AP BC Calculus Free Response 2016

5. (NO CALC) The inside of a funnel of height 10 inches has circular cross sections, as shown in the figure above. At height h, the radius of the funnel is given by r =(1/20)(3+h²), where 0≤h≤10. The units of r and h are inches. (a) Find the average value of the radius of the funnel.

(a) *Average Value of a Function* *A = [1/(b-a)]∫{a,b}f(x)dt* A = (1/10)∫{0,10}(1/20)(3+h²)dh =(1/200)[3h+h³/3]{0,10} =(1/200)[(30+(1000/3))-0] = 109/60 in

4. (NO CALC) Consider the differential equation dy/dx = x²-½y. (a) Find d²y/dx² in terms of x and y.

(a) *Chain Rule* d²y/dx²=2x-½(dy/dx) *plug in dy/dx given* =2x-½(x²-½y)

6. (NO CALC) The function f has a Taylor series about x=1 that converges to f(x) for all x in the interval of convergence. It is known that f(1)=1, f′(1)= −½, and the nth derivative of f at x=1 is given by fⁿ(1)=(-1)ⁿ(n-1)!/2ⁿ for n≥2 (a) Write the first four nonzero terms and the general term of the Taylor series for f about x=1.

(a) *Taylor Series* [fⁿ(a)/n!](x-a)ⁿ where f(1)=1 f′(1)= −(1/2) f''(1)=(1/2²) f'''(1)=-(2/2³) fⁿ(1)=[(-1)ⁿ(n-1)!]/2ⁿ where a=1 f(x)=1-(1/2)(x-1)+(1/(2²(2)))(x-1)²-(1/(2³(3)))(x-1)³+...+[(-1)ⁿ/(2ⁿn)](x-1)ⁿ+...

1. Water is pumped into a tank at a rate modeled by Wt 2000e^(-t²/20) liters per hour for 0≤t≤8, where t is measured in hours. Water is removed from the tank at a rate modeled by R(t) liters per hour, where R is differentiable and decreasing on 0≤t≤8. Selected values of R(t) are shown in the table above. At time t=0, there are 50,000 liters of water in the tank. t | 0 | 1 | 3 | 6 | 8 R(t) | 1340 | 1190 | 950 | 740 | 700 (a) Estimate R'(2). Show the work that leads to your answer. Indicate units of measure.

(a) R'(2) ≈ [R(3)-R(1)] / [3-1] = [950-1190] / [3-1] =-120 liters/hr² *m=∆y/∆x*

3. (NO CALC) The figure above shows the graph of the piecewise-linear function f. For 4≤x≤12, the function g is defined by g(x)=∫{2,x}f(t)dt. (a) Does g have a relative minimum, a relative maximum, or neither at x=10 ? Justify your answer.

(a) The function g has neither a relative minimum nor a relative maximum at x = 10 since g′(x) = f (x) and f(x)≤0 for 8≤x≤12. *f(x) does not change sign* *critical point when g'(x) changes sign*

2. At time t, the position of a particle moving in the xy-plane is given by the parametric functions (x(t), y(t)), where (dx/dt)=t²+sin(3t²) . The graph of y, consisting of three line segments, is shown in the figure above. At t =0, the particle is at position (5,1). (a) Find the position of the particle at t=3.

(a) x(3)=x(0)+∫{0,3}x'(t)dt=5+9.37705=14.377 y(3)=-½ The position of the particle at t=3 is (14.377, −0.5). *f(x)=∫f'(x)dx+C* where C was x(0)

4. (NO CALC) Consider the differential equation dy/dx = x²-½y. (b) Let y=f(x) be the particular solution to the given differential equation whose graph passes through the point (-2, 8). Does the graph of f have a relative minimum, a relative maximum, or neither at the point (-2, 8)? Justify your answer.

(b) *Second Derivative Test* *If f'(c)=0 and f''(c)<0 then f has a local maximum at c* plug in point to given equation f' = dy/dx = x²-½y ]{(x,y)=(-2,8)} = (-2)²-½(8)=4-4=0 plug the point into the equation found in Part A f'' =2x-½(x²-½y) = 2(-2)-½((-2)²-½(8))=-4<0 Thus, the graph of f has a relative maximum at the point (−2, 8).

5. (NO CALC) The inside of a funnel of height 10 inches has circular cross sections, as shown in the figure above. At height h, the radius of the funnel is given by r =(1/20)(3+h²), where 0≤h≤10. The units of r and h are inches. (b) Find the volume of the funnel.

(b) *Volume=π∫r²* V=π∫{0,10}[(1/20)(3+h²)]²dh =(π/400)∫{0,10}(9+6h²+h⁴)dh =(π/400)(9h+2h³+h⁵/5)]{0,10} =(π/400)[(90+2000+(100000/5))-0] =2209π/40 in³

2. At time t, the position of a particle moving in the xy-plane is given by the parametric functions (x(t), y(t)), where (dx/dt)=t²+sin(3t²) . The graph of y, consisting of three line segments, is shown in the figure above. At t =0, the particle is at position (5,1). (b) Find the slope of the line tangent to the path of the particle at t 3.

