)2n dx.
We can integrate by parts to see that π/2
I k =
(cos x)
−π/2
= (2k
2k
π/2
− 1)
2k−1
dx = (cos x) (cos x)2k−2 (1
−π/2
sin x
−
−π/2
+ (2k
− 1)
(cos x)2k−2 sin2 x dx
−π/2
− cos2 x) dx = (2k − 1)(I k−1 − I k ).
Therefore, 2k 1 I k = I k−1 = 2k
π/2
π/2
n
⇒
I n =
k=1
−
2k 1 2k
n
I 0 = π
k=1
2k 1 . 2k
−
As n , the spikes get sharper and sharper; this means that the denominator 2x of the integrand gets concentrated at x = 0, π, 2π , . . . . Therefore, we expect that as n ,
→ ∞ n
k=1
→∞
2k 2k 1
−
∞ (cos x)2n 2x
−1
n
dx
→
k=1
2k 2k 1
−
∞
k=0
I n 1 2π = π = π π . 2kπ 1 2−π 2 1
−
−
Solution 2: We present a more rigorous approach here. First, rewrite the problem into the following form:
√ n(cos x) √ n ∞ For a positive integer n, let a n = −1 dx. Also let c = lim n →∞ 2 1 1 ), which is a positive finite constant. Evaluate 2k c lim n→∞ an .
2n
x
n k=1 (1
−
Let B = 0, π, 2π , . . . . The idea is that the numerator of the integrand approaches a function with cπ area concentrated infinitely closely to each point in B. Therefore the limit should be
{
}
lim
n→∞
∞ √ n(cos x)2n
−1
2x
dx =
x∈B
cπ 2π = cπ , 2x 2π 1
−
where the last equality follows by the formula for summing geometric series. We will soon get to a more precise way of thinking about the area being concentrated infinitely closely to points in B, but first let’s see why the numerator should have cπ area around each
RMT 2013
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
February 2, 2013
point in B . Since the area around each point in B is the same (cos2n is periodic), we need only consider the area around 0. We can apply integration by parts to find a formula for the area around 0 in each term of the sequence. The recurrence is π/2
−π/2
√ n(cos x)2ndx = (1 −
π/2 1 ) 2n −π/2
√ n(cos x)2(n−1)dx.
Repeatedly applying this formula, we get π/2
√ n(cos x)2ndx = π √ n
−π/2
n
(1
k=1
− 2k1 ).
Taking the limit as n , the area around 0 goes to cπ. So the answer makes sense. Now we will prove it more rigorously.
→∞
√ n(cos x)
2n
For x / B, the integrand goes to 0 as n 2 S sufficiently disjoint from B , we might guess that
∈
x
lim
→ ∞ because cos x < 1. So for any open set
√
n(cos x)2n dx = 0. 2x
n→∞ S
If we require that every point in S is at least > 0 away from any point in B, then this is indeed true. There are two ways to see this. The fanciest way to see it is to use the “dominated convergence theorem” which says that if a sequence of functions f n converges pointwise to a funciton f and if there is some function φ with f n (x) < ϕ(x) for all x S and S ϕ < , then limn→∞ S f n = S f . To apply this theorem, we let f n be the integrand of the n-th term of the sequence. To construct ϕ, notice that since every point in S is at least away from every point in B , there is some δ < 1 so that cos x < δ for all x S . So n(cos x)2n is bounded by nδ 2n for all x S . Since nδ 2n has a finite limit as n , n(cos x)2n is bounded by some finite number B for all x S . So we can let ϕ(x) = B/2x . Then f n (x) < ϕ(x) for all x S and S ϕ < , just as we need to apply the theorem. So we apply the theorem to get
|
|
∈
∈ √ √ →∞ |
√
|
n→∞ S
n(cos x)2n dx = 2x
lim
√ n(cos x)2n
S n→∞
√ ∈
∈ ∞
∈
√
lim
∞
2x
dx =
|
|
0dx = 0.
S
But we of course don’t expect you to know the dominated convergence theorem, so we can also prove this using a “bare hands” method that is actually easier. (Bare hands is usually much harder than the dominated convergence theorem proof. That is why people use the dominated convergence theorem. But we have arranged for this problem to work with bare hands.) As we argued above, there is some δ < 1 so that cos x < δ for all x S . Then n(cos x)2n < nδ 2n for all x S . So we have a bound
∈
∈
√ S
Since the integral
1 2
n(cos x)2n dx 2x
≤ √ nδ 2n
√
√
1 dx. x S 2
→ ∞ so we again have the desired result. The upshot of all this is that we can now define B to be the points in [−1, ∞) that are within of B and have √ n(cos x)2n ∞ √ n(cos x)2n x
converges, this goes to 0 as n
lim
n→∞
RMT 2013
−1
2x
dx = lim
n→∞ B
Calculus Test Solutions
2x
dx.
February 2, 2013
RMT 2013
Calculus Test Solutions
February 2, 2013
To calculate the integral on the right, notice that it is just the sum over all integers k kπ+
√ n(cos x)2n 2x
kπ −
√
1 dx = kπ 2 −
n(cos x)2n dx. 2x
By the formula for summing geometric series, the sum of this over all integers k lim
n→∞
∞ √ n(cos x)2n
2x
−1
√
2π
dx =
2π
−1
≥ 0 of
≥ 0 is
n(cos x)2n dx. 2x
−
(1)
So we have reduced the problem to calculating the following limit:
√
lim
n→∞
n(cos x)2n dx. 2x
−
To do this, bound the limit above and below by taking the highest and lowest possible values of 2x out of the integral:
√
n(cos x)2n dx 2x
√ n(cos x)2ndx ≤ lim 2− lim n→∞ − n→∞ −
≤2
lim
n→∞
√
n(cos x)2n dx.
−
By a very similar argument as above, nothing outside ( , ) contributes to the integrals in our bounds and therefore
−
lim
n→∞
√
π/2
2n
n(cos x) dx = lim
n→∞
−
√ n(cos x)2ndx.
−π/2
We have already calculated the right hand side: it is cπ. So we can plug this back into our bounds to get n(cos x)2n 2− cπ lim dx 2 cπ x n→∞ − 2
√
≤
≤
Plugging this bound into (1) gives
∞ √ n(cos x)2n
2π − 2 cπ
→∞ −1 2x − 1 ≤ nlim Since was arbitrary, taking → 0 forces ∞ √ n(cos x)2n 2π
lim
n→∞
−1
dx
dx = cπ
2x
2π
≤ 2 cπ 2π − 1 . 2π
2π
− 1,
as desired. Finally, you might be interested in knowing why c is a positive finite constant. (This is not necessary to solve the problem, but it is necessary to be sure that the problem makes sense. 1 And it is interesting.) To see this, let b n = n nk=1 (1 2k ). Then
√
1 log bn = log n + 2
n
k=1
log(1
−
−
1 1 ) = log n + 2k 2
n
− (
k=1
1 1 + O( 2 )). 2k k
Here O( k1 ) denotes some function of k whose absolute value is always less than C k1 for some 1 1 C big enough. The fact that log(1 2k ) = 2k + O( k1 ) follows from the taylor expansion of log 2
2
−
RMT 2013
−
2
Calculus Test Solutions
February 2, 2013
RMT 2013
Calculus Test Solutions
around 1. It is well known that this in gives 1 log bn = log n 2
−
1 log n 2
−
n 1 k=1 k =
log n + γ + O( n1 ) where γ is some constant. Plugging
1 1 γ + O( ) + 2 n
1 Since ∞ k=1 k converges to some constant, 1 n . Therefore log bn 2 γ +α as n is a positive finite number as desired.
→∞
RMT 2012
2
→ −
February 2, 2013
n
k=1
1 O( 2 ) = k
n 1 k=1 O( k2 )
→∞
−
1 1 γ + O( ) + 2 n
n
k=1
O(
1 ). k2
converges to some finite constant α as . This is some finite number, so c = exp( 12 γ +α)
Calculus Test
−
February 18, 2012
RMT 2012
1. What is
Calculus Test
10 0
(
x
February 18, 2012
− 5) + (x − 5)2 + (x − 5)3 dx?
