AP Calculus BC Practice Quiz: Defining and Differentiating 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 9 questions to check your progress.

Question 1 of 9

A curve is defined by the parametric equations $x(t) = 4t^3$ and $y(t) = 6t^2$. What is $\frac{dy}{dx}$ in terms of $t$?

A)$\frac{1}{t}$B)$t$C)$\frac{1}{2}t$D)$2$

All Questions (9)

A curve is defined by the parametric equations $x(t) = 4t^3$ and $y(t) = 6t^2$. What is $\frac{dy}{dx}$ in terms of $t$?

A) $\frac{1}{t}$

B) $t$

C) $\frac{1}{2}t$

D) $2$

Correct Answer: A

To find $\frac{dy}{dx}$, we calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ first. We have $\frac{dy}{dt} = 12t$ and $\frac{dx}{dt} = 12t^2$. Then, using the formula for the derivative of a parametric function, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{12t}{12t^2} = \frac{1}{t}$, provided $t \neq 0$.

A particle moves along a curve defined by the parametric equations $x(t) = 5\cos(t)$ and $y(t) = 5\sin(t)$. What is the slope of the line tangent to the curve at $t = \frac{\pi}{6}$?

A) $\sqrt{3}$

B) $-\sqrt{3}$

C) $\frac{1}{\sqrt{3}}$

D) $-\frac{1}{\sqrt{3}}$

Correct Answer: B

The slope of the tangent line is given by $\frac{dy}{dx}$. First, find the derivatives with respect to $t$: $\frac{dx}{dt} = -5\sin(t)$ and $\frac{dy}{dt} = 5\cos(t)$. Then, $\frac{dy}{dx} = \frac{5\cos(t)}{-5\sin(t)} = -\cot(t)$. Evaluating at $t = \frac{\pi}{6}$, the slope is $-\cot(\frac{\pi}{6}) = -\sqrt{3}$.

Which of the following expressions correctly represents $\frac{dy}{dx}$, the slope of the tangent line to a curve defined by the parametric equations $x = f(t)$ and $y = g(t)$, provided the denominator is not zero?

A) $\frac{g'(t)}{f'(t)}$

B) $\frac{f'(t)}{g'(t)}$

C) $f'(t) \cdot g'(t)$

D) $f'(g(t)) \cdot g'(t)$

Correct Answer: A

For a curve defined parametrically by $x = f(t)$ and $y = g(t)$, the derivative $\frac{dy}{dx}$ is determined by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$. Since $f'(t) = \frac{dx}{dt}$ and $g'(t) = \frac{dy}{dt}$, the correct expression is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$.

What is the equation of the line tangent to the curve given by $x(t) = t^2$ and $y(t) = t^3 - 3t$ at the point where $t = 2$?

A) $y - 2 = \frac{9}{4}(x - 4)$

B) $y - 4 = \frac{9}{4}(x - 2)$

C) $y - 2 = \frac{3}{2}(x - 4)$

D) $y - 2 = 2(x - 4)$

Correct Answer: A

First, find the point of tangency by evaluating $x(2)$ and $y(2)$. $x(2) = 2^2 = 4$ and $y(2) = 2^3 - 3(2) = 8 - 6 = 2$. The point is $(4, 2)$. Next, find the slope $\frac{dy}{dx}$. $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2 - 3$. So, $\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$. At $t=2$, the slope is $\frac{3(2^2) - 3}{2(2)} = \frac{12 - 3}{4} = \frac{9}{4}$. Using the point-slope form, the equation of the tangent line is $y - 2 = \frac{9}{4}(x - 4)$.

For the curve defined by the parametric equations $x(t) = t^3 - 12t$ and $y(t) = t^2 - 1$, at which value of $t$ is the tangent line horizontal?

A) $t = 2$

B) $t = -2$

C) $t = 0$

D) $t = 1$

Correct Answer: C

A horizontal tangent line occurs when the slope $\frac{dy}{dx}$ is zero. This happens when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. Here, $\frac{dy}{dt} = 2t$. Setting $\frac{dy}{dt} = 0$ gives $2t = 0$, so $t=0$. We must check that $\frac{dx}{dt}$ is not zero at $t=0$. $\frac{dx}{dt} = 3t^2 - 12$. At $t=0$, $\frac{dx}{dt} = 3(0)^2 - 12 = -12 \neq 0$. Therefore, the tangent line is horizontal at $t=0$.

For the curve defined by the parametric equations $x(t) = t^3 - 12t$ and $y(t) = t^2 - 1$, at which positive value of $t$ is the tangent line vertical?

A) $t = 0$

B) $t = 1$

C) $t = 2$

D) $t = \sqrt{12}$

Correct Answer: C

A vertical tangent line occurs when the slope $\frac{dy}{dx}$ is undefined. This happens when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$. Here, $\frac{dx}{dt} = 3t^2 - 12$. Setting $\frac{dx}{dt} = 0$ gives $3t^2 = 12$, so $t^2 = 4$, which means $t = 2$ or $t = -2$. The question asks for the positive value, which is $t=2$. We must check that $\frac{dy}{dt}$ is not zero at $t=2$. $\frac{dy}{dt} = 2t$. At $t=2$, $\frac{dy}{dt} = 2(2) = 4 \neq 0$. Therefore, the tangent line is vertical at $t=2$.

The derivative $\frac{dy}{dx}$ for a parametrically defined curve is found using the formula $\frac{dy/dt}{dx/dt}$. The condition $\frac{dx}{dt} \neq 0$ is required because it ensures that:

A) the curve is continuous at that point.

B) the denominator is not zero, avoiding an undefined slope associated with a vertical tangent.

C) the relationship between $x$ and $y$ can be expressed as a function.

D) a horizontal tangent line exists at that point.

Correct Answer: B

The formula for the slope of a parametric curve is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Division by zero is undefined. The condition $\frac{dx}{dt} \neq 0$ is necessary to ensure the denominator is not zero. When $\frac{dx}{dt} = 0$ (and $\frac{dy}{dt} \neq 0$), the curve has a vertical tangent, and the slope is considered undefined.

A curve is defined by the parametric equations $x(t) = e^t$ and $y(t) = t^2 + 1$. Find $\frac{dy}{dx}$ in terms of $t$.

A) $2te^t$

B) $\frac{2t}{e^t}$

C) $\frac{e^t}{2t}$

D) $t^2 e^t$

Correct Answer: B

To find $\frac{dy}{dx}$, we first compute the derivatives of $x$ and $y$ with respect to $t$. $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = 2t$. Using the formula for parametric derivatives, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{e^t}$.

The method for calculating the derivative $\frac{dy}{dx}$ for a parametric function as $\frac{dy/dt}{dx/dt}$ is an extension and application of which fundamental calculus rule?

A) The Product Rule

B) The Quotient Rule

C) The Chain Rule

D) The Power Rule

Correct Answer: C

The formula for the derivative of a parametric function is derived from the Chain Rule. According to the Chain Rule, $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$. By algebraically rearranging this equation to solve for $\frac{dy}{dx}$, we get $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. This shows that the methods for real-valued functions are extended to parametric functions.