AP Precalculus Quiz: Parametrization Of Implicitly Defined Functions

What this quiz covers

This quiz focuses on Parametrization Of Implicitly Defined Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\vec r(t)=\langle 16\cos t,\,9\sin t\rangle$
2. $\vec r(t)=\langle 4\cos t,\,3\sin t\rangle$ (correct answer)
3. $\vec r(t)=\langle 4\sin t,\,3\sin t\rangle$
4. $\vec r(t)=\langle 4\cos t,\,9\sin t\rangle$

Explanation: This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on ellipse parametrization from standard form. Parametrization involves expressing an implicitly defined function in terms of parameters, with ellipses requiring scaling of the basic circular parametrization. In this scenario, the ellipse x²/16 + y²/9 = 1 has semi-major axis a = 4 (horizontal) and semi-minor axis b = 3 (vertical). Choice B is correct because x = 4cos(t) and y = 3sin(t) satisfy the ellipse equation: (4cos(t))²/16 + (3sin(t))²/9 = 16cos²(t)/16 + 9sin²(t)/9 = cos²(t) + sin²(t) = 1. Choice A is incorrect because it uses 16 and 9 directly instead of their square roots, giving x = 16cos(t) and y = 9sin(t), which would trace a much larger ellipse. To help students: For an ellipse x²/a² + y²/b² = 1, the parametrization is x = a·cos(t), y = b·sin(t). Always take the square root of the denominators to find the semi-axes lengths.

Question 2

1. $(x+2)^2 - \frac{(y-4)^2}{9} = 1$ (correct answer)
2. $\frac{(y-4)^2}{9} - (x+2)^2 = 1$
3. $(x-2)^2 - \frac{(y+4)^2}{9} = 1$
4. $(x+2)^2 + \frac{(y-4)^2}{9} = 1$

Explanation: To find the rectangular equation, we eliminate the parameter $t$. Isolate the trigonometric functions: $\sec(t) = x+2$ and $\tan(t) = \frac{y-4}{3}$. Using the identity $\sec^2(t) - \tan^2(t) = 1$, we get $(x+2)^2 - (\frac{y-4}{3})^2 = 1$, which simplifies to $(x+2)^2 - \frac{(y-4)^2}{9} = 1$.

Question 3

1. $y = (x-1)^2 + 3$
2. $x = (y+1)^2 + 3$
3. $y = x^2+2x+4$
4. $y = (x+1)^2 + 3$ (correct answer)

Explanation: To eliminate the parameter, we solve one equation for $t$ and substitute into the other. From $x(t) = t-1$, we find $t = x+1$. Substituting this into the equation for $y(t)$ gives $y = (x+1)^2 + 3$.

Question 4

1. $x(t) = r\cos(t), y(t) = r\sin(t)$ for $0 \le t < 2\pi$
2. $x(t) = r\sin(t), y(t) = r\cos(t)$ for $0 \le t < 2\pi$
3. $x(t) = r\cos(t), y(t) = -r\sin(t)$ for $0 \le t < 2\pi$ (correct answer)
4. $x(t) = -r\cos(t), y(t) = r\sin(t)$ for $0 \le t < 2\pi$

Explanation: At $t=0$, we need the point to be $(r,0)$. For choice C, $x(0) = r\cos(0) = r$ and $y(0) = -r\sin(0) = 0$, so it starts at the correct point. As $t$ increases from 0, $x(t)$ decreases and $y(t)$ becomes negative, which corresponds to clockwise motion from $(r,0)$.

Question 5

1. $x(t) = t, y(t) = e^{2t}+1$
2. $x(t) = e^{2t}+1, y(t) = t$ (correct answer)
3. $x(t) = t, y(t) = \frac{\ln(t-1)}{2}$
4. $x(t) = t, y(t) = e^{-2t}-1$

Explanation: If a function $y=f(x)$ is invertible, a parametrization for its inverse $f^{-1}$ is given by $x(t) = f(t)$ and $y(t) = t$. In this case, $f(t) = e^{2t}+1$. So, the parametrization for $f^{-1}$ is $x(t) = e^{2t}+1, y(t) = t$.

