UTM Midterm Solutions MAT 232 Jacob

P 1 Consider the polar equation:

() = sin(/2) (a) Sketch the polar curve defined by the above equation (make sure to draw all the curve!) (b) Find all points of vertical or horizontal tangency; (c) Find the point(s) of intersection intersection of the curve with the vertical vertical line passing through the pole (d) Find the tangent slope(s) at points found at the previous step (e) Compute Compute the length length of the curve H: you might find the following identities of some use:

cos  sin (/2) = 1 − 2cos

cos  cos (/2) = 1 + 2cos

2

S: Observe that we need to take

2

 0 ≤  ≤ 4

. Te sketch looks like

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

Let us write the equation for the slope of the curve at angle :

=  =

d d

 sin  sin  +  cos cos  =  cos  cos  −  sin   cos(/2 os(/2)) sin sin  + sin(/2 n(/2)) cos cos   cos(/2 os(/2)) cos cos  − sin(/2 n(/2)) sin sin 

d d d d 1 2 1 2

2

Let us look for horizontal tangencies: we need to find zeros of the numerator

cos(/2)sin + 2sin(/2)cos = 0  cos = 2cos (/2) − 1  sin = 2sin(/2)cos(/2) 2sin(/2)cos (/2) + 2sin(/2)(2cos (/2) − 1) = 0.  2sin(/2)  = 0   = 2 =0 3cos (/2) = 1 cos(/2) = ± 1√ 3 2

Using the formula provided we have . We gather 2

we can factor (which give

 and

2

, which gives us the solutions  and : both correspond to the pole); the other solutions are 2

which correspond to the other four solutions. Now let us find vertical tangencies; we need to find zeros of the equation

cos(/2)cos − sin(/2)sin = 0 Using the previous formulas we have:

cos(/2)(2cos (/2) − 1) − 4sin (/2)cos(/2) = 0  cos(/2) /2 = /2 /2 = 3/2   = ,3 =1 2cos (/2) − 1 − 4sin (/2) = 0 6cos (/2) = 5 2

2

We can now factor , which gives and that is  (for which ). Ten we obtain the equation:

,

2

2

2

which gives

cos(/2) = ± 56 which corresponds to the other four solutions. In order to find points of intersection with the indicated line we need to take , (which give )and (that is, the pole). We already found that the pole is a point of horizontal tangency. Plugging in the values of   in the formula for the slope, we obtain that the slope is

 = /2,3/2,5/2,7/2   ±1/2

|| = 1/√ 2  = 0

3

Te length of the curve can be computed using the formula:

 = ∫  14 cos (/2) + sin (/2) = = 2√ 1 2 ∫  1 + cos + 4(1 − cos) = = 2√ 1 2 ∫ √ 5 − 3cos. 4

2

0

4

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4

0

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