(b) Slope=y'(3)/x'(3)=0.5/9.956376=0.05 *dy/dx=[dy/dt]/[dx/dt]*

3. (NO CALC) The figure above shows the graph of the piecewise-linear function f. For 4≤x≤12, the function g is defined by g(x)=∫{2,x}f(t)dt. (b) Does the graph of g have a point of inflection at x=4 ? Justify your answer.

(b) The graph of g has a point of inflection at x=4 since g′(x)=f(x) is increasing for 2≤x≤ 4 and decreasing for 4≤x≤8. *inflection point when g'(x) changes direction*

1. Water is pumped into a tank at a rate modeled by Wt 2000e^(-t²/20) liters per hour for 0≤t≤8, where t is measured in hours. Water is removed from the tank at a rate modeled by R(t) liters per hour, where R is differentiable and decreasing on 0≤t≤8. Selected values of R(t) are shown in the table above. At time t=0, there are 50,000 liters of water in the tank. t | 0 | 1 | 3 | 6 | 8 R(t) | 1340 | 1190 | 950 | 740 | 700 (b) Use a left Riemann sum with the four subintervals indicated by the table to estimate the total amount of water removed from the tank during the 8 hours. Is this an overestimate or an underestimate of the total amount of water removed? Give a reason for your answer.

(b) The total amount of water removed is given by ∫{0,8} R(t) dt ≈ 1R(0)+2R(1)+3R(3)+2R(6) =1(1340)+2(1190)+3(950)+2(740) =8050 liters *Left Riemann sum* is an *overestimate* since R is a *decreasing function*. *∫f(x)=∆xy₀+∆xy₁+∆xy₂+...+∆xyₙ*

6. (NO CALC) The function f has a Taylor series about x=1 that converges to f(x) for all x in the interval of convergence. It is known that f(1)=1, f′(1)= −½, and the nth derivative of f at x=1 is given by fⁿ(1)=(-1)ⁿ(n-1)!/2ⁿ for n≥2 (b) The Taylor series for f about x=1 has a radius of convergence of 2. Find the interval of convergence. Show the work that leads to your answer.

(b) given R=2 *|x-a|<R* |x-1|<2 -2<(x-1)<2 −1<x<3 The series converges on the interval (−1, 3). When x=−1, the series is 1+1+1/2+1/3+1/4+.... Since the harmonic series diverges, this series diverges. When x = 3, the series is 1−1+1/2−1/3+1/4+... Since the alternating harmonic series converges, this series converges. Therefore, the interval of convergence is −1<x≤3.

5. (NO CALC) The inside of a funnel of height 10 inches has circular cross sections, as shown in the figure above. At height h, the radius of the funnel is given by r =(1/20)(3+h²), where 0≤h≤10. The units of r and h are inches. (c) The funnel contains liquid that is draining from the bottom. At the instant when the height of the liquid is h=3 inches, the radius of the surface of the liquid is decreasing at a rate of 1/5 inch per second. At this instant, what is the rate of change of the height of the liquid with respect to time?

(c) *(dr/dt)=(dr/dh)(dh/dt)* given forumla to find dr/dh given dr/dt as *decreasing* 1/5 so *-*1/5 (dr/dt)=(1/20)(2h)(dh/dt) given h=3 (-1/5)=(3/10)(dh/dt) (dh/dt)=(-1/5)(10/3)=(-2/3) in/sec

2. At time t, the position of a particle moving in the xy-plane is given by the parametric functions (x(t), y(t)), where (dx/dt)=t²+sin(3t²) . The graph of y, consisting of three line segments, is shown in the figure above. At t =0, the particle is at position (5,1). (c) Find the speed of the particle at t=3.

(c) Speed=√[(x'(3))²+(y'(3))²] = 9.969 (or 9.968) *x²+y²=r²* *speed=(position)' *

1. Water is pumped into a tank at a rate modeled by Wt 2000e^(-t²/20) liters per hour for 0≤t≤8, where t is measured in hours. Water is removed from the tank at a rate modeled by R(t) liters per hour, where R is differentiable and decreasing on 0≤t≤8. Selected values of R(t) are shown in the table above. At time t=0, there are 50,000 liters of water in the tank. t | 0 | 1 | 3 | 6 | 8 R(t) | 1340 | 1190 | 950 | 740 | 700 (c) Use your answer from Part B to find an estimate of the total amount of water in the tank, to the nearest liter, at the end of 8 hours.

(c) Total ≈ 50000 + ∫{0,8} W(t)dt-8050 =50000+7836.195325-8050≈49786 liters Use fnInt ONCALC *Initial + Rate In - Rate Out*

6. (NO CALC) The function f has a Taylor series about x=1 that converges to f(x) for all x in the interval of convergence. It is known that f(1)=1, f′(1)= −½, and the nth derivative of f at x=1 is given by fⁿ(1)=(-1)ⁿ(n-1)!/2ⁿ for n≥2 (c) The Taylor series for f about x=1 can be used to represent f (1.2) as an alternating series. Use the first three nonzero terms of the alternating series to approximate f(1.2).

