BC Calc Final FRQ's

2019 question 3 c: Let f be the function defined above: f(x) = { rad (9-x^2) for -3 < (or equal to) x < (or equal to) 0 and -x + 3cos (pi x/ 2) for 0 < x < (or equal to) 4. Find the average value of f on the interval -3 < (or equal to) x < (or equal to) 4.

1/4--3 * the integral of f(x)dx from -3 to 4. Split into two which are (1/7 * the integral of rad (9-x^2) from -3 to 0) + (1/7 * the integral of -x + 3cos (pi x/2) from 0 to 4). Using calculator you get 1.0098 for the integral of rad (9-x^2) * 1/7 and -1.1429 for the integral of -x + 3cos (pi x/2) * 1/7. 1.0098-1.1429 = -0.1331

2017 question 4 d: The function f satisfies f(0) = 20. The first derivative of f satisfies the inequality 0 < f'(x) < 7 for all x in the closed interval [0, 6]. Selected values of f' are shown in the table above. The function f has a continuous second derivative for all real numbers. Find the limit as x approaches 0 of (f(x) - 20e^x) / (0.5f(x) - 10)

Apply L'Hopitals Rule. Limit as x approaches 0 of (f'(x) - 20e^x) / (0.5f'(x)). f'(0) - 20e^0 / (0.5 f'(0)). 4-20 / (0.5 * 4). -16 / 2 = -8

2018 question 2 a: A particle moving in the xy-plane has position (x(t), y(t)) at time t > 0, where dx/dt = cos(t^2) and dy/dt = e^t sin (t^2). At time t=0, the particle is at position (1, 2). The figure above shows the path of the particle for 0 < t < 2. Find the position of the particle at time t = 2.

Find the position of x(2) = 1 + integral of dx/dt from 0 to 2 = 1.461. Find the position of y(2) = 2 + integral of dy/dt from 0 to 2 = 4.268. Position is (1.461, 4.268)

2017 question 2 d: During the time interval 0 < t < 4.5 hours, water flows into tank A at a rate of a(t)= (2t-5) + 5e^2 sin t liters per hour. During the same time interval, water flows into tank B at a rate of b(t) liters per hour. Both tanks are empty at time t = 0. The graphs of y = a(t) and y = b(t), shown in the figure above, intersect at t = k and t = 2.416. During the time interval 2.7 < t < 4.5 hours, the rate at which water flows into tank B is modeled by w(t) = 21 - 30t/ (t-8)^2 liters per hour. Is the difference w(t) - a(t) increasing or decreasing at time 3.5? Show the work that leads to your answer

Find w'(3.5) - a'(3.5) using math 8 on calc. Answer is -1.14298. Since it is negative, the difference w(t) - a(t) is decreasing at time t = 3.5

2017 question 2 c: During the time interval 0 < t < 4.5 hours, water flows into tank A at a rate of a(t)= (2t-5) + 5e^2 sin t liters per hour. During the same time interval, water flows into tank B at a rate of b(t) liters per hour. Both tanks are empty at time t = 0. The graphs of y = a(t) and y = b(t), shown in the figure above, intersect at t = k and t = 2.416. The area of the region bounded by the graphs of y = a(t) and y = b(t) for k < t < 2,416 is 14.470. How much water is in tank B at time t = 2.416 ?

First you do the integral of a(t) from k (.892) to 2.416 - the integral of b(t) from k to 2.416 = 14.470. The integral of a(t) from k to 2.416 = 44.498 then you subtract 14.470 from it to find the integral of b(t) from k to 2.416 which = 30.028. Next you need to find the integral of b(t) from 0 to k by multiplying 20.5 by k which equals 18.286. 30.028 + 18.286 = 48.314 liters of water

2019 question 3 d: Let f be the function defined above: f(x) = { rad (9-x^2) for -3 < (or equal to) x < (or equal to) 0 and -x + 3cos (pi x/ 2) for 0 < x < (or equal to) 4. Must there be a value of x at which f(x) attains an absolute maximum on the closed interval -3 < (or equal to) x < (or equal to) 4? Justify your answer.

Limit as x approaches 0 from the left of f(x) = f(0) which = 3. The limit as x approaches 0 from the right of f(x) = 3. Since both are the same that means f is continuous. Since f is continuous on [-3. 4], the Extreme Value Theorem guarantees that f attains an absolute maximum on [-3, 4]

2018 question 4 a: During a chemical reaction, the function y = f(t) models the amount of a substance present, in grams, at time t seconds. At the start of the reaction (t=0), there are 10 grams of the substance present. The function y= f(t) satisfies the differential equation dy/dt = -0.02y^2. Use the line tangent to the graph of y= f(t) at t = 0 to approximate the amount of the substance remaining at time t = 2 seconds.

Plug 10 into dy/dt to get the slope which is -2. Then using point-slope formula make the formula y-10= -2 (t-0). Plug in 2 for t and get y = 6 grams.