2. Find the maximum value of
3π/2
sin(x)f (x) dx π/2
−
subject to the constraint f (x)
|
| ≤ 5.
3. Calculate
35
1
− x3/5 dx. √ 4. Compute the x -coordinate of the point on the curve y = x that is closest to the point (2 , 1). x
25
5. Let f (x) = x +
and set g (x) = f
1
−
x2
2
+
x3
3
+
x4
4
+
x5
5
,
(x). Compute g(3) (0).
6. Compute lim
sin x
x→0
1 1−cos
x
.
x
7. A differentiable function g satisfies x
0
for all x
(x
− t + 1)g(t) dt = x4 + x2
≥ 0. Find g(t).
8. Compute
∞
0
ln x dx. x2 + 4
√ 1!2! ·· · n! 2
n
9. Find the ordered pair ( α, β ) with non-infinite β = 0 such that lim
10. Find the maximum of
1
n→∞
nα
= β holds.
f (x)3 dx
0
given the constraints
1
−1 ≤ f (x) ≤ 1,
RMT 2012
f (x) dx = 0.
0
Calculus Test Solutions
February 18, 2012
RMT 2012
1. Answer:
Calculus Test Solutions
February 18, 2012
250 3
Solution: This integral is equal to 5
2
5
3
x + x + x dx =
−5
2
x dx =
−5
53 3
( 5)3 250 = . 3 3
− −
2. Answer: 20 Solution: Clearly we want to maximize f (x) when sin(x) 0 and minimize f (x) when sin (x) < 0. We do this by setting f (x) = 5 in the first case and f (x) = 5 in the second case. Noting that the bounds of integration cover precisely one full period of sin, we see that the integral becomes equivalent to twice the integral of 5 sin (x) over the half period where sin (x) 0. This results in 20 .
− ≥
3. Answer:
5 2
ln
≥
8 3
Solution: Note that we can write the integral as 35
25
We solve via u-substitution. Let u = x 2/5 du = The integral becomes 5 2 which evaluates to
1 x3/5 (x2/5
− 1:
2 −3/5 x dx = 5
⇒
32 1
−
22 1
−
x3/5 x3/5
5 (ln8 2 4. Answer:
− 1) dx.
·
dx =
5 du = 2 u
− ln3) =
5 3/5 x du. 2 8
3
1 du, u
5 8 ln . 2 3
√
2+ 3 2
Solution: We want to minimize the distance between the points ( a2 , a) and (2, 1). We can equivalently minimize the square of the distance between those two points, which is
− a2 )2 + (1 − a)2 = a4 − 3a2 − 2a + 5. The derivative of this function is 4a3 − 6a − 2, which can be factored as 2( a +1)(2a2 − 2a − 1). The roots √ of this cubic are therefore a = −1, 1±2 3 . Two of the roots are negative and therefore invalid, so therefore √ a = 1+2 3 and √ (2
a2 = x =
2+ 3 2
5. Answer: 1 Solution: We begin by observing that due to the inverse function rule, the first three derivatives of g are ˆ and gˆ(x) = f ˆ−1 (x) determined by the first three derivatives of f . Additionally, f (0) = g(0) = 0. Let f (x) be new functions whose first three derivatives at zero equal those of f and g respectively. ˆ = ln(1 x) is a suitable choice. Then gˆ(x) = 1 e−x and By Taylor Series expansion, we see that f (x) (3)
gˆ
(0) = g
(3)
−
0
(0) = e = 1 .
−
−
6. Answer: e−1/3 Solution: We can approximate sin x and cos x by their Taylor series. 3
sin x
RMT 2012
≈ x − x6 ,
2
cos x
≈ 1 − x2
Calculus Test Solutions
February 18, 2012
RMT 2012
Calculus Test Solutions
February 18, 2012
Substituting the limit becomes lim
→0
x
x2 6
− 1
2/x2
1 = lim 1 + 2 x→∞ 6x
−2x
2
1 = lim 1 + 2 x→∞ x
−x
2
/3
1 = lim 1 + x→∞ x
−x/3
= e−1/3
1 x . x
because e = limx→∞ 1 +
7. Answer: 12x2 − 24x + 26 − 26e−x Solution: First differentiate the equation with respect to x: x
g(x) +
g(t) dt = 4x3 + 2x.
0
Differentiate again to obtain A particular solution 12x2 general solution will be
g (x) + g(x) = 12x2 + 2.
− 24x + 26 can be found using the method of undetermined coefficient, so the g(x) = 12x2 − 24x + 26 + Ce −x
for some constant C . By substituting x = 0 into the first equation, we see that g(0) = 0. We therefore find that C = 8. Answer:
−26, making the answer
12x2
− 24x + 26 − 26e−x .
π ln 2 4
Solution: Substitute x = 2 tan θ to get
∞ ln x
1 dx = x2 + 4 2
0
π/2
1 π 1 ln(2 tan θ) dθ = ln 2 + 2 2 2
·
0
π/2
We will now show that this final integral is zero by substituting θ = π/2 0
π/2
π 2
− − −
ln(tan θ) dθ =
ln tan
0
π/2
π/2
=
ln
0
which gives us what we wanted, so the answer is
φ
ln(tan θ) dθ.
0
− φ to yield
dφ
π/2
1 dφ = tan φ
ln(tan φ) dφ,
0
π ln 2 . 4
9. Answer: (1/2, e−3/4 ) Solution: Take the logarithm and approximate using Stirling’s approximation 1 . Stirling’s approximation says that ln(n!)
√ 2
n
ln
1!2! nα
· · · n!
1 = 2 ln(1!2! n
≈ n12
≈ n ln n − n in the limit of large n. Using this, we have
· · · n!) −
n
(k ln k
k=1
1 α ln n = 2 n
n
ln(k!)
k=1 n
− k) − α ln n = n12
− α ln n
(k ln k)
k=1
1) − n12 n(n + − α ln n 2
Approximate n(n + 1) with n 2 and approximate the infinite sum by an integral
≈ 1
1 n2
n
1
x ln x dx
− 12 − α ln n
http://en.wikipedia.org/wiki/Stirling’s_approximation
RMT 2012
Calculus Test Solutions
February 18, 2012
RMT 2012
Calculus Test Solutions
February 18, 2012
Integrating by parts 1 n2
≈ Therefore,
n2 ln n 2
√ 2
1!2! nα
n
lim ln
n
→∞
−
n2 1 + 4 4
· · · n!
− − 1 2
= lim n
→∞
α ln n
≈ 12 ln n − 34 − α ln n.
− 1 2
α ln n
−
3 , 4
− α = 0, in which case α = 21 and the limit evaluates to 43 . Therefore, we wish √ √ 1!2! · · · n! 1!2! · · · n! lim = exp lim ln = e3/4 .
which is finite only when to compute
1 2
2
n
n
n1/2
→∞
n
→∞
2
n
nα
10. Answer: 1/4 Solution: Consider the following expression 1
(f (x)
0
Since f (x)
−
1 1) f (x) + 2
2
dx.
