AP Calculus BC Practice Quiz: Second Derivatives of Parametric Equations

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

A curve is defined by the parametric equations x = f(t) and y = g(t). Which of the following expressions represents the second derivative, d²y/dx²?

A)d/dt(dy/dx) / (dx/dt)B)(d²y/dt²) / (d²x/dt²)C)d/dt(dy/dx)D)(d/dt(dy/dx)) / (dy/dt)

All Questions (7)

A curve is defined by the parametric equations x = f(t) and y = g(t). Which of the following expressions represents the second derivative, d²y/dx²?

A) d/dt(dy/dx) / (dx/dt)

B) (d²y/dt²) / (d²x/dt²)

C) d/dt(dy/dx)

D) (d/dt(dy/dx)) / (dy/dt)

Correct Answer: A

The formula for the second derivative of a parametric equation is d²y/dx² = [d/dt(dy/dx)] / (dx/dt). This is found by taking the derivative of the first derivative (dy/dx) with respect to t, and then dividing by the derivative of x with respect to t (dx/dt). Option B is a common misconception, while options C and D are incomplete or incorrect variations of the formula.

A curve is defined by the parametric equations x(t) = t² and y(t) = t³ - 3t. Find the second derivative, d²y/dx².

A) (3t² - 3) / (2t)

B) 6t

C) 3 / (4t)

D) (3t² + 3) / (4t³)

Correct Answer: D

First, find the first derivative dy/dx. dx/dt = 2t and dy/dt = 3t² - 3. So, dy/dx = (dy/dt) / (dx/dt) = (3t² - 3) / (2t) = (3/2)t - (3/2)t⁻¹. Next, find the derivative of dy/dx with respect to t: d/dt(dy/dx) = d/dt((3/2)t - (3/2)t⁻¹) = 3/2 + (3/2)t⁻² = (3t² + 3) / (2t²). Finally, divide this by dx/dt: d²y/dx² = [d/dt(dy/dx)] / (dx/dt) = [(3t² + 3) / (2t²)] / (2t) = (3t² + 3) / (4t³).

A particle moves along a curve such that its position is given by x(t) = eᵗ and y(t) = t² + t. What is the value of d²y/dx² at t = 1?

A) -1 / e²

B) 1 / e

C) 3 / e²

D) 1 - 2e

Correct Answer: A

First, find dy/dx. dx/dt = eᵗ and dy/dt = 2t + 1. So, dy/dx = (2t + 1) / eᵗ. Next, find d/dt(dy/dx) using the quotient rule: d/dt(dy/dx) = [2(eᵗ) - (2t + 1)(eᵗ)] / (eᵗ)² = (2 - 2t - 1) / eᵗ = (1 - 2t) / eᵗ. Now, find d²y/dx²: d²y/dx² = [d/dt(dy/dx)] / (dx/dt) = [(1 - 2t) / eᵗ] / eᵗ = (1 - 2t) / e²ᵗ. Finally, evaluate at t = 1: (1 - 2(1)) / e²⁽¹⁾ = -1 / e².

A curve is defined by the parametric equations x(t) = 3cos(t) and y(t) = 2sin(t). Find the second derivative, d²y/dx².

A) -(2/3)cot(t)

B) (2/3)csc²(t)

C) -(2/9)csc³(t)

D) -(2/9)tan(t)

Correct Answer: C

First, find the derivatives with respect to t: dx/dt = -3sin(t) and dy/dt = 2cos(t). The first derivative is dy/dx = (dy/dt) / (dx/dt) = 2cos(t) / (-3sin(t)) = -(2/3)cot(t). Next, differentiate dy/dx with respect to t: d/dt(dy/dx) = d/dt(-(2/3)cot(t)) = -(2/3)(-csc²(t)) = (2/3)csc²(t). Finally, calculate the second derivative: d²y/dx² = [d/dt(dy/dx)] / (dx/dt) = [(2/3)csc²(t)] / [-3sin(t)] = -2 / (9sin(t)csc²(t))⁻¹ = -2 / (9sin³(t)) = -(2/9)csc³(t).

For what values of t is the curve defined by the parametric equations x(t) = t² and y(t) = t³ - 3t concave up?

A) t > 0

B) t < 0

C) -1 < t < 1

D) t > 1

Correct Answer: A

Concavity is determined by the sign of the second derivative, d²y/dx². First, find d²y/dx². We have dx/dt = 2t and dy/dt = 3t² - 3. Then dy/dx = (3t² - 3) / (2t). Differentiating dy/dx with respect to t gives d/dt(dy/dx) = (3t² + 3) / (2t²). Finally, d²y/dx² = [d/dt(dy/dx)] / (dx/dt) = [(3t² + 3) / (2t²)] / (2t) = (3t² + 3) / (4t³). The curve is concave up when d²y/dx² > 0. The numerator, 3t² + 3, is always positive. Therefore, the sign is determined by the denominator, 4t³. The expression 4t³ is positive when t > 0.

A curve is defined parametrically. For this curve, it is known that dy/dx = 2t² - 1 and dx/dt = 3t. What is d²y/dx² in terms of t?

A) 4/3

B) 4t

C) 4t / (2t² - 1)

D) 12t²

Correct Answer: A

The formula for the second derivative is d²y/dx² = [d/dt(dy/dx)] / (dx/dt). We are given dy/dx and dx/dt. First, we need to find the derivative of dy/dx with respect to t: d/dt(dy/dx) = d/dt(2t² - 1) = 4t. Now, we plug this into the formula: d²y/dx² = (4t) / (3t). For t ≠ 0, this simplifies to 4/3.

Find the second derivative, d²y/dx², for the curve defined by the parametric equations x(t) = t ln(t) and y(t) = t³.

A) 3t / (ln(t) + 1)²

B) (6t ln(t) + 3t) / (ln(t) + 1)³

C) 6t / (1/t)

D) (3t² (ln(t) + 1)) / (6t ln(t) + 3t)

Correct Answer: B

First, find the derivatives with respect to t. Using the product rule for x(t), dx/dt = 1 * ln(t) + t * (1/t) = ln(t) + 1. For y(t), dy/dt = 3t². The first derivative is dy/dx = (dy/dt) / (dx/dt) = 3t² / (ln(t) + 1). Next, find d/dt(dy/dx) using the quotient rule: d/dt(dy/dx) = [6t(ln(t) + 1) - 3t²(1/t)] / (ln(t) + 1)² = [6t ln(t) + 6t - 3t] / (ln(t) + 1)² = (6t ln(t) + 3t) / (ln(t) + 1)². Finally, d²y/dx² = [d/dt(dy/dx)] / (dx/dt) = [(6t ln(t) + 3t) / (ln(t) + 1)²] / (ln(t) + 1) = (6t ln(t) + 3t) / (ln(t) + 1)³.