2017 question 2 b: During the time interval 0 < t < 4.5 hours, water flows into tank A at a rate of a(t)= (2t-5) + 5e^2 sin t liters per hour. During the same time interval, water flows into tank B at a rate of b(t) liters per hour. Both tanks are empty at time t = 0. The graphs of y = a(t) and y = b(t), shown in the figure above, intersect at t = k and t = 2.416. During the time interval of 0 < t < k hours, water flows into tank B at a constant rate of 20.5 liters per hour. What is the difference between the amount of water in tank A and the amount of water in tank B at time t = k?

Plug 20.5 into Y1 and a(t) into Y2 and then calculate the intersection to find K. K =0.892. Next do the integral of (20.5 - a(t)) from 0 to K (.892) and get 10.599 liters

2019 question 6 d: Consider the series Σ at n=1 to infinity of ( (-1)^n+1 (x-3)^n )/ (5^n * n^P), where p is a constant and p > 0. When p= 1 and x= 3.1, the series converges to a value S. Use the first two terms of the series to approximate S. Use the alternating series error bound to show that this approximation differs from S by less than 1/ 300,000.

Plug in p=1 and x=3.1 getting (-1)^n+1 * (0.1)^n / 5^n * n. Use the first two terms which are 0.1/5 and 0.1^2/5^2 *2 and subtract them getting 99/5000. By the alternating series error bound, the approximation S = 99/5000 has absolute error bounded by the magnitude of the third term, 1/ (5^3 * 10^3 * 3). Meaning | S -99/5000 | < | 1/ (5^3 * 10^3 *3) | = 1/375,000 < 1/300,000

2019 question 6 b: Consider the series Σ at n=1 to infinity of ( (-1)^n+1 (x-3)^n )/ (5^n * n^P), where p is a constant and p > 0. For p=1 and x=8, does the series converge absolutely, converge conditionally, or diverge? Explain your reasoning

Plug in p=1 and x=8 getting (-1)^n+1 * 5^n / 5^n * n. Simplify. (-1)^n+1 / n. Using the alternating series test this converges however 1/n is the harmonic series which diverges meaning it converges conditionally

2019 question 6 a: Consider the series Σ at n=1 to infinity of ( (-1)^n+1 (x-3)^n )/ (5^n * n^P), where p is a constant and p > 0. For p=3 and x=8, does the series converge absolutely, converge conditionally, or diverge? Explain your reasoning.

Plug in p=3 and x=8 getting (-1)^n+1 * 5^n / 5^n * n^3. Simplify getting (-1)^n+1/ n^3. Ignoring the alternating 1/n^3 is a p-series with p=3 > 1, which converges. Meaning the series converges absolutely

2019 question 6 c: Consider the series Σ at n=1 to infinity of ( (-1)^n+1 (x-3)^n )/ (5^n * n^P), where p is a constant and p > 0. When x= -2, for what values of p does the series converge? Explain your reasoning.

Plug in x= -2. (-1)^n+1 * (-5)^n / 5^n * n^p. Simplify getting -1/n^p. 1/ n^p is a p-series which converges if and only if p>1. Therefore the series converges for p > 1.

2018 question 2 c: A particle moving in the xy-plane has position (x(t), y(t)) at time t > 0, where dx/dt = cos(t^2) and dy/dt = e^t sin (t^2). At time t=0, the particle is at position (1, 2). The figure above shows the path of the particle for 0 < t < 2. Find the speed of the particle at time t = 2. Find the acceleration vector of the particle at time t = 2.

Speed: rad (dx/dt)^2 + (dy/dt)^2 at time t=2 = 5.630. Acceleration Vector: x''(2) and y''(2) using math 8 on calc and putting in dx/dt and dy/dt for them. For x''(2)= 3.027. For y''(2)= -24.911. < 3.027, -24.911 >

2017 question 4 c: The function f satisfies f(0) = 20. The first derivative of f satisfies the inequality 0 < f'(x) < 7 for all x in the closed interval [0, 6]. Selected values of f' are shown in the table above. The function f has a continuous second derivative for all real numbers. Evaluate the integral of f''(x) from 2 to 4

The integral of f''(x) from 2 to 4 = f'(4) - f'(2). 1.7 -2 = -0.3

2017 question 4 a: The function f satisfies f(0) = 20. The first derivative of f satisfies the inequality 0 < f'(x) < 7 for all x in the closed interval [0, 6]. Selected values of f' are shown in the table above. The function f has a continuous second derivative for all real numbers. Use midpoint Riemann sum with three subintervals of equal length indicated by the data in the table to approximate the value of f(6).

The integral of f'(x)dx from 0 to 6 = f(6) - f(0). Then using midpoint sum do 2 (3.5) + 2 (0.8) + 2 (5.8) = f(6) - 20. Finally you get 20.2 = f(6) - 20. Then add 20 to 20.2 and get 40.2 for f(6).