≤ 1 this expression is less than or equal to 0. Meanwhile expanding the integrand gives (f (x)
−
1 1) f (x) + 2
2
= f (x)3
− 34 f (x) − 14 ,
so its integral is 1
0
(f (x)
− 1)
f (x)
−
1 2
2
1
dx =
3
f (x) dx
0
1
=
f (x)3 dx
0
−
3 4
1
0
− 14 ,
f (x) dx
−
1 4
1
dx
0
proving that the answer is at most 1 /4. Equality occurs when f (x) =
−
1/2 if 0 x 2/3 , 1 if 2/3 < x 1
≤ ≤
≤
so 1/4 is indeed the maximum.
RMT 2011
Calculus Test
February 19, 2011
RMT 2011
1. Find
Calculus Test
February 19, 2011
x+2 (x−1)2 (x−2) dx.
2. Tangent lines are drawn at the points of inflection for the function f (x) = cos x on [0, 2π]. The lines intersect with the x-axis so as to form a triangle. What is the area of this triangle? 3. Let f be one of the solutions to the differential equation f (x) − 2xf (x) − 2f (x) = 0.
Supposing that f has Taylor expansion f (x) = 1 + x + ax2 + bx3 + cx4 + dx5 + · · ·
near the origin, find ( a,b,c,d). 4. What is the value of the alternating harmonic series 1 − 21 + 31 − 41 + . . .? 5. Solve the integral equation x
f (x) =
ex−y f (y )dy − (x2 − x + 1)ex
0
for f (x). 6. Evaluate the integral
π
|sin(2x) − sin(3x)| dx
0
√
a+b c , d
and express your answer in the 7. Let f (x) =
2
x3 e( ) 1 x2 . x
−
where a, b, c, and d are integers.
Find f (7) (0), the 7th derivative of f evaluated at 0.
8. For the curve sin(x) + sin(y ) = 1 lying on the first quadrant, find the constant α such that lim xα
x
→0
d2 y dx2
exists and is nonzero. 9. Evaluate the integral
0
π
2
dx
1 + (tan x)π
.
10. Evaluate the following integral:
0
RMT 2011
1
dx x(x + 1) ln 1 + 1
2011
x
Calculus Solutions
February 19, 2011
RMT 2011
Calculus Solutions
February 19, 2011
1. Answer: 4 ln(x − 2) − 4ln(x − 1) + 3/(x − 1) or equivalent forms Expand by partial fractions: x + 2 1)2 (x
(x
=
A
+
B
+
C . (x 1)2
− − 2) x − 2 x − 1 − Then x + 2 = A(x − 1)2 + B(x − 2)(x − 1) + C (x − 2). If x = 2, then 4 = A. If x = 1, then C = −3. If x = 0, then 2 = 4 + 2B + 6 ⇒ B = −4. So
2. Answer:
π
(x
−
x + 2 1)2 (x
− 2) dx
=
A
x = 4 ln(x
B
C + + dx 2 x 1 (x 1)2 2) 4ln(x 1) + 3/(x 1).
− − − − − −
−
2
4
By taking the second derivative, we see that the points of inflection occur when cos x = 0, so π x = π2 , 3π 2 . At 2 , the slope of the tangent line is the derivative evaluated at this point, which is sin π2 = 1. So the tangent line has equation y = x + b, and since cos π2 = 0, 0 = π2 + b and so the tangent line is given by y = x + π2 . Similarly, we can find that the tangent line at 3π 2 is y = x 3π . The intersection of the two tangent lines forms the vertex of the triangle, which occurs 2 π when x + π2 = x 3π 2 , or at x = π. At this point, both tangent lines equal 2 . Hence the triangle π formed has height 2 . The tangent lines cross the x-axis when y = 0, giving π2 and 3π 2 , which are 1 π π separated by π. Hence the area of the triangle is 2 2 π = 4 .
−
−
−
−
− −
−
−
−
−
· ·
2
3. Answer: (1, 2/3, 1/2, 4/15) Just compute the Taylor expansion of f (x)
− 2xf (x) − 2f (x) to the third term, which is (2a + 6bx + 12cx2 + 20dx3 + · · · ) − (2x + 4ax2 + 6bx3 + ·· · ) − (2 + 2x + 2ax2 + 2bx3 + · ·· ). All coefficients should be zero, so 2a − 2 = 0, 6b − 4 = 0, 12c − 6a = 0 and 20d − 8b = 0. Solving these equations gives the answer. 4. Answer: ln(2) 1 1 Note that 1+x = 1 x + x2 x3 + . . . and that ln(x + 1) = 1+x dx = C + x Since ln(1) = 0 = C , we set C = 0. Evaluating at x = 1, we get that ln(2) = 1
−
−
5. Answer: f (x) = (2x − 1)e
2
3
3
− x2 + x3 − x4 + . . .. − 21 + 31 − 41 + . . ..
x
Differentiate both sides to get x
f (x) = f (x) +
ex−y f (y)dy
0
But
x
ex−y f (y)dy = f (x) + (x2
0
so by substituting it we get
f (x) + (x2 and f (x) = (2x 6. Answer:
− (x2 + x)ex. − x + 1)ex
− x + 1)ex − (x2 + x)ex = 0,
− 1)ex.
√ 5 5+4 6
First, we eliminate the absolute value by finding the intervals on which the expression inside is positive and negative. Sketching sin(2x) and sin(3x) suggests that there are two points 0 < x < π such that sin(2x) = sin(3x). The identity sin(x) = sin(π x) gives us the solution x = π5 , and sin(x) = sin(3π x)
−
RMT 2011
Calculus Solutions
−
February 19, 2011
RMT 2011
Calculus Solutions
February 19, 2011
π 3π gives us x = 3π 5 . Then we have sin(3x) > sin(2x) on the intervals (0, 5 ) and ( 5 , π) but sin(2x) > sin(3x) on ( π5 , 3π 5 ). So the integral becomes: π
π
|
sin(2x)
0
− sin(3x)| dx =
5
[sin(3x)
0
− sin(2x)]dx 3π 5
+
π
[sin(2x)
π
5
Next, we let F (x) equal cos(2x) 2 to:
− sin(3x)]dx +
3π F 5
F (0)
π F 5
π 3π 2F 5 5 1 1 2π 2 3π = + + cos cos 2 3 5 3 5 2 5 2π 5 π = + cos + cos . 3 3 5 3 5 = F (π) + 2F
−
Now we use the identities cos
[sin(3x)
3π 5
− sin(2x)]dx
− cos(3x) , so that F (x) = sin(3x) − sin(2x). Then the integral simplifies 3
− − − − − π F 5
π 5
=
√
5+1 4
+ F (π)
− F
3π 5
F (0)
− cos 6π5 + 23 cos 9π5 − 12 + 13 √
5−1 4
and cos 2π 5 =
2 5 2π 5 π 2 5 + cos + cos = + 3 3 5 3 5 3 3
to simplify this further:
√ −
√ 5 5
5 1 + 4
√ 5 + 1 4
2 = + 3 6 5 5+4 = . 6
√
7. Answer: 12600 Since f (n) (0) = a n n!, where a n is the nth Taylor series coefficient, we just need to find the Taylor series of f and read off the appropriate coefficient. The Taylor series is given by f (x) = x The coefficient of x 7 is 8. Answer:
1 1 2! + 1! +
3
x 2 x 4 1+ + + 1! 2!
· ··
1 = 25 , so f (7) (0) = 7!
1 + x2 + x4 +
···
· 25 = 12600.
.
3 2
Differentiate the equation to get cos(x) +
dy cos(y) = 0 dx
and again
−
d2 y sin(x) + 2 cos(y) dx
− dy dx
2
sin(y) = 0.
By solving these we have dy = dx and
RMT 2011
− cos(x) cos(y)
d2 y sin(x)cos2 (y) + sin(y)cos2 (x) = . dx2 cos3 (y)
Calculus Solutions
February 19, 2011
RMT 2011
Calculus Solutions
Let sin(x) = t, then sin(y) = 1 Substituting it gives
− t.
