Test de práctica AP Calculus BC: solo BC

Marca todas las que correspondan.

• A. dy/dx = 9/4 when t = 2
• B. The curve has a horizontal tangent when t = 1
• C. The curve has a vertical tangent when t = 0
• D. The curve has a horizontal tangent at exactly one value of t
• E. dy/dx = 3t^2 - 3 for all t

Respuesta: A. dy/dx = 9/4 when t = 2, B. The curve has a horizontal tangent when t = 1, C. The curve has a vertical tangent when t = 0

For a parametric curve, dy/dx = (dy/dt)/(dx/dt) = (3t^2 - 3)/(2t). At t = 2 that is 9/4. A horizontal tangent needs dy/dt = 0 with dx/dt not 0: dy/dt = 0 at t = 1 and t = -1, and dx/dt is 2 and -2 there, so there are two such values, not one. A vertical tangent needs dx/dt = 0 with dy/dt not 0: at t = 0, dx/dt = 0 and dy/dt = -3. The expression 3t^2 - 3 is dy/dt, not dy/dx.

• A. 8
• B. 10
• C. 14
• D. 100

Respuesta: B. 10

The velocity vector is (x'(t), y'(t)) = (6, 8t), which is (6, 8) at t = 1. Speed is the magnitude of the velocity vector: sqrt(6^2 + 8^2) = sqrt(100) = 10. The answer 14 adds the components, 100 forgets the square root, and 8 is only the vertical component of the velocity.

• A. pi
• B. 2 pi
• C. 3 pi / 2
• D. 3 pi

Respuesta: C. 3 pi / 2

The area in polar coordinates is (1/2) times the integral of r^2 d(theta). Here r^2 = 1 + 2 cos(theta) + cos^2(theta). Over a full turn, the integral of 1 is 2 pi, the integral of 2 cos(theta) is 0 and the integral of cos^2(theta) is pi, so the total is 3 pi. Multiply by 1/2 to get 3 pi / 2. The answer 3 pi forgets the factor of 1/2, which is the most common slip on polar area.

• A. 3/4
• B. 3/2
• C. 3
• D. 3/8

Respuesta: A. 3/4

First dy/dx = (3t^2)/(2t) = 3t/2. For the second derivative, differentiate dy/dx with respect to t and then divide by dx/dt again: d^2y/dx^2 = (3/2)/(2t) = 3/(4t). At t = 1 that is 3/4. The answer 3/2 is d/dt of dy/dx, which skips the second division by dx/dt, and that skipped division is the trap in this topic.

Marca todas las que correspondan.

• A. The sum from n = 1 to infinity of 1/n
• B. The sum from n = 1 to infinity of 1/n^2
• C. The sum from n = 1 to infinity of (-1)^n / n
• D. The sum from n = 1 to infinity of n/(n + 1)
• E. The sum from n = 0 to infinity of (2/3)^n

Respuesta: B. The sum from n = 1 to infinity of 1/n^2, C. The sum from n = 1 to infinity of (-1)^n / n, E. The sum from n = 0 to infinity of (2/3)^n

The sum of 1/n^2 is a p-series with p = 2, which is greater than 1, so it converges. The sum of (-1)^n / n passes the alternating series test, because the terms shrink to 0 and decrease in size, so it converges, but only conditionally. The geometric series with ratio 2/3 converges because the ratio is between -1 and 1. The harmonic series 1/n is a p-series with p = 1 and diverges. The terms n/(n + 1) approach 1, not 0, so the series diverges by the nth term test.

• A. [-1, 5)
• B. (-1, 5)
• C. (-1, 5]
• D. [-1, 5]

Respuesta: A. [-1, 5)

The ratio test gives |x - 2|/3 < 1, so the radius is 3 and the series is centered at 2, which makes the open interval (-1, 5). Now check each endpoint separately. At x = 5 the terms are 3^n / (n * 3^n) = 1/n, the harmonic series, which diverges. At x = -1 the terms are (-3)^n / (n * 3^n) = (-1)^n / n, the alternating harmonic series, which converges. So the interval includes -1 and excludes 5.

• A. 1/6
• B. 8
• C. 4/3
• D. 2/3

Respuesta: C. 4/3

Start from e^u = 1 + u + u^2/2! + u^3/3! + ... and substitute u = 2x. The x^3 term is (2x)^3 / 3! = 8x^3 / 6 = (4/3) x^3. The coefficient 1/6 forgets to cube the 2, and 8 is f'''(0), the third derivative at 0, before dividing by 3 factorial. Maclaurin coefficients are f^(n)(0) / n!.

Substituting -x^2 into the geometric series gives 1/(1 + x^2) = 1 - x^2 + _____ - x^6 + ... , and integrating term by term gives the series arctan x = x - _____ + x^5/5 - ... .

Opciones para el hueco 1: x^3, x^4, x^6, x^2

Opciones para el hueco 2: x^3/2, x^3, x^3/3, x^2/2

Respuesta: x^4, x^3/3

The geometric series 1/(1 - u) = 1 + u + u^2 + u^3 + ... with u = -x^2 gives 1 - x^2 + x^4 - x^6 + ..., so the missing term is x^4. Integrating term by term raises each power by one and divides by the new power: x - x^3/3 + x^5/5 - ... . That is the series for arctan x, because the derivative of arctan x is 1/(1 + x^2) and arctan 0 = 0 fixes the constant of integration.