Question 6

1. $x(t) = 2\cos(t), y(t) = 2\sin(t)$
2. $x(t) = 4\cos^3(t), y(t) = 4\sin^3(t)$
3. $x(t) = 8\cos^3(t), y(t) = 8\sin^3(t)$ (correct answer)
4. $x(t) = 8\cos(t), y(t) = 8\sin(t)$

Explanation: To check the parametrization, substitute $x(t)$ and $y(t)$ into the equation. For choice C, we have $(8\cos^3(t))^{2/3} + (8\sin^3(t))^{2/3} = 8^{2/3}(\cos^3(t))^{2/3} + 8^{2/3}(\sin^3(t))^{2/3} = (\sqrt[3]{8})^2\cos^2(t) + (\sqrt[3]{8})^2\sin^2(t) = 2^2\cos^2(t) + 2^2\sin^2(t) = 4(\cos^2(t) + \sin^2(t)) = 4(1) = 4$. This matches the equation.

Question 7

1. $x(t) = t, y(t) = \sqrt{t}$ for $t \ge 0$
2. $x(t) = t^2, y(t) = t$ for all real $t$ (correct answer)
3. $x(t) = \cos(t), y(t) = \sqrt{\cos(t)}$
4. $x(t) = t^2, y(t) = t$ for $t \ge 0$

Explanation: Let $y(t) = t$. Since $y$ can take any real value, $t$ must range over all real numbers. Substituting $y=t$ into the equation $x=y^2$ gives $x(t) = t^2$. This parametrization traces the entire parabola because as $t$ goes from $-\infty$ to $+\infty$, $y$ covers all real numbers, and $x$ covers all non-negative numbers.

Question 8

1. $x(t) = 3\tan(t), y(t) = 4\sec(t)$ for $-\pi/2 < t < \pi/2$
2. $x(t) = 3\sec(t), y(t) = 4\tan(t)$ for $-\pi/2 < t < \pi/2$
3. $x(t) = 3\tan(t), y(t) = -4\sec(t)$ for $-\pi/2 < t < \pi/2$ (correct answer)
4. $x(t) = 3\tan(t), y(t) = 4\sec(t)$ for $\pi/2 < t < 3\pi/2$

Explanation: The hyperbola is vertical, so we use $x(t) = a\tan(t)$ and $y(t) = k \pm b\sec(t)$. Here $a=3, b=4$. The lower branch means $y$ values are negative. For $-\pi/2 < t < \pi/2$, the value of $\sec(t)$ is positive. Therefore, $y(t) = -4\sec(t)$ will produce negative $y$ values, corresponding to the lower branch of the hyperbola.

Question 9

1. $\sec^2(t) - \tan^2(t) = 1$
2. $\sin(2t) = 2\sin(t)\cos(t)$
3. $\tan(t) = \frac{\sin(t)}{\cos(t)}$
4. $\sin^2(t) + \cos^2(t) = 1$ (correct answer)

Explanation: To convert the parametric equations of an ellipse to rectangular form, we isolate $\cos(t)$ and $\sin(t)$ to get $\cos(t) = \frac{x-h}{a}$ and $\sin(t) = \frac{y-k}{b}$. Then we substitute these into the Pythagorean identity $\sin^2(t) + \cos^2(t) = 1$, which results in the standard rectangular equation for an ellipse.

Question 10

1. $x(t) = -1 + 3\cos(t), y(t) = 4 + 5\sin(t)$ (correct answer)
2. $x(t) = 1 + 3\cos(t), y(t) = -4 + 5\sin(t)$
3. $x(t) = -1 + 9\cos(t), y(t) = 4 + 25\sin(t)$
4. $x(t) = -1 + 5\cos(t), y(t) = 4 + 3\sin(t)$

Explanation: The standard parametrization for an ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ is $x(t) = h + a\cos(t)$ and $y(t) = k + b\sin(t)$. Here, the center is $(h,k) = (-1, 4)$. The horizontal semi-axis is $a = \sqrt{9} = 3$, and the vertical semi-axis is $b = \sqrt{25} = 5$. So, $x(t) = -1 + 3\cos(t)$ and $y(t) = 4 + 5\sin(t)$.