(c) f(1.2)≈1-½(0.2)+(1/8)(0.2)² =1-(1/10)+(1/200)=1-0.1+0.005=0.905

3. (NO CALC) The figure above shows the graph of the piecewise-linear function f. For 4≤x≤12, the function g is defined by g(x)=∫{2,x}f(t)dt. (c) Find the absolute minimum value and the absolute maximum value of g on the interval −4≤x≤12. Justify your answers.

(c) g′(x) = f (x) changes sign only at x=−2 and x=6. x | g(x) −4 | -4 −2 |−8 6 | 8 12 | −4 On the interval −4 ≤ x ≤ 12, the absolute minimum value is g(−2) =− 8 and the absolute maximum value is g(6) = 8. *critical point when g'(x) changes sign* *check endpoints too*

4. (NO CALC) Consider the differential equation dy/dx = x²-½y. (c) Let y=g(x) be the particular solution to the given differential equation with g(1)=2. Find lim{x→-1} [g(x)-2]/[3(x+1)²] Show the work that leads to your answer.

(c) lim{x→-1}(g(x)-2)=0 and lim{x→-1}3(x+1)²=0 *indeterminate form 0/0* Using *L'Hospital's Rule*, lim{x→-1} [g(x)-2]/[3(x+1)²] =lim{x→-1} [g'(x)]/[6(x+1)] lim{x→-1} g'(x)=0 and lim{x→-1} 6(x+1)=0 *indeterminate form 0/0* Using *L'Hospital's Rule AGAIN*, lim{x→-1} [g'(x)]/[6(x+1)] =lim{x→-1} [g''(x)]/[6]=-2/6=-1/3 *find g' and g'' using formula given and Part A*

6. (NO CALC) The function f has a Taylor series about x=1 that converges to f(x) for all x in the interval of convergence. It is known that f(1)=1, f′(1)= −½, and the nth derivative of f at x=1 is given by fⁿ(1)=(-1)ⁿ(n-1)!/2ⁿ for n≥2 (d) Show that the approximation found in Part C is within 0.001 of the exact value of f 1.2.

(d) *Alternating Series Error* *if bₙ₊₁≤bₙ* *if lim{n→∞}=0* *then |Rₙ|=|s-sₙ|≤bₙ₊₁* |f(1.2)-T₂(1.2)|≤|-(1/(2³(3)))(0.2)³| =1/3000≤0.001

4. (NO CALC) Consider the differential equation dy/dx = x²-½y. (d) Let y = h(x) be the particular solution to the given differential equation with h(0) = 2. Use Euler's method, starting at x = 0 with two steps of equal size, to approximate h(1).

(d) *Euler's Method* *yₙ=yₙ-₁+∆x(y'ₙ-₁(xₙ-₁))* using the point (0,2) step size=∆x=½ h(½)≈h(0)+h'(0)½=2+(-1)½=3/2 h(1)≈h(½)+h'(1)½=(3/2)+(-½)½=5/4 *get h' from the formula given* h'(0) = 0²-½(2)=-1 h'(1) = (1/2)² - ½(3/2)= (1/4)-(3/4) =-(2/4)=-½

1. Water is pumped into a tank at a rate modeled by Wt 2000e^(-t²/20) liters per hour for 0≤t≤8, where t is measured in hours. Water is removed from the tank at a rate modeled by R(t) liters per hour, where R is differentiable and decreasing on 0≤t≤8. Selected values of R(t) are shown in the table above. At time t=0, there are 50,000 liters of water in the tank. t | 0 | 1 | 3 | 6 | 8 R(t) | 1340 | 1190 | 950 | 740 | 700 (d) For 0≤t≤8, is there a time t when the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank? Explain why or why not.

(d) *Intermediate Value Theorem* W(0)-R(0)>0, W(8)-R(8)<0, and W(t)-R(t) is continuous Therefore, the Intermediate Value Theorem guarantees at least one time 0≤t≤8, for which W(t)-R(t)=0, or W(t)=R(t) For this value of t, the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank *using f(x)=W(t)-R(t) and c=0*

2. At time t, the position of a particle moving in the xy-plane is given by the parametric functions (x(t), y(t)), where (dx/dt)=t²+sin(3t²) . The graph of y, consisting of three line segments, is shown in the figure above. At t =0, the particle is at position (5,1). (d) Find the total distance traveled by the particle from t=0 to t=2.

(d) Distance=∫{0,2} √[(x'(3))²+(y'(3))²] dt = ∫{0,1}√[(x'(3))²+(-2)²] dt + ∫{0,1}√[(x'(3))²+(0)²] dt =2.237871 = 2.112003 = 4.350 (or 4.349) *distance is the area under the speed/velocity curve* *use the graph to get the slopes of y (values of y') * *Use CALC fnInt* with Y₁=t²+sin(3t²) Y₂=√[Y₁²+4] Y₃=√[Y₁²]

3. (NO CALC) The figure above shows the graph of the piecewise-linear function f. For 4≤x≤12, the function g is defined by g(x)=∫{2,x}f(t)dt. (d) For −4≤x≤12, find all intervals for which g(x)≤0.

(d) g(x)≤0 for −4≤x≤2 and 10≤x≤12. *g(x) is area under the graph*