The alternating series for the sum from n = 1 to infinity of (-1)^(n+1) / n^2 is approximated by its first 3 terms. The alternating series error bound says the error is at most _____, the size of the _____.

Opciones para el hueco 1: 1/9, 1/16, 1/4, 1/3

Opciones para el hueco 2: last term used, first omitted term, second omitted term, sum of all omitted terms

Respuesta: 1/16, first omitted term

The terms 1/n^2 decrease to 0, so the alternating series error bound applies: the error from stopping after a term is no larger than the absolute value of the first omitted term. After 3 terms, the first omitted term is the fourth, 1/4^2 = 1/16. The value 1/9 is the last term used, not the first one left out. The true error is the sum of all omitted terms, and the bound says that sum is no bigger than the first one in size.

• A. 2
• B. 2.25
• C. about 3.44
• D. 2.5

Respuesta: D. 2.5

Step 1: the slope at (0, 1) is 0 + 1 = 1, so y(0.5) is about 1 + 0.5(1) = 1.5. Step 2: the slope at (0.5, 1.5) is 0.5 + 1.5 = 2, so y(1) is about 1.5 + 0.5(2) = 2.5. The answer 2 is a single step of size 1, and 2.25 reuses the old y-value for the second slope. The exact solution, y = 2e^x - x - 1, gives about 3.44 at x = 1. Euler's method underestimates here because the solution is concave up, so each tangent line lies below the curve.

Marca todas las que correspondan.

• A. The carrying capacity is 500
• B. The population grows fastest when P = 250
• C. P(t) is increasing for all t >= 0
• D. The graph of P is concave up for all t > 0
• E. dP/dt is largest at t = 0

Respuesta: A. The carrying capacity is 500, B. The population grows fastest when P = 250, C. P(t) is increasing for all t >= 0

This is a logistic equation, dP/dt = kP(1 - P/K), so K = 500 is the carrying capacity. As a function of P, the rate 0.2P(1 - P/500) is a downward parabola with its peak halfway to K, at P = 250, where the rate is 25. The population stays between 50 and 500, so dP/dt is positive and P is always increasing. The graph is concave up while P is below 250 and concave down after that, so it is not concave up for all t. At t = 0 the rate is only 0.2(50)(0.9) = 9, far below the maximum.

• A. (e^2 - 1)/4
• B. e^2 / 2
• C. (e^2 + 1)/2
• D. (e^2 + 1)/4

Respuesta: D. (e^2 + 1)/4

Use integration by parts with u = ln x and dv = x dx, so du = dx/x and v = x^2/2. The antiderivative is (x^2/2) ln x - the integral of x/2 dx = (x^2/2) ln x - x^2/4. At x = e that is e^2/2 - e^2/4 = e^2/4. At x = 1 it is 0 - 1/4 = -1/4. Subtracting the lower value gives e^2/4 + 1/4 = (e^2 + 1)/4. The answer (e^2 - 1)/4 comes from dropping the minus sign when subtracting -1/4.

• A. (1/3) ln|(x - 1)/(x + 2)| + C
• B. ln|(x - 1)(x + 2)| + C
• C. (1/3) ln|(x - 1)(x + 2)| + C
• D. 3 ln|(x - 1)/(x + 2)| + C

Respuesta: A. (1/3) ln|(x - 1)/(x + 2)| + C

Split the fraction: 1/((x - 1)(x + 2)) = A/(x - 1) + B/(x + 2), so 1 = A(x + 2) + B(x - 1). Setting x = 1 gives 1 = 3A, so A = 1/3. Setting x = -2 gives 1 = -3B, so B = -1/3. Integrating gives (1/3) ln|x - 1| - (1/3) ln|x + 2| + C = (1/3) ln|(x - 1)/(x + 2)| + C. The log of a product would come from adding the two terms, but the second term has a minus sign, so the logs combine as a quotient.

The improper integral of 1/x^2 dx from x = 1 to infinity equals _____, and the improper integral of 1/x^p dx from x = 1 to infinity converges exactly when p is _____ 1.

Opciones para el hueco 1: 0, 1, 2, The integral diverges

Opciones para el hueco 2: less than, at most, equal to, greater than

Respuesta: 1, greater than

Write the improper integral as a limit: the limit as b approaches infinity of the integral from 1 to b, which is the limit of (1 - 1/b) = 1. For general p, the antiderivative x^(1-p)/(1-p) goes to 0 as b grows only when 1 - p is negative, that is, when p is greater than 1. At p = 1 the antiderivative is ln b, which grows without bound, so the integral diverges, just like the harmonic series.

• A. 3
• B. 14/3
• C. 7/3
• D. 16/3

Respuesta: B. 14/3

Arc length is the integral of sqrt(1 + (dy/dx)^2) dx. Here dy/dx = x^(1/2), so 1 + (dy/dx)^2 = 1 + x. The integral of sqrt(1 + x) from 0 to 3 is (2/3)(1 + x)^(3/2) evaluated from 0 to 3, which is (2/3)(8 - 1) = 14/3. The curve starts at the origin and moves 3 units to the right, so the length has to be more than 3, and 14/3 is about 4.67.