Also cos(x) =
d2 y t2 (2 t) + (1 t)(1 = dx2 t3/2 (1 2t)3/2
−
−
−
February 19, 2011
√ 1 − t2 and cos(y) =
1
(1
− t)2 =
− t(2
t).
− t2) = t−3/2 t2 − t + 1 . (1 − 2t)3/2 √
2
Since limx→0 xt = 1, α = 23 should give the limit lim x→0 xα ddxy = this limit undefined or zero.
2 4 .
2
9. Answer:
−
All other values of α will make
π
4
We make the substitution, x = leave it as x). Then we have: π
0
2
dx = 1 + (tan x)π
π 2
− y (note that the actual variable of integration is irrelevant so we
0
π
2
−dx = 1 + tan ( π2 − x)π
π
2
0
dx = 1 + (cot x)π
π
0
2
(tan x)π dx . (tan x)π + 1
Then we add the original integral to both sides: π
2
2
0
dx = 1 + (tan x)π
π
0
So the integral we want is 10. Answer:
2
1 (tan x)π + dx = 1 + (tan x)π 1 + (tan x)π
π
0
2
1 + (tan x)π dx = 1 + (tan x)π
π
0
2
dx =
π . 2
π 4.
1 2010(ln 2)2010
Make the substitution 1
0
1 x
= e t
t
− 1, so that − (ee −dt1) t
= dx. This transforms the original integral into
∞ dt − 1)2 dt dt = dt 2011 = t t 2 2011 2011 ln 2 e (e − 1) t ln 2 t ∞ 1 1 =− = 2010 . 2010 2010t 2010(ln2) ∞ et (et
dx 1 x
x(x + 1) ln 1 +
2
ln 2
RMT 2009
Calculus Test
February 21, 2009
RMT 2009
Calculus Test
1. Find the exact value of 1 −
February 21, 2009
1 1 + − . . .. 3! 5!
2. At RMT 2009 is a man named Bill who has an infinite amount of time. This year, he is walking continuously at a speed of 1+1t , starting at time t = 0. If he continues to walk for an infinite amount of time, how far will he walk? 2
10x2 . 0 sin2 (3x)
3. Evaluate lim x→
1
4. Compute
tan
1
−
(x)dx
0
5. Let a(t) = cos2 (2t) be the acceleration at time t of a point particle traveling on a straight line. Suppose at time t = 0, the particle is at position x = 1 with velocity v = −2. Find its position at time t = 2. 6. Find
∞
n
d
dxn n=2
−ax
(e
)
for | a| < 1. 7. Compute
n
4k n − k lim cos .
n→∞
k =1
∞
4x + 7e 1 9. Compute n . 8. Evaluate
n
dx. Remember to express your answer as a single fraction.
n
∞
n=0
2x
−
0
n2
5
∞
1 , as a decimal to the nearest tenth. 50 + n /80000 n=1
10. Evaluate
RMT 2009
2
Calculus Solutions
February 21, 2009
RMT 2009
Calculus Solutions
February 21, 2009
1. Answer: sin 1 3
x
By Taylor Expansion, sin x = x − 2. Answer: ∞
d = 0
+
x
5
7
x
−
5!
+ . . . . Let x=1, and the desired value equals sin 1.
7!
π
2 1 1+t2 dt =
3. Answer:
3!
tan
1
−
(t)|0 = limt
tan
1
−
∞
→∞
(t) − tan
1
−
(0) =
π
2.
10 9
By l’Hˆopital’s rule, 10x2 0 sin2 (3x)
20x 0 6 sin(3x)cos(3x) x 20x = lim x 0 3 sin(6x) 20 = lim x 0 18 cos(6x) 10 = . 9
lim
=
x→
lim →
→
→
4. Answer:
π
−2 ln(2) 4
or equivalent expression
We integrate by parts: 1
1 · tan
1
−
1
x tan (x) − x dx x +1 π 1 1 0
1
(x)dx =
−
0
0
2
1
− 0 −
=
4 2 π ln(2) − . 4 2
= 5. Answer: − cos(8) + 32
v(t) =
ln(x2 + 1)
0
33 32
a(t)dt =
2
cos (2t)dt =
1 + cos(4t) 2
dt =
sin(4t) t + + c1 , 8 2
t) where c 1 is a constant. Plug in t = 0 to find v (0) = c 1 = − 2. So v(t) = sin(4 + 2t − 2. 8
x(t) =
v(t)dt =
sin(4t) 8
t cos(4t) t 2 + − 2dt = − + − 2t + c2 . 2 32 4
1 Plug in t = 0 to get x(0) = − 16 + c2 = 1, so c 2 =
x(2) = − 6. Answer: Since
n
d dxn
2 a
1+a
(e
e
−ax
33 32 .
Thus,
cos(8) 33 + . 16 32
−ax
) = (−a)n e
−ax
, ∞
∞
n
d
dxn n=2
−ax
(e
) = e
−ax
(−a) . n
n=2
This forms a geometric series with common ratio −a and first element a2 , which converges since |a| < 1. a Thus the answer is 1+ e ax . a 2
RMT 2009
−
Calculus Solutions
February 21, 2009
RMT 2009
Calculus Solutions
February 21, 2009
1−cos(4) 16
7. Answer:
Define a partition on [0, 1] with n elements by setting xi = ni for 0 ≤ i ≤ n. Then x i − xi all i. If we let f (y) = (1 − y)cos(4y) and put y k = nk for 1 ≤ k ≤ n, then we have
1
−
n
=
1 n
for
n
n − k cos 4k = f (y )(x − x
k =1
n2
k
n
i
i−1
).
k=1
Thus, we may conclude that n
f (y )(x − x lim k
n→∞
i
1
f (y)dy (1 − y) cos(4y) 1 − y
1) =
i−
0
k =1
1
=
0
cos(4y) sin(4y) − 4 16 − cos(4) 1 + . 16 16
= = 14e2 −12 , 2 e −1
8. Answer:
1
0
14−12e−2 1−e−2
or
To evaluate the floor function, split the integral into unit intervals: k+1
∞
∞
4x + 7e
2x
−
dx =
0
k=0
4(k + 7)e
2x
−
dx
k
= (14e 0 − 14e 2 ) + (16e 2 − 16e 4 ) + (18e = 12 + 2(e 0 + e 2 + e 4 + . . .) 2 14 − 12e 2 14e2 − 12 = 12 + = = . 1 − e 2 1 − e 2 e2 − 1 −
−
−
−
−
−
4
−
− 18e
6
−
) + . . .
−
−
9. Answer: Let S =
−
−
∞
∞
5 16 n
∞
n=0
1 S = 5
5
n
. Then
∞
n=0
n n+1
5
∞
n + 1 − 1 = n + 1 − = n=0
5
⇒
n+1
n=0
5
n+1
n=0
1 5
n+1
∞
n− = n
n=1
5
1 5
1 −
1 5
= S −
1 4
S 1 4S 1 5 = S − ⇒ = ⇒ S = . 5 4 5 4 16
10. Answer: 62.8 This sum is difficult to evaluate exactly. However, it can be closely approximated by the improper integral of the same function, which is easily evaluated using u-substitution.
RMT 2009
Calculus Solutions
February 21, 2009
RMT 2009
Calculus Solutions
∞
0
dx 50 + x2 /80000
February 21, 2009
1 dx 50 0 1 + (x/20000)2 1 2000du 50 0 1 + u2 2000 tan 1 u 0 50 40 lim (tan 1 (b) − tan 1 (0)) ∞
=
∞
= = =
∞
−
−
−
b→∞
= 40 lim tan
1
−
b→∞
(b)
π 2 = 20π ≈ 62.83. = 40
To see that this integral is correct to the nearest tenth, we observe that since the integrand is a monotonic function, we can bound it above and below by Riemann sums. More precisely:
∞
∞
1 ≤ 20π ≤ 2 50 + n /80, 000 n=1
n=0
1 50 + n2 /80, 000
.
