2 Geometrical Ap A pplications of Calculus TERMINOLOGY TERMINOLO GY
Anti-dierentiation: The process o fnding a primitive Anti-dierentiation: (original) unction rom the derivative. It is the inverse operation to dierentiation
Minimum turning point: point: A local stationary point (where the frst derivative is zero) and where the curve is concave up. The gradient o the tangent is zero
Concavity: The shape o a curve as it bends around (it can Concavity: be concave up or concave down)
Monotonic increasing or decreasing unction: unction: A unction is always increasing or decreasing
Dierentiation: The process o fnding the gradient o a Dierentiation: tangent to a curve or the derivative
Point o infexion: infexion: A point at which the curve is neither concave upwards nor downwards, but where the concavity changes
Gradient o a secant: secant: The gradient (slope) o a line between two points that lie close together on a curve
Primitive unction: unction: The original unction ound by working backwards rom the derivative. Found by antidierentiation
Gradient o a tangent: tangent: The gradient (slope) o a line that is a tangent to a curve at a point on a unction. It is the derivative o the unction
Rate o change: change: The rate at which the dependent variable changes as the independent variable changes
Horizontal point o infexion: infexion: A stationary point (where the frst derivative is zero) where the concavity o the curve changes Instantaneous rate o change: change: The derivative o a unction
Stationary (turning) point: point: A local point at which the gradient o the tangent is zero and the tangent is horizontal. The frst derivative is zero
Maximum turning point: point: A local stationary point (where the frst derivative is zero) and where the curve is concave down. The gradient o the tangent is zero
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Chapter 2 Geometrical Applications of Calculus
51
INTRODUCTION YOU LEARNED ABOUT differentiation in the Preliminary Course. Course. This is the
process of finding the gradient of a tangent to a curve. This chapter looks at how the gradient of a tangent can be used to describe the shape of a curve. Knowing this will enable us to sketch various curves and find their maximum and minimum values. The theory also allows us to solve various problems involving maximum and minimum values.
DID YOU KNOW? Although Newton and Leibniz are said to have discovered calculus, elements o calculus were around beore then. It was Newton and Leibniz who perfected the processes and developed the notation associated with calculus. Pierre de Fermat (1601–65) used coordinate geometry to fnd maximum and minimum values o unctions. His method is very close to calculus. He also worked out a way o fnding the tangent to a curve. The 17th-century mathematicians who developed calculus knew that it worked, but it was not ully understood. Limits were not introduced into calculus until the nineteenth century.
Gradient of a Curve To learn about the shape of a curve, we first need to revise what we know about the gradient of a tangent. The gradient (slope) of a straight line measures the rate of change of y with respect to the change in x.
Since the gradient of a curve varies, we find the gradient of the tangent at each point along the curve.
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Maths In Focus Mathematics Extension 1 HSC Course
EXAMPLES 1.
2.
3.
4.
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Chapter 2 Geometrical Applications of Calculus
53
In the examples on the previous page, where the gradient is positive, the curve is going up, or increasing (reading from left to right). Where the gradient is negative, the curve is going downwards, or decreasing. The gradient is zero at particular points of the curves. At these points the curve isn’t increasing or decreasing. We say the curve is stationary at these stationary at points.
]g ]g ]g
l
If f x If f x If f x l
l
0, the curve is increasing < 0, the curve is decreasing = 0, the curve is stationary >
A curve is monotonic increasing or decreasing if it is always increasing or decreasing; that is, if ff x > 0 for all x (monotonic increasing) if or f x < 0 for all x (monotonic decreasing)
]g ]g
l
l
EXAMPLES
]g
1. Find all x values for which the curve f x
=
x2 − 4x + 1 is increasing.
Solution
]g ]g
l
f x = 2x − 4 f x > 0 for increasing curve i.e. 2x − 4 > 0 2x > 4 x>2 So the curve is increasing for x l
>
2.
This unction is a parabola.
CONTINUED
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Maths In Focus Mathematics Extension 1 HSC Course
2. Find the stationary point on the parabola y parabola y = x2 − 6x + 3.
Solution dy dx
=
2x − 6 dy
For statio stationar nary y point points, s,
=0 dx i.e. 2x − 6 = 0 2x = 6 x=3 2 When x = 3, y = 3 − 6 3 + 3 = −6
]g
^
h
So the stationary point is 3, − 6 .