2017 question 2 a: During the time interval 0 < t < 4.5 hours, water flows into tank A at a rate of a(t)= (2t-5) + 5e^2 sin t liters per hour. During the same time interval, water flows into tank B at a rate of b(t) liters per hour. Both tanks are empty at time t = 0. The graphs of y = a(t) and y = b(t), shown in the figure above, intersect at t = k and t = 2.416. How much water will be in tank A at time t = 4.5

You do the integral of a(t) from o to 4.5 and get 66.532 liters of water

2018 question 4 d: During a chemical reaction, the function y = f(t) models the amount of a substance present, in grams, at time t seconds. At the start of the reaction (t=0), there are 10 grams of the substance present. The function y= f(t) satisfies the differential equation dy/dt = -0.02y^2. Determine whether the amount of the substance is changing at an increasing or decreasing rate. Explain your reasoning.

d^2y/ dt^2 = -0.04y dy/dt. -0.04y (-0.02y^2). -0.0008y^3. Since 0.008y^3 > 0 meaning y >0, the substance is changing at an increasing rate.

2018 question 2 b: A particle moving in the xy-plane has position (x(t), y(t)) at time t > 0, where dx/dt = cos(t^2) and dy/dt = e^t sin (t^2). At time t=0, the particle is at position (1, 2). The figure above shows the path of the particle for 0 < t < 2. Find the slope of the line tangent to the particle's path at time t = 2

dy/dt / dx/dt = 8.555

2018 question 4 b: During a chemical reaction, the function y = f(t) models the amount of a substance present, in grams, at time t seconds. At the start of the reaction (t=0), there are 10 grams of the substance present. The function y= f(t) satisfies the differential equation dy/dt = -0.02y^2. Using the given differential equation, determine whether the graph of f could resemble the following graph. Give a reason for your answer.

dy/dt = -0.02y^2 which has a negative slope. It cannot because the graph should only decrease due to its negative slope but the given graph increases for some values of x.

2018 question 4 c: During a chemical reaction, the function y = f(t) models the amount of a substance present, in grams, at time t seconds. At the start of the reaction (t=0), there are 10 grams of the substance present. The function y= f(t) satisfies the differential equation dy/dt = -0.02y^2. Find an expression for y = f(t) by solving the differential equation dy/dt = -0.02y^2 with the initial condition f(0) = 10

dy/dt = -0.02y^2. Separate the variables. dy/y^2 = -0.02dt. y^-2dy = -0.02dt. Then do antiderivative. -1/y = -0.02t + c. Solve for c by plugging in initial condition. -1/10 = -0.02(0) + c. -1/10 = c. Then plug into equation. -1/y = -0.02t - 1/10. Multiply both side by y. -1 = (-.02t - 1/10) y. Solve for y. 1/(.02t - 1/10)=y

2019 question 3 a: Let f be the function defined above: f(x) = { rad (9-x^2) for -3 < (or equal to) x < (or equal to) 0 and -x + 3cos (pi x/ 2) for 0 < x < (or equal to) 4. Find the average rate of change of f on the interval -3 < (or equal to) x < (or equal to) 4.

f'(c) = ( f(b) - f(a) ) / (b-a). ( f(4) - f(-3) ) / (4--3). (-1 - 0) / (4--3) = 1/7

2017 question 4 b: The function f satisfies f(0) = 20. The first derivative of f satisfies the inequality 0 < f'(x) < 7 for all x in the closed interval [0, 6]. Selected values of f' are shown in the table above. The function f has a continuous second derivative for all real numbers. Determine whether the actual value of f(6) could be 70. Explain your reasoning

f'(x) < or equal to 7 meaning that the integral of f'(x) from 0 to 6 has to be < or equal to 6 * 7 (6 is farthest it can go out). 6 * 7 = 42. So then f(6) - f(0) < or equal to 42. f(6) - 20 < or equal to 42. f(6) < or equal to 62. The actual value therefore cannot be 70 because it cannot be greater than 62.

2019 question 3 b: Let f be the function defined above: f(x) = { rad (9-x^2) for -3 < (or equal to) x < (or equal to) 0 and -x + 3cos (pi x/ 2) for 0 < x < (or equal to) 4. Write an equation for the line tangent to the graph of f at x = 3.

f(3) = -3 + 3cos (3pi/2). f(3)= -3. f'(x) = -1 - 3pi/2 sin (pi x/2). f'(3) = -1 -3pi/2 sin(3pi/2). -1 + 3pi/2. Make equation using point-slope. y--3 = (-1 + 3pi/2) (x-3). y= (-1 + 3pi/2) (x-3) -3

2018 question 2 d: A particle moving in the xy-plane has position (x(t), y(t)) at time t > 0, where dx/dt = cos(t^2) and dy/dt = e^t sin (t^2). At time t=0, the particle is at position (1, 2). The figure above shows the path of the particle for 0 < t < 2. Consider a rectangle with vertices at points (0,0), (x(t), 0), (x(t), y(t)), and (0, y(t)) at time t > 0. For 0 < t < 2, at what time t is the perimeter of the rectangle a maximum? Justify your answer.

p'(t) = 2(dx/dt + dy/dt). Set p'(t)=0 and use calc to find intersect which is 1.722. Since p'(t) changes from positive to negative at this point, it has a maximum at t= 1.722