By rearranging terms, this implies that:
∞
20π −
1 1 ≤ ≤ 20π. 50 n=1 50 + n2 /80, 000
From this it follows that 62.8 is indeed correct to the nearest tenth.
RMT 2008
Calculus Test
February 23, 2008
RMT 2008
Calculus Test
February 23, 2008
π/2
1. Compute
sin x cos x dx.
0
2. Evaluate: lim
10x2 − 21 x3
x→0
e
1 3
x2
−1
3. Find the area enclosed by the graph given by the parametric equations y = sin(2t) x = sin(t)
4. Find the value of the nth derivative of f (x) = sin n (x) at x = 0. 5. Water flows into a tank at 3 gallons per minute. The tank initially contains 100 gallons of water, with 50 pounds of salt. The tank is well-mixed, and drains at a rate of 2 gallons per minute. How many pounds of salt are left after one hour?
e sin(x)dx. 2 . 7. Compute 6. Evaluate
3x
∞
n=0
n−1
n!
h2 f (x+2x) −2f (x+h)+f (x) h→0
8. Find f (x) such that lim
3
= − x2 − x −
1 2x .
9. Suppose x (t) + x (t) = t 5 x(t). Let the power series representation of x be x(t) = terms of an−1 and an−7 , where n > 7. 10. Evaluate:
x
a t . Find a n
n
n
in
t2t et dt
−∞
RMT 2008
Calculus Solutions
February 23, 2008
RMT 2008
Calculus Solutions
February 23, 2008
1 2
1. Answer:
Using the product to sum trigonometric identity, we get: π/2
π/2
1 (sin2x − sin0) dx 2 sin2x dx − cos 2x
sin x cos x dx =
0
0
1 2
=
π/2
0
π/2
1 = 21 2 0 = 21 (1/2 − (−1/2)) = 21 (1) = 1/2
2. Answer: 30 Two applications of L’Hˆopital’s rule are necessary to obtain a fraction which does not divide by zero; the result is 20 − 3x 20 − 0 = 30 2 x 4 2 x = 2 + 9x e 3e 3 + 0 2
3. Answer:
2
8 3
The graph is a figure-8. By symmetry, the area it encloses is four times the area under the curve √ that we π get from just considering 0 ≤ t ≤ 2 . For these values of t, we can write y = 2 sin(t) cos(t) = 2x 1 − x2 for 0 ≤ x ≤ 1. But this is integrable easily enough by letting u = 1 − x2 . The antiderivative that we get is − 23 u , which gives an area of 32 . Thus the total area enclosed by the parametric graph is 38 . 3 2
4. Answer:
n!
You can think of taking the derivatives this way, in an extended version of the product rule: For each j ≤ n, let f j (x) = sin(x). Then f (x) = f 1 (x)f 2 (x)...f n (x). Each time you take the derivative, you pick one of the f j ’s and replace it with its derivative. Some f j ’s might be differentiated more than once in this process. To find the derivative of f , take the results from each way of choosing the f j ’s to differentiate and add them all up. That said, since sin(0) = 0, if any f j doesn’t get differentiated, it’ll make the whole product zero when x = 0, so these terms don’t matter. But we’re only taking the nth derivative, so if every f j gets differentiated at least once, then they all have to be differentiated exactly once. There are n! remaining terms (the number of ways to choose what order we differentiate the f j ’s in), each of which is cosn (0) = 1, so the value we want is just n!. 5. Answer:
625 32
Let w(t) and s(t) denote the amounts of water and salt, respectively, in the tank at time t. We can immediately see that w(t) = t + 100. Since the tank is constantly mixed, we know that ds s(t) = −2 dt w(t) ds dt = −2 s t + 100 ln(s) = −2ln(C (t + 100)) C s = (t + 100)2 Since s(0) = 50, C = 500000, so s(60) = 625/32. 6. Answer:
RMT 2008
1 3x e 10
(3 sin(x) − cos(x))
Calculus Solutions
February 23, 2008
RMT 2008
Calculus Solutions
February 23, 2008
Let I be the answer. Integrating by parts twice: 1 1 3x I = e3x sin(x) − e cos(x)dx 3 3 1 1 1 = e3x sin(x) − e3x cos(x) − I 3 9 9 9 1 3x 1 I = e sin(x) − e3x cos(x) 10 3 9
7. Answer:
1 2 e 2 ∞
n
2 . Multiply the sum by 2 and you get n=0
n!
Now notice that the nth derivative of e2x is 2n e2x , which gives 2 n evaluated at x = 0. The Taylor ∞ 2n n 2x series for e around x = 0 is thus x . Evaluated at x = 1, this gives the sum above, so the sum n! n=0 we want is equal to
1 2 2e .
8. Answer: tan−1 x or tan−1 (x) + ax + b 1 f (x) .
The limit is, by definition,
Therefore:
1 x4 + 2x2 + 1 (1 + x2 )2 − − = = f (x) 2x 2x 2x f (x) = − (1 + x2 )2 1 f (x) = 1 + x2 f (x) = tan −1 x 9. Answer:
an−7
−(n−1)a n(n−1)
−1
n
Plugging the power series into the differential equation gives:
n(n − 1)a t + na t n(n − 1)a t + (n − 1)a t n
n
n−2
n
n−2
n−1
n−1 n−2
a t a t = =
n
n+5
n−7
n−2
n(n − 1)an tn−2 + (n − 1)an−1 tn−2 = a n−7 tn−2 n(n − 1)an + (n − 1)an−1 = a n−7
10. Answer:
(2e) 1+ln2
x
x
(2e) ) (1+ln 2)2 x
−
x
1 t2 e dt = t (2e)t ln(2e) −∞
t t
x
x
0
−∞
t x(2e) = − 1 + ln 2 (1 + ln 2) x
=
1 (2e)t dt ln 2e −∞
− (2e) x
x(2e) (2e) − 1 + ln 2 (1 + ln 2)2
x −∞
Calculus Test 2007 Rice Math Tournament February 24, 2007 −1+cos x 2 3 . x→0 3x +4x
1. Find lim
2. A line through the origin is tangent to y = x 3 + 3x + 1 at the point ( a, b). What is a ? 3. A boat springs a leak at time t = 0, with water coming in at constant rate. At a time t = τ > 0 hours, someone notices that there is a leak and starts to record distance the boat travels. The boat’s speed is inversely related to the amount of water in the boat. If the boat travels twice as far in the first hour as in the second hour, what is τ ? 4. Let I (n) =
π
∞
0 sin(nx)dx. Find
I (5n ).
n=0
d 5. Let Θk (x) be 0 for x < k and 1 for x k. The Dirac delta “function” is defined to be δ k (x) = dx Θk (x). d (It’s really called a distribution, and we promise it makes sense.) Suppose dx f (x) = δ 1 (x) + δ 2 (x) and f (0) = f (0) = 0. What is f (5)?
≥
2
2
6. Point A is chosen randomly from the circumference of the unit circle, while point B is chosen randomly in the interior. A rectangle is then constructed using A, B as opposite vertices, with sides parallel or perpendicular to the coordinate axes. What is the probability that the rectangle lies entirely inside the circle?
