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Calc 1: Practice Exam 2 Solutions Name: (1) Find ... - Westmont

Calc 1: Practice Exam 2 Solutions Name: (1) Find ... - Westmont

The derivative always

The derivative always exists, so the only critical points arewhere v ′ (r) = 0:2kr 0 r − 3kr 2 = 0 ⇒r(2kr 0 − 3kr) = = ⇒r = 0 or r = 2r 03Now, when r = 0 the person has suffocated, so we need onlyevaluate v(r) when r = 2r 0and when r equals one of the endpoints:36v( 2r 0) = 4kr3 0− 8r3 0= (12k+8)r3 03 9 27 27v( 1 2 r 0) = kr3 04 − kr3 08=kr 3 04v(r 0 ) = kr 3 0 − kr 3 0 = 0Since 12k > k it is straightforward to see that (12k+8)r3 0> kr3 027 4 27Thus, r = 2r 0gives the maximum velocity.3(9) Find the speed of the following curve at the point when t = 0:{ x(t) = 2 sin(4t)y(t) = 3 sin(5t)4 .Answer: The speed is defined to be √ (x ′ ) 2 + (y ′ ) 2 . Differentiatingx and y we obtain: x ′ (t) = 8 cos(4t) and y ′ (t) =15 cos(5t). When t = 0 we have x ′ (0) = 8 and y ′ (0) = 15. Sothe speed when t = 0 is √ 8 2 + 15 2 .(10) If two resistors with resistances R 1 and R 2 are placed in parallelin a circuit, the total resistance of the circuit, R is related toR 1 and R 2 be the equation:1R = 1 R 1+ 1 R 2If R 1 is increasing at a rate of .4 ohms per second and R 2is increasing at a rate of .6 ohms per second, how fast is Rchanging when R 1 = 20 ohms and R 2 = 30 ohms?

Answer: We know that dR 1= .4 and dR 2= .6. We wish todt dtfind dR . Differentiating the given equation with respect to t wedtobtain:− 1 dRR 2 dt = − 1 dR 1− 1 dR 2R12 dt R22 dtPlugging in the values for dR 1and dR 2and dividing the wholedt dtthing by (−1) we obtain:1 dRR 2 dt = 1 (.4) + 1 (.6)R1 2 R22When R 1 = 20 and R 2 = 30 we solve and find out thatR = 12. Thus:So:1 dR144 dt = 1400 (.4) + 1900 (.6)dRdt = ( 11000 + 2 )144 = .2430007(11) A baseball diamond is a square with sides 90 ft. A batter hitsthe ball and runs toward first base with a speed of 24 feet persecond. At what rate is his distance from 3rd base increasing,when he is halfway to 1st base? (The bases are located at thecorners of the square).Answer: Draw a picture. The path from 1st base to 3rd base(straight across the field) to home is a right triangle. When theplayer is x feet away from home the distance to 3rd base isd(x) = √ x 2 + 90 2 . Rewrite this as:d 2 = x 2 + 90 2Taking the derivative with respect to t:Thus:2d dddt = 2xdx dtdddt = x dxd dt

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