Math 107. Rumbos Fall Solutions to Review Problems for Exam 3

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1 Math 17. umbos Fall 29 1 Solutions to eview Problems for Eam 3 1. Consider a wheel of radius a which is rolling on the ais in the plane. Suppose that the center of the wheel moves in the positive direction and a constant speed v o. Let P denote a fied point on the rim of the wheel. (a) Give a path σ(t) ((t), (t)) giving the position of the P at an time t, if P is initiall at the point (, 2a). Solution: Let θ(t) denote the angle that the ra from the center '$ r )&% (, r Figure 1: Circle of the circle to the point ((t), (t)) makes with a vertical line through the center. Then, v o t aθ(t); so that θ(t) v o t and a and (t) v o t + a sin(θ(t)) (t) a + a cos(θ(t)) (b) Compute the velocit of P at an time t. When is the velocit of P horizontal? What is the speed of P at those times? Solution: The velocit vector is σ (t) ( (t), (t)) (v o + aθ (t) cos(θ(t)), aθ (t) sin(θ(t))) where We then have that θ (t) v o a. σ (t) (v o + v o cos(θ(t)), v o sin(θ(t))).

2 Math 17. umbos Fall 29 2 The velocit of P is horizontal when or where n is an integer; and when sin(θ(t)), θ(t) nπ, cos(θ(t)) 1. We then get that the velocit of P is horizontal when θ(t) 2kπ where k is an integer. The speed at the points where the velocit if horizontal is then equal to 2v o. 2. Let C {(, ) , }; i.e., C is the upper unit semi circle. C can be parametrized b σ(τ) (τ, 1 τ 2 ) for 1 τ 1. (a) Compute s(t), the arclength along C from ( 1, ) to the point σ(t), for t 1. ( ) Solution: Compute σ τ (τ) 1,. for all τ ( 1, 1). 1 τ 2 Then, It then follows that s(t) σ (τ) t τ τ 2 1 τ τ 2. dτ for 1 t 1. (b) Compute s (t) for 1 < t < t and sketch the graph of s as function of t. Solution: B the Fundamental Theorem of Calculus, s (t) 1 1 t 2 for 1 < t < 1.

3 Math 17. umbos Fall 29 3 Note then that s (t) > for all t ( 1, 1) and therefore s is strictl increasing on ( 1, 1). Net, compute the derivative of s (t) to get the second derivative of s(t): s t (t) for 1 < t < 1. (1 t 2 ) 3/2 It then follows that s (t) < for 1 < t < and s (t) > for < t < 1. Thus, the graph of s s(t) is concave down on ( 1, ) and concave up on (, 1). Finall, observe that s( 1), s() π/2 (the arc length along a quarter of the unit circle), and s(1) π (the arc length along a semi circle of unit radius). We can then sketch the graph of s s(t) as shown in Figure 2. s Figure 2: Sketch of s s(t) t (c) Show that cos(π s(t)) t for all 1 t 1, and deduce that sin(s(t)) 1 t 2 for all 1 t 1. Solution: Figure 3 shows the upper unit semicircle and a point σ(t) on it. Putting θ(t) π s(t), then θ(t) is the angle, in radians, that the ra from the origin to σ(t) makes with the positive ais. It then follows that cos(θ(t)) t

4 Math 17. umbos Fall 29 4 and Since the result follows. sin(θ(t)) 1 t 2. sin(θ(t)) sin(π s(t)) sin(s(t), s σ(t) θ(t) Figure 3: Sketch of Semi circle 3. Let C denote the unit circle traversed in the counterclockwise direction. Evaluate the line integral 2 d 2 d. C Solution: Let F (, ) 2 ˆi + 2 ˆj. Then, C 2 d 2 d F ˆn ds. C Thus, b Green s Theorem in divergence form, 2 d 2 d divf d d, C where is the unit disc bounded b C, and divf (, ) ( ) + ( 2 2) Consequentl, C 2 d 2 d d d area() π.

5 Math 17. umbos Fall Let F (, ) 2 ˆi ˆj and be the square in the plane with vertices (, ), (2, 1), (3, 1) and (1, 2). Evaluate F n ds. Solution: Appl Green s Theorem in divergence form, F n ds divf d d, where Thus, divf (, ) (2) + ( ) F n ds d d area(). To find the area of the region, shown in Figure 4, observe that is w v Figure 4: Sketch of egion in Problem 4 a parallelogram determined b the vectors v 2 ˆi ˆj and w ˆi+2 ˆj. Thus, area() v w 5. It the follows that F n ds d d 5.

6 Math 17. umbos Fall Evaluate the line integral rectangular region ( 4 + ) d + (2 4 ) d, where is the {(, ) 2 1 3, 2 1}, and is traversed in the counterclockwise sense. Solution: Write ( 4 + ) d + (2 4 ) d so that where F is the vector field ( 4 + ) d + (2 4 ) d (2 4 ) d [ ( 4 + )] d, F (, ) (2 4 ) ˆi ( 4 + ) ˆj. Then, b Green s Theorem in divergence form, ( 4 + ) d + (2 4 ) d divf where F n ds, d d, divf (, ) (2 4 ) (4 + ) It then follows that ( 4 + ) d + (2 4 ) d d d area() Integrate the function given b f(, ) 2 over the region,, defined b: {(, ) 2, 4 2 }.

7 Math 17. umbos Fall Figure 5: Sketch of egion in Problem 8 Solution: The region,, is sketched in Figure 5. We evaluate the double integral, 2 d d, as an iterated integral 2 d d d d d d d (4 2 ) 3 d. To evaluate the last integral, make the change of variables: u 4 2. We then have that du 2 d and 2 d d d d 1 6 u 3 4 du u 3 du.

8 Math 17. umbos Fall 29 8 Thus, 2 d d Let denote the region in the plane defined b inside of the ellipse for a > and b >. (a) Evaluate the line integral 2 a + 2 1, (1) 2 b2 traversed in the positive sense. d d, where is the ellipse in (1) Solution: A sketch of the ellipse is shown in Figure 6 for the case a < b. b a Figure 6: Sketch of ellipse A parametrization of the ellipse is given b a cos t, b sin t, for t 2π.

9 Math 17. umbos Fall 29 9 We then have that d a sin t dt and d b cos t dt. Therefore 2π d d [a cos t b cos t b sin t ( a cos t)] dt 2π [ab cos 2 t + ab sin 2 t] dt ab 2π (cos 2 t + ab sin 2 t) dt 2π ab dt 2πab. (b) Use our result from part (a) and the divergence form of Green s theorem to come up with a formula for computing the area of the region enclosed b the ellipse in (1). Solution: Let F (, ) ˆi + ˆj. Then, d d F n ds. Thus, b Green s Theorem in divergence form, d d divf d d, where Consequentl, It then follows that Thus, divf (, ) () + () 2. d d 2 area() 1 2 b the result in part (a). area() πab, d d 2 area(). d d.

10 Math 17. umbos Fall Evaluate the double integral plane sketched in Figure 7. e 2 d d, where is the region in the 2 Figure 7: Sketch of egion in Problem 8 Solution: Compute e 2 d d 2 2 e 2 d d 2 2e 2 [ e 2] 2 d 1 e 4.