√ ± −
7. A balloon in cross-section has the equation y = 2x x2 e−x/2 , with the x -axis beginning at the top of the balloon pointing toward the knot at the bottom. What is its volume? 8. Silas does nothing but sleep, drink coffee, and prove theorems, and he never more than one at a time. It takes 5 minutes to drink a cup of coffee. When doing math, Silas proves s + ln c theorems per hour, where c is the number of cups of coffee he drinks per day, and s is the number of hours he sleeps per day. How much coffee should Silas get in a day to prove the most theorems? 2n
1 . n→∞ k=n+1 k
9. Evaluate lim
10. Find the 10th nonzero term of the power series for f (x) =
x (x2 −1)2 (expanding
about x = 0).
Calculus Solutions 2007 Rice Math Tournament February 24, 2007 1. Answer: − 16 Use l’Hopital’s rule: lim
−1 + cos x = lim − sin x
x→0 3x2
x→0 6x + 12x2
= lim
x→0
− cos x
6 + 12x
√ 3
2. Answer:
+ 4x3
4
2
y = 3a2 + 3 =
a3 + 3a + 1 y = a x
2a3 1 =0 a 1 a3 = 2
−
3. Answer:
√
5
−1
2
The speed will cancel out so assume it is 1. We then have: t+1
τ +2 1 1 dt = 2 dt t τ τ +1 t τ + 1 τ + 2 ln = 2 ln τ τ + 1 2 τ + 1 τ + 2 = τ τ + 1
−1 ± τ =
√
5
2
4. Answer:
5 2
For odd n, I (n) =
−
cos(nx) n
5. Answer: 7
π 0
∞
= 2/n, so
∞
I (5n ) =
n=0
2/5n = 5/2
n=0
We have f (x) = (δ 1 (x) + δ 2 (x)) dx = Θ1 (x) + Θ2 (x) + C , and f (0) = 0 so C = 0. Integrating up to f is most easily accomplished graphically; the region under the curve from 0 to 5 is a 1 4 rectangle from x = 1 to x = 5 with a 1 3 rectangle from x = 2 to x = 5 on top. 6. Answer:
×
×
4 π
2
Suppose A lies at polar coordinate 0 < θ < π/2. For the rectangle to lie within the circle, B must lie in the rectangle with vertices at A, A reflected over the x-axis, A reflected over the y-axis, and A reflected over both axes. Thus for this fixed A, the probability is (2 sin θ)(2cos θ)/π = 2 sin(2θ)/π. The total π/2 probability is then π2 0 π2 sin(2θ)dθ. (Integrating over the circle requires taking the absolute value of the expression for area, which then splits up into four sections identical to the one considered here.)
1
7. Answer:
4π 2 e
2
V = π
− 2x
x2 e−x/2
0
2
2
dx = π
(2x
0
− x2 )e
x
−
dx = πx 2 e−x
8. Answer: 12 cups of coffee
2 0
The number of theorems proven is (s + ln c)(24 s c/12). Differentiating with respect to s gives c c 1 24 12 2s ln c = 0, so s = 12 24 2 ln c. This is a maximum in s since the second derivative c is 2. Plugging this back in and simplifying gives (12 24 + ln2c )2 = f (c)2 theorems proven. This differentiates to 2f (c)f (c), so the derivative will be zero when either f (c) or f (c) is zero. f (c) = 0 1 1 is difficult to solve, involving both a logarithm and a binomial, but f (c) = 2c 24 , so c = 12 is a solution. It is a maximum in c since the second derivative is 2f (c)2 + 2f (c)f (c), with f (12) < 0, f (12) > 0, and f (12) = 0.
− − − −
− −
− −
−
−
9. Answer: ln 2 2n
k=n+1
This is a Riemann sum: 10. Answer: 10x19
Note that f (x)dx =
2 1 1 x dx =
1
2(1−x2 )
=
n
k=1
1 = k + n
n
k=1
1 1 n1+
k n
ln2.
1 4
1 n = k n
−
1 1 1+x + 1−x
f (x)dx = =
. These are geometric sums, so we have
∞
1 4 1 2
∞
xk +
k=0
∞
x2k
k=0
∞
f (x) =
kx 2k−1
k=0
2
( x)k
k=0
Calculus Test 2006 Rice Math Tournament February 25, 2006 1. Evaluate: lim
d dx
sin x
x
x
x→0
2. Given the equation 4y + 3y − y = 0 and its solution y = e λt , what are the values of λ?
1 3. Find the volume of an hourglass constructed by revolving the graph of y = sin2 (x) + 10 from − π2 to about the x-axis.
4. Evaluate
1
x→
6. Evaluate
2
ln(x + 1) 0 x · ((1 + x ) − e)
lim
5. Evaluate:
π
(
x
x tan−1 x)dx π/ 2
0
sin3 x dx sin3 x + cos3 x
. 7. Find H n+1 (x) in terms of H n (x), H n (x), H n (x), . . . for
H n (x) = (−1)n ex
dn
2
dxn
2
e−x
8. A unicorn is tied to a cylindrical wizard’s magic tower with an elven rope stretching from the unicorn to the top of the tower. The tower has radius 2 and height 8; the rope is of length 10. The unicorn begins as far away from the center of the tower as possible. The unicorn is startled and begins to run as close to counterclockwise as possible; as it does so the rope winds around the tower. Find the area swept out by the shadow of the rope, assuming the sun is directly overhead. Also, you may assume that the unicorn is a point on the ground, and that the elven rope is so light it makes a straight line from the unicorn to the tower. 9. Define the function tanh x = simplify:
e e
x x
e−
−
x
+e−x
Let tanh d dx
−1
denote the inverse function of tanh. Evaluate and
tanh
−1
tan x
10. Four ants Alan, Bill, Carl, and Diane begin at the points (0, 0), (1, 0), (1, 1), and (0, 1), respectively. Beginning at the same time they begin to walk at constant speed so that Alan is always moving directly toward Bill, Bill toward Carl, Carl toward Diane, and Diane toward Alan. An approximate solution finds that after some time, Alan is at the point (0 .6, 0.4). Assuming for the moment that this approximation is correct (it is, to better than 1%) and so the pont lies on Alan’s path, what is the radius of curvature at that point. In standard Cartesian coordinates, the radius of curvature of a function y (x) is given by: R
1+ =
dy dx
d2 y dx2
1
2
3/2
Calculus Solutions 2006 Rice Math Tournament February 25, 2006 1. Answer: − 13 lim
d sin x dx x
x cos x − sin x x 0 x3 cos x − x sin x − cos x = lim x 0 3x2 sin x cos x 1 = lim − = lim − = − x 0 x 0 3x 3 3 = lim
x
x→0
→
→
→
→
2. Answer: 14 , −1 You substitute the solution of y = e λt into the differential equation. You then get 4λ2 eλt +3λeλt −eλt = 0 where you can divide through by e λt and end up with 4λ2 + 3λ − λ = 0. You get (4λ − 1)(λ + 1) = 0 where λ must be equal to 14 , −1. 3. Answer:
2
97π 200
The volume is π
2
−
π
2
1 π sin x + 10 2
π
dx = π
2
−
π
2
π2 = + π 100
1 1 sin x + sin 2 x + 5 100 4
dx
π
1 sin x(1 − cos x) + sin x dx 5 π 6 1 = + π sin x − sin 2x dx 100 5 4 π 6 π 1 π 97π 2
−
2
π
2
π
2
2
−
2
2
π
2
2
= 4. Answer:
100
2
+ π
52
−
1
200
0
ln(x+1) x
1
1 (1+x) 2 − e
. Let y = ln(x + 1) . Using L’Hopital’s rule, limx x
→
y
0 y
→
42
e
→
y
=
1
Rearrange to get lim x limx
2
2
1 0 y.
