AP BC Calculus Free Response 2014

2. The graphs of the polar curves r=3 and r=3-2sin(2θ) are shown in the figure above for 0≤θ≤π. (a) Let R be the shaded region that is inside the graph of r=3 and inside the graph of r=3-2sin(2θ). Find the area of R.

(a) *Area=½∫f(θ)²dθ* =¼ of circle + area under r₂ =(9π/4)+½∫{0,π/2} (3−2sin(2θ))²dθ = 9.708 (or 9.707)

1. Grass clippings are placed in a bin, where they decompose. For 0≤t≤30, the amount of grass clippings remaining in the bin is modeled by A(t)= 6.687(0.931)^t , where A(t) is measured in pounds and t is measured in days. (a) Find the average rate of change of A(t) over the interval 0≤t≤30. Indicate units of measure.

(a) *∆y/∆x* [A(30) − A(0)] / [30-0] = -0.197 (or 0.196) lbs/day *Indicate units of measure*

4. (NO CALC) Train A runs back and forth on an east-west section of railroad track. Train A's velocity, measured in meters per minute, is given by a differentiable function v_A(t), where time t is measured in minutes. Selected values for v_A(t) are given in the table above. (a) Find the average acceleration of train A over the interval 2 ≤ t ≤ 8.

(a) *∆y/∆x* [v(8)-v(2)] / [8-2] =[-120-100] / [6] =-110/3 m/min²

5. (NO CALC) Let R be the shaded region bounded by the graph of y=xe^(x²), the line y=-2x, and the vertical line x=1, as shown in the figure above. (a) Find the area of R.

(a) Area=∫{0,1} (xe^(x²)-(-2x))dx *U-Substitution* on first term let u=x² du=2x dx dx=(1/2)(1/x) (1/2)∫{0,1} e^(u) du (1/2)e^(u) (1/2)e^(x²) =[½e^(x²)+x²] ]{x=0,x=1} (½e+1)-½=(e+1)/2

6. (NO CALC) The Taylor series for a function f about x=1 is given by ∑{n=1,∞}(−1)ⁿ⁺¹(2ⁿ/n)(x − 1)ⁿ and converges to f (x) for |x − 1|<R, where R is the radius of convergence of the Taylor series. (a) Find the value of R.

(a) Let aₙ be the nth term of the Taylor series. * lim{n→∞} (aₙ-₁/aₙ)<1* (aₙ-₁/aₙ)=[(−1)ⁿ⁺²(2ⁿ⁺¹)(x − 1)ⁿ⁺¹/(n+1)] [n/(−1)ⁿ⁺¹(2ⁿ)(x − 1)ⁿ] =[-2(x-1)]/[n+1] lim{n→∞} |[-2(x-1)]/[n+1]|=2|x-1| 2|x-1|<1 |x-1|<½ The radius of convergence is R=½

3. (NO CALC) The function f is defined on the closed interval [-5, 4]. The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by g(x)=∫{-3,x} f(t) dt. (a) Find g(3).

(a) g(3) = ∫{-3,3} f(t) dt = 6 + 4 −1 = 9

5. (NO CALC) Let R be the shaded region bounded by the graph of y=xe^(x²), the line y=-2x, and the vertical line x=1, as shown in the figure above. (b) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y =−2.

(b) *Volume= π∫f(x)²dx* *Volume= π∫[(f(x)-R)²-(g(x)-r)²]dx* π∫{0,1}[(xe^(x²) + 2)² − (-2x + 2)² ] dx

2. The graphs of the polar curves r=3 and r=3-2sin(2θ) are shown in the figure above for 0≤θ≤π. (b) For the curve r=3-2sin(2θ), find the value of dx/dθ at θ=π/6.

(b) *x=r cos(θ)* x=(3-2sin(2θ))cosθ dx/dθ ]{θ=π/6} = −2.366

1. Grass clippings are placed in a bin, where they decompose. For 0≤t≤30, the amount of grass clippings remaining in the bin is modeled by A(t)= 6.687(0.931)^t , where A(t) is measured in pounds and t is measured in days. (b) Find the value of A'(15). Using correct units, interpret the meaning of the value in the context of the problem.

(b) A′(15) = −0.164 (or −0.163) *nDeriv* on CALC The amount of grass clippings in the bin is *decreasing* at a rate of 0.164 (or 0.163) *lbs/day* at time t = 15 days.