3. Find any stationary points on the curve y curve y = x3 − 48x − 7.
Solution l
y
=
3x2 − 48 l
For statio stationar nary y points points,, y = 0 i.e. 3 x 2 − 48 = 0 3x2 = 48 x2 = 16 ` x = ±4 When x
You will use stationary points to sketch curves later in this chapter.
When x
=
4, y = 43 − 48 (4) − 7 = − 135
= − 4,
y = =
] 4g
3
−
−
48 (− 4) − 7
121
^
h
So the stationary points are 4, −135 and
^
− 4,
h
121 .
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Chapter 2 Geometrical Applications of Calculus
55
PROBLEM What is wrong with this working out ? Find the stationary point on the curve y curve y = 2x2 + x − 1.
Solution l
y = 4x + 1 For statio stationar nary y point points, s, y = 0 i.e. 4x + 1 = 0 4x = − 1 x = − 0.25 l
When x = − 0.25, y = 4 (− 0.25) + 1 = −1 + 1 =0 So the stationary point is − 0.25 25, 0 .
^
Can you fnd the correct answer?
h
2.1 Exercises 1.
Find the parts of each curve where the gradient of the tangent is positive, negative or zero. Label each curve with +, − or 0.
(d)
(a)
2.
Find all values of x of x for which the curve y curve y = 2x 2 − x is decreasing.
3.
Find the domain over which the function f x = 4 − x 2 is increasing.
4.
Find values of x of x for which the 2 y = x − 3x − 4 is curve y curve (a) decreasing (b) increasing (c) stationary.
5.
Show that the function f x = − 2x − 7 is always (monotonic) decreasing.
(b)
]g
(c)
]g
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Maths In Focus Mathematics Extension 1 HSC Course
6.
7.
Prove that y that y = x 3 is monotonic increasing for all x ≠ 0. Find the stationary point on the curve f (x) = x 3.
8.
Find all x values for which the curve y curve y = 2x 3 + 3x 2 − 36x + 9 is stationary.
9.
Find all stationary points on the curve (a) y = x 2 − 2x − 3 (b) f x = 9 − x 2 (c) y = 2x 3 − 9x 2 + 12x − 4 (d) y = x 4 − 2x 2 + 1.
]g
]g ]g
l
17. Sketch a function with f x for x < 2, f 2 = 0 and f x when x > 2.
]g
l
l
0 <0 >
18. Draw a sketch showing a curve dy dy with < 0 for x < 4, = 0 when dx dx dy x = 4 and > 0 for x > 4. dx 19. Sketch a curve with dy
x ≠ 1 and
dx
=
dy dx
>
0 when x
0 for all =
1.
10. Find any stationary points on the y = x − 2 4. curve y curve
20. Draw a sketch of a function that has f x > 0 for x < −2, x > 5, f x = 0 for x = − 2, 5 and f x < 0 for − 2 < x < 5.
11. Find all values of x of x for which the curve f (x) = x 3 − 3x + 4 is decreasing.
21. A function has f (3) = 2 and f 3 < 0. Show this information on a sketch.
12. Find the domain over which the curve y curve y = x 3 + 12x 2 + 45x − 30 is increasing.
22. The derivative is positive at the point (−2, −1). Show this information on a graph.
13. Find any values of x of x for which the 3 curve y curve y = 2x − 21x 2 + 60x − 3 is (a) stationary (b) decreasing (c) increasing.
23. Find the stationary points on the y = 3x − 1 x − 2 4. curve y curve
]
g
]g
14. The function f x = 2x 2 + px + 7 has a stationary point at x = 3. p.. Evaluate p Evaluate 15. Evaluate a and b if y = x 3 − ax 2 + bx − 3 has stationary points at x = − 1 and x = 2. 16. (a) Find the derivative of y = x 3 − 3x 2 + 27x − 3. (b) Show that the curve is monotonic increasing for all values of x of x.
]g
l
]g
l
]g
l
]g
l
g]
]
g
24. Differentiate y Differentiate y = x x + 1 . Hence find the stationary point on the curve, giving the exact value. 25. The curve f (x) = a ax x 4 − 2x 3 + 7x 2 − x + 5 has a stationary point at x = 1. Find the value of a of a. 26. Show that f (x) = x has no stationary points.
]g
1 has no x3 stationary points.
27. Show that f x
=
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