= limx
→
To evaluate the limit in the denominator, 1
ln (1+x) x
1
lim (1 + x) = lim e x
x→0
so the answer is
0
ln(1+x)
→
x
= e
limx
0
→
1 1+x 1
= e
1 e
5. Answer: − 2 + x
= e
x→0
limx
2
1+x 2
1
−
(x tan
−1
tan
x
x x)dx = − 2 x = − 2 x = − 2 x = − 2
+ + + +
1 1 + 2x tan 1 x dx 2 1 1 + x2 + 2x tan 1 x dx 2 1 + x2 1 d d (1 + x2 ) (tan 1 x) + (1 + x2 )tan 2 dx dx 1 + x2 tan 1 x 2
−
−
−
−
1
1
−
x dx
ln(y ) 0 y −e
=
6. Answer: 4 Let u = π2 − x. Substituting then changing u to x gives π
π/ 2
0
π/ 2
sin3 x dx = sin3 x + cos3 x
0
cos3 x dx cos3 x + sin3 x
Adding the two integrals π/ 2
sin3 x dx = sin3 x + cos3 x
2 ·
0
so
π/ 2
0
π/ 2
dx
0
sin3 x π 1 π dx = · = 2 2 4 sin3 x + cos3 x
7. Answer: 2xH (x) − H (x)
n
n
x2
H n+1 (x) = −e
d e dx
x2
−
−2xe
2
= −ex
H (x)
x2
−
n
H n (x) + e
x2
−
H n (x)
= 2xH n (x) − H n (x) 8. Answer: 18 + 9π
The shadow of the rope has length 6 by the Pythagorean theorem; we can work the rest of the problem pretending we have a horizontal rope of length 6 from the unicorn to the tower. The unicorn travels a quarter-circle before the rope begins to wrap around the tower; from then on, if the rope has wrapped around an angle θ of the tower, the rope remaining is 6 − 2θ long. Using the Pythagorean theorem, we find that if r is the unicorn’s distance from the center of the tower, r2 = 22 + (6 − 2θ)2 . The area swept out is the initial quarter-circle travelled by the unicorn plus the area swept out by the line from the unicorn to the center of the tower minus the area of the tower covered in the angle traversed: π62 1 + 4 2
6/2
(22 + (6 − 2θ)2 )dθ − π22
0
6 62Θ 2 Θ
2
6/2 2π
9. Answer: sec 2x 1
−
First we solve for the inverse function; let x = tanh
y.
ex − e x ex + e x e2x − 1 y = 2x e +1 2x e + 1 y = e 2x − 1 −
y = tanh x =
−
e2x (y − 1) = y + 1 1 y + 1 x = ln 2 y − 1
Now we return to the given expression: d tanh dx
1
−
tan x = = = = = =
10. Answer:
d 1 tan x + 1 ln dx 2 tan x − 1 1 d (ln (tan x + 1) − ln(tan x − 1)) 2 dx 1 sec2 x sec2 x − 2 tan x + 1 tan x − 1 1 (tan x − 1) − (tan x + 1) sec 2 x 2 tan2 x − 1 1 sec2 x cos2 x = sin2 x 1 − tan2 x 1 − cos 2x 1 = sec 2x cos2 x − sin2 x
1 5
By symmetry, Bill’s path must be the same as Alan’s, rotated 90 counter-clockwise and shifted to start at (1, 0). If Alan’s position is given by (x, y), Bill’s is then given by (1 − y, x). The slope of Alan’s path at a point (x, y) is then y = 1 x y y x . This is undefined at the given point, so noting that the sign of the derivatives in the radius of curvature does not matter, we switch to the function x(y), with x = 1 x y y x = 1 0.06.6 0.04.4 = 0 at the given point. Differentiating, x = ( 1 x )(x y(x) (yx)2 1)(1 y x) = ◦
−
−
−
−
−
−
−
−
−
(−1)(.6−.4)−(−1)(1−.6−.4) (.6−.4)2
−
−
−
−
−
−
−
= −5. The radius of curvature is then
1+ R =
dx dy
3/2
2
d2 x dy 2
(1 + 0)3/2 1 = = |−5| 5
The answer can also found using the parametric form of the radius curvature with ( x (t), y (t))
3
−
Calculus Test 2005 Rice Math Tournament February 26, 2005 1. Mathisgreatco, Inc. can produce at most 24 spherical cow statues each week. Experience has shown that the demand for spherical cows sets the price at D = 110 − 2n where n is the number of statues produced that week. Producing n statues costs 600 + 10n + n2 dollars. How many statues should be made each week to maximize profit? 2. f (x) = 21 f (2x + 1) and f (1) = 2005. What is lim x→−1+ f (x) assuming it exists? 3. Sammy the Owl wants to design a window that is a rectangle with a semicircle on top. If the total perimeter is constrained to be 24 feet, what dimensions should Sammy pick so that the window admits the greatest amount of light? Give the radius of the semicircular region and the height of the rectangular portion. 4. Given f (x) = x 5 + 2x3 + 2x, find (f −1 ) (−5). 5. Find the average value of f (x) = | x3 | on the interval [−1, 4]. 6. Sammy the Owl wants to watch his home movie of a trip to Rice University on his movie screen. The lower edge of the screen, which is 30 feet high, is 6 feet above eye level. How far from the screen should the observer sit to obtain the most favorable view, i.e. to maximize the visual angle (the observed angle between the top and bottom of the screen)? 7. Find the value of x for which the function f (x) = x x achieves a minimum on the positive real line (i.e. for all real numbers x such that x > 0)? 8. Evaluate
2
lim
n→∞
1+
0
t n + 1
n
dt
9. Define a new ”function” δ (x) with the properties δ (x) = 0 for all x = 0 and ∞
δ (x)f (x)dx = f (0)
−∞
for every function f(x). Compute
∞
δ (x)sin(x)dx.
−∞
For the purposes of this problem, you may assume the existence of δ (x) ≡
d δ (x) . dx
10. Compute ∞
( + 1) i i
i=0
for 0 < p < 1.
1
pi−2
Calculus Solutions 2005 Rice Math Tournament February 26, 2005 1. Answer: 17 The profit is P is revenue minus cost. Thus P = n D (600 + 10n + n2 ) = 3n2 + 100n 600. Consider the smoothed version of P , and we can find the critical point of 16 23 . Thus either n = 16 or n = 17 is the largest profit. Quick computation yields n = 17.
∗ −
−
−
2. Answer: 0 f (0) = 12 f (1), f (
− 12 ) = 12 f (0), f (− 34 ) = 12 f (− 12 ), and so on. So the limit is 0.
24 3. Answer: ( 4+ , π
24 ) 4+π
We want to maximize the area of the window. Let x be the radius of the semicircular region and y the height of the rectangular portion of the window. The area of the window is A = 12 πx2 + 2xy. The perimeter is P = πx + 2x + 2y . Eliminating y in A yields A = 24x (2 + π2 )x2 . We can take the 24 derivative and find the critical point of x = 4+ . Solving for y yields the same answer. π
−
4. Answer:
1 13
Applying the chain rule to f (f −1 (x)) = x yields f (f −1 (x)) (f −1 ) (x) = 1. Rearranging yields (f −1 ) (x) = f (f 1 (x)) . f −1 ( 5) is the solution to 5 = x 5 + 2x3 + 2x which has the unique solution of x = f ( 1) = 5 + 6 + 2 = 13. 1 Thus 13 is the answer.
·
−1
−
−
5. Answer:
−
−1.