6. (NO CALC) The Taylor series for a function f about x=1 is given by ∑{n=1,∞}(−1)ⁿ⁺¹(2ⁿ/n)(x − 1)ⁿ and converges to f (x) for |x − 1|<R, where R is the radius of convergence of the Taylor series. (b) Find the first three nonzero terms and the general term of the Taylor series for f′, the derivative of f, about x = 1.

(b) The first three nonzero terms are 2-4(x-1)+8(x-1)². The general term is (−1)ⁿ⁺¹2ⁿ(x−1)ⁿ⁻¹ for n≥1. Using ∑{n=1,∞}(−1)ⁿ⁺¹(2ⁿ/n)(x − 1)ⁿ f=2(x-1)-2(x-1)²+(8/3)(x-1)³ f'=2-4(x-1)+8(x-1)²

3. (NO CALC) The function f is defined on the closed interval [-5, 4]. The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by g(x)=∫{-3,x} f(t) dt. (b) On what open intervals contained in −5 < x < 4 is the graph of g both increasing and concave down? Give a reason for your answer.

(b) g′(x) = f (x) The graph of g is increasing and concave down on the intervals −5<x<−3 and 0<x<2 because g′=*f is* *positive* and *decreasing* on these intervals.

4. (NO CALC) Train A runs back and forth on an east-west section of railroad track. Train A's velocity, measured in meters per minute, is given by a differentiable function v_A(t), where time t is measured in minutes. Selected values for v_A(t) are given in the table above. (b) Do the data in the table support the conclusion that train A's velocity is −100 meters per minute at some time t with 5 < t < 8? Give a reason for your answer.

(b) v_A is *differentiable* ⇒ v_A is *continuous* v(8)=-120<-100<40=v(5) Therefore, by the *Intermediate Value Theorem*, there is a time t, 5<t<8, such that v_A(t)=−100.

1. Grass clippings are placed in a bin, where they decompose. For 0≤t≤30, the amount of grass clippings remaining in the bin is modeled by A(t)= 6.687(0.931)^t , where A(t) is measured in pounds and t is measured in days. (c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0≤t≤30.

(c) *Average Value of a Function* *[∫ f(x) dx]/[b-a]* A(t)= (1/30)∫{0,30} A(t) dt =-0.197 *Solver on CALC* ⇒ t =12.415 (or 12.414)

6. (NO CALC) The Taylor series for a function f about x=1 is given by ∑{n=1,∞}(−1)ⁿ⁺¹(2ⁿ/n)(x − 1)ⁿ and converges to f (x) for |x − 1|<R, where R is the radius of convergence of the Taylor series. (c) (c) The Taylor series for f′ about x=1, found in Part B, is a geometric series. Find the function f ′ to which the series converges for |x-1|<R. Use this function to determine f for |x-1|<R.

(c) *Geometric Series Sum* *a/(1-r)* found a=2 The common ratio is -2(x-1). found r=-2(x-1) *Find C* f'(x)=[2]/[1-(-2(x-1))]=[2]/[2x-1] for |x-1|<½ since [2]/[2x-1]<1 becomes |x-1|<½ f(x)=∫[2]/[2x-1]dx=ln|2x-1|+C found f(1)=0 ln|1|+C=0 so C=0 f(x)=ln|2x-1| for *|x-1|<½*

3. (NO CALC) The function f is defined on the closed interval [-5, 4]. The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by g(x)=∫{-3,x} f(t) dt. (c) The function h is defined by h(x) = g(x)/5x. Find h′(3).

(c) *Quotient Rule* *[vu'-uv'] / v²* h′(x) = [5xg′(x) − g(x)5] / (5x)² = [5xg′(x) − 5g(x)] / [25x²] h′(3) = [(5)(3)g′(3) − 5g(3)] / [25·3²] found g(3)=9 found g'(3)=f(3)=-2 =[15(-2)-5(9)] / [225] -75/225 -1/3

4. (NO CALC) Train A runs back and forth on an east-west section of railroad track. Train A's velocity, measured in meters per minute, is given by a differentiable function v_A(t), where time t is measured in minutes. Selected values for v_A(t) are given in the table above. (c) At time t = 2, train A's position is 300 meters east of the Origin Station, and the train is moving to the east. Write an expression involving an integral that gives the position of train A, in meters from the Origin Station, at time t = 12. Use a trapezoidal sum with three subintervals indicated by the table to approximate the position of the train at time t = 12.