257 20
The average value is
R 4
−1
f (x)dx
4−(−1)
=
1 5
√
4
0 −x3dx + 4 x3dx) = 257 . 1 3| = ( | x dx 5 20 1 1 0
−
−
6. Answer: 6 6 feet From the diagram below, we can see we want to maximize θ. Note that tan θ = tan(A B ) = − x −42x +9072 tan A−tan B 2 = x 42 1+tan A tan B = 1+ +216 . Using implicit differentiation, we can find that θ = cos θ (x +216) . 36
6
x
·
x
36 6
2
√ The maximum occurs at x = 6 6. x
7. Answer:
x
1 e
f (x) = x x
=
⇒ ln f (x) = x ln x f (x) =⇒ = 1 + ln x f (x)
=
⇒ f (x) = f (x)[1 + ln x] = x
x
[1 + ln x]
Critical points at f (x) = 0 =
⇒ 0 = x
=
⇒ 0 = 1 + ln x
x
[1 + ln x]
(xx always positive for x > 0)
=
⇒ ln x = −1
=
⇒ x = e
1
−
= 1
1 e
(critical point)
−
2
2
2
Furthermore, f (x) = f (x)[1 + ln x] +
f (x) x
f = x x (1 + ln x)2 + xx−1 > 0 (for x > 0)
So, f (x) > 0 for all x > 0. Hence, the function is concave up everywhere, and so x = 1e must be a minimum. 8. Answer: limn→∞
2
−1 1 + e
2 0
n
t
n+1
dt
t
n+1
= limn→∞
2 n+1
= e 2 9. Answer:
n+1
1 +
= limn→∞ 1 +
− 1 (since lim
|20 +1 − lim
n
n→∞
−1
Use integration by parts:
1+
n+1
1 + 0 n+1
n→∞ t n = et ) n
∞
δ (x) sin(x)dx
−∞ ∞
∞
= δ (x) sin(x)dx −∞
| −
=0
− cos(0)
± ∞)
−1
2 p(1−p)3 ∞
∞
i(i + 1) p
i−2
i=0
∞
i(i − 1) p
i−2
i=0
d2 = 2 dp
∞
ip
i−2
i=0
Then
δ (x) cos(x)dx
−∞
(δ (x) = 0“at =
10. Answer:
2 (1− p)3
2 2 + p(1− = p(1− p) p) 2
=
1 p
∞
i(i − 1) p =
i−2
i=0
∞
2
p = d i
1
dp2
i=0 ∞
ip
i−1
i=0
3
2
=
1
− p
ip +2
i−2
i=0
=
1 2 d = 2 (1 p)3 dp (1 p)
1 d
1
p dp 1
− p
−
=
−
1 p(1
− p)2
Calculus Test 2004 Rice Math Tournament February 28, 2004 1. Evaluate lim
√
( 4x2 + 7x
x→∞
− 2x).
2. Suppose the function f (x) f (2x) has derivative 5 at x = 1 and derivative 7 at x = 2. Find the derivative of f (x) f (4x) at x = 1.
−
−
3. An object moves along the x-axis with its position at any given time t During what time interval is the object slowing down?
4
5
≥ 0 given by x(t) = 5t − t .
4. For x > 0, let f (x) = x . Find all values of x for which f (x) = f (x). x
5. The highway department of North Eulerina plans to construct a new road between towns Alpha and Beta. Town Alpha lies on a long abandoned road running east west. Town Beta lies 3 miles north and 5 miles east of Alpha. Instead of building a road directly between Alpha and Beta, the department proposes renovating part of the abandoned road (from Alpha to some point P ) and then bulding a new road from P to Beta. If the cost of restoring each mile of old road is $200,000 and the cost per mile of a new road is $400,000, how much of the old road should be restored in order to minimize costs? 6. Consider the two graphs y = 2x and x 2 xy + 2y 2 = 28. What is the absolute value of the tangent of the angle between the two curves at the points where they meet?
−
7. A mouse is sitting in a toy car hooked to a spring launching device on a negligibly small turntable. The car has no way to turn, but the mouse can control when the car is launched and when the car stops (the car has brakes). When the mouse chooses to launch, the car will immediately leave the turntable on a straight trajectory at 1 m/s. Suddenly someone turns on the turntable; it spins at 30 rpm. Consider the set of points the mouse can reach in his car within 1 second after the turntable is set in motion. What is the area of this set? 8. A spherical cow is being pulled out of a deep well. The bottom of the well is 100 feet down and the cow and all the water on him weighs 200 pounds. He is b eing hauled up at a constant rate with a chain which weighs 2 pounds per foot. The water on the cow drips off the cow at the rate of 21 pounds per foot as he is being hauled up. How much work is required to rescue the cow in foot-pounds? Remember work is force times distance. 9. The base of a solid is the region between the parabolas x = y 2 and 2y 2 = 3 x. Find the volume of the solid if the cross-sections perpendicular to the x-axis are equilateral triangles.
−
10. Find the positive constant c0 such that the series
∞
n=0
converges for c > c0 and diverges for 0 < c < c 0 .
1
n! (cn)
n
Calculus Solutions 2004 Rice Math Tournament February 28, 2004 1. Answer:
7 4
√ √ √ (√ 4 2 2 lim →∞ ( 4x + 7x − 2x) = lim →∞ ( 4x + 7x − 2x) · ( 4 lim →∞ √ 7 = 74 . ( 4+ +2) x
x2 +7x+2x) = x2 +7x+2x)
x
x
7x limx→∞ (√ 4x2 +7 x+2x)
·
1 x
1
=
x
7
x
2. Answer: 19 The derivative of f (x)
− f (2x) is f (x) − 2f (2x). So f (1) − 2f (2) = 5, f (2) − 2f (4) = 7. Thus f (1) − 4f (4) = (f (1) − 2f (2)) + 2(f (2) − 2f (4)) = 5 + 2 · 7 = 19,
the answer. 3. Answer: [3,4] or (3,4) or from t=3 to t=4 The velocity of the object is given by v(t) = x (t) = 20t3 5t4 , and the acceleration function is a(t) = v (t) = 60t2 20t3 . The object is slowing down when the velocity is positive and the acceleration is negative, or vice versa. v(t) is positive from t = 0 to t = 4 and is negative after that. a(t) is positive from t = 0 to t = 3 and negative afterward. These only differ in sign from t = 3 to t = 4.
−
−
4. Answer: 1 Let g(x) = log f (x) = x log x. Then 1 + log x = 1, that is, when x = 1. 5. Answer: 5 −
f (x) f (x)
= g (x) = 1 + log x. Therefore f (x) = f (x) when
√
3 miles
Let x be the amount of old road restored. Then the length of the new road is 9 + (5 x)2 using the Pythagorean Theorem. Thus the total cost of the plan is C (x) = 200000x + 400000 x2 10x + 34. The minimum cost occurs at one of the critical points which are x = 5 3. Clearly 5 + 3 is not a valid answer and one can check 5 3 is indeed a minimum.
± √
− √
√ − − √
6. Answer: 2 The two graphs intersect at x2 2x2 + 8x2 = 28 or rather x = 2 with y = 4. At x = +2, m1 = 2 and m 2 = y (2). Using implicit differentiation on the second graph, we find y (x) = y4y−−2xx and plugging m2 −m1 in (2, 4) gives a slope of 0. If α is the angle between the graphs then tan(α) = 1+ . Plugging m1 m2 in the values yields the answer 2. x = 2 yields the same value.
−
±
±
|
−
| |
|
7. Answer: π/6 The mouse can wait some amount of time while the table rotates and then spend the remainder of the time moving along that ray at 1 m/s. He can reach any point between the starting point and the furthest reachable point along the ray, (1 θ/π) meters out. So the area is
−
π
π
(1/2)(1
0
− θ/π)
2
dθ = (1/2)(1/π)
2
θ 2 dθ = π/6.
0
8. Answer: 27500 foot-pounds Let x indicate the distance the cow has yet to travel. Then the work for a distance dx is (2x + 200 100 1 x))dx. Thus the total work is 0 ( 52 x + 150)dx = 27500 foot-pounds. 2 (100
−
1
−