(c) *x(t)=∫v(t)dt + C* s(12)=s(2)+∫{2,12} v(t) dt 300+∫{2,12} v(t) dt where ∫{2,12} v(t) dt≈3[(100+40)/2]+3[(40-120)/2]+4[(-120-150)/2]=-450 s(12)≈300-450=-150 The position of Train A at time t=12 minutes is approximately 150 meters west of Origin Station.

5. (NO CALC) Let R be the shaded region bounded by the graph of y=xe^(x²), the line y=-2x, and the vertical line x=1, as shown in the figure above. (c) Write, but do not evaluate, an expression involving one or more integrals that gives the perimeter of R.

(c) Perimeter = Length of Curve+Side1+Side2 *Length of a Curve* *∫√[1+(dy/dx)²] dx* Find dy/dx y'=(d/dx)(xe^(x²)) *Product Rule* u=x and u'=1 v=e^(x²) and v'=2x²e^(x²) y'=e^(x²)(1+2x²) For y₁=-2x let x=1 *x²+y²=r²* 1²+(-2)²=r² r=√[5] y=-2 For y₂=xe^(x²) x=1 y=e Perimeter = ∫√[1+(e^(x²)(1+2x²))²] dx+√[5]+(e+2)

2. The graphs of the polar curves r=3 and r=3-2sin(2θ) are shown in the figure above for 0≤θ≤π. (c) The distance between the two curves changes for 0≤θ≤π/2 . Find the rate at which the distance between the two curves is changing with respect to θ when θ=π/3.

(c) The distance between the two curves is *D=y₁-y₂* D=3− (3− 2sin(2θ))=2sin(2θ) *dD/dθ* ]{θ=π/3} =4cos(2θ) ]{θ=π/3} =−2

3. (NO CALC) The function f is defined on the closed interval [-5, 4]. The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by g(x)=∫{-3,x} f(t) dt. (d) The function p is defined by p(x) = f(x²-x). Find the slope of the line tangent to the graph of p at the point where x=−1.

(d) *Chain Rule* p′(x)=f′(x²−x)(2x−1) p′(-1) = f′(2)(-3)=(-2)(-3)= 6

1. Grass clippings are placed in a bin, where they decompose. For 0≤t≤30, the amount of grass clippings remaining in the bin is modeled by A(t)= 6.687(0.931)^t , where A(t) is measured in pounds and t is measured in days. (d) For t>30, L(t), the linear approximation to A at t=30, is a better model for the amount of grass clippings remaining in the bin. Use L(t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

(d) *Linear Approximation* *f(a)+f'(a)(x-a)* L(t)=A(30)+A′(30)(t-30) *nDeriv on CALC* to find A'(t) A'(30)= -0.055976 A(30)=0.782928 0.783-0.0559(t-30)=0.5 L(t)=0.5 ⇒ t=35.054

4. (NO CALC) Train A runs back and forth on an east-west section of railroad track. Train A's velocity, measured in meters per minute, is given by a differentiable function v_A(t), where time t is measured in minutes. Selected values for v_A(t) are given in the table above. (d) A second train, train B, travels north from the Origin Station. At time t the velocity of train B is given by v_B(t)=-5t²+60t+25, and at time t=2 the train is 400 meters north of the station. Find the rate, in meters per minute, at which the distance between train A and train B is changing at time t=2.

(d) Let x be train A's position, y train B's position, and z the distance between train A and train B. z²=x²+y² ⇒2z(dz/dt)=2x(dx/dt)+2y(dy/dt) *implicit differentiation* x=300 and dx/dt=125 (from formula) v_B(2)=-20+120+25=125 y=400 and dy/dt=100 (from chart) z=500 and find dz/dt 500(dz/dt)=(300)(100) + (400)(125) dz/dt=80000/500=160 meters per minute

2. The graphs of the polar curves r=3 and r=3-2sin(2θ) are shown in the figure above for 0≤θ≤π. (d) A particle is moving along the curve r=3-2sin(2θ) so that dθ/dt=3 for all times t≥0. Find the value of dr/dt at θ=π/6.

(d) dr/dt=(dr/dθ)(dθ/dt)=(dr/dθ)3 given dθ/dt=3 use *nDeriv on CALC* to find dr/dθ=-2 dr/dt ]{θ=π/6}=(-2)(3)=-6