Question 11

1. $y = 2x^2 - 1$ for all real $x$
2. $y = 2x - 1$ for $x>0$
3. $y = x^2 - 1$ for all real $x$
4. $y = 2x^2 - 1$ for $x > 0$ (correct answer)

Explanation: From $x(t) = e^t$, we can see that $x$ must be positive. We can also write $e^{2t} = (e^t)^2 = x^2$. Substituting this into the equation for $y(t)$ gives $y = 2x^2 - 1$. Since $x=e^t$, the domain of the rectangular equation is restricted to $x>0$.

Question 12

1. $\frac{(x+5)^2}{4} + \frac{(y-1)^2}{36} = 1$
2. $\frac{(x-5)^2}{2} + \frac{(y+1)^2}{6} = 1$
3. $\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1$ (correct answer)
4. $(x-5)^2 + (y+1)^2 = 1$

Explanation: To eliminate the parameter $t$, we isolate the trigonometric functions. From $x = 5 + 2\cos(t)$, we get $\cos(t) = \frac{x-5}{2}$. From $y = -1 + 6\sin(t)$, we get $\sin(t) = \frac{y+1}{6}$. Using the identity $\cos^2(t) + \sin^2(t) = 1$, we substitute to get $(\frac{x-5}{2})^2 + (\frac{y+1}{6})^2 = 1$, which simplifies to $\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1$.

Question 13

1. $x(t) = 1 + 2\tan(t), y(t) = -3 + 6\sec(t)$
2. $x(t) = 1 + 6\sec(t), y(t) = -3 + 2\tan(t)$
3. $x(t) = -1 + 36\tan(t), y(t) = 3 + 4\sec(t)$
4. $x(t) = 1 + 6\tan(t), y(t) = -3 + 2\sec(t)$ (correct answer)

Explanation: The standard parametrization for a vertical hyperbola $\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1$ is $x(t) = h + a\tan(t)$ and $y(t) = k + b\sec(t)$. Here, the center is $(h,k) = (1, -3)$, $a = \sqrt{36} = 6$, and $b = \sqrt{4} = 2$. So, $x(t) = 1 + 6\tan(t)$ and $y(t) = -3 + 2\sec(t)$.

Question 14

1. A circle
2. A parabola
3. An ellipse
4. A hyperbola (correct answer)

Explanation: To identify the conic, we eliminate the parameter $t$. From the given equations, we have $\tan(t) = \frac{x+2}{3}$ and $\sec(t) = \frac{y-1}{5}$. Using the Pythagorean identity $\sec^2(t) - \tan^2(t) = 1$, we get $\left(\frac{y-1}{5}\right)^2 - \left(\frac{x+2}{3}\right)^2 = 1$. This equation, $\frac{(y-1)^2}{25} - \frac{(x+2)^2}{9} = 1$, is the standard form of a hyperbola.

Question 15

1. $x(t) = 3 + 16\cos(t), y(t) = -2 + 16\sin(t)$
2. $x(t) = -3 + 4\cos(t), y(t) = 2 + 4\sin(t)$
3. $x(t) = 3 + 4\cos(t), y(t) = -2 + 4\sin(t)$ (correct answer)
4. $x(t) = 3 + 4\cos(t), y(t) = -2 + 4\sin(t)$ for $0 \le t < \pi$

Explanation: The standard parametrization for a circle with center $(h, k)$ and radius $r$ is $x(t) = h + r\cos(t)$ and $y(t) = k + r\sin(t)$. The given equation has center $(3, -2)$ and radius $r = \sqrt{16} = 4$. Therefore, the correct parametrization is $x(t) = 3 + 4\cos(t)$, $y(t) = -2 + 4\sin(t)$ for a full cycle, $0 \le t < 2\pi$.

Question 16

1. $x(t) = 3\cos(t), y(t) = 3\sin(t)$ for $0 \le t \le 2\pi$
2. $x(t) = 3\cos(t), y(t) = 3\sin(t)$ for $-\pi/2 \le t \le \pi/2$
3. $x(t) = 3\sin(t), y(t) = 3\cos(t)$ for $0 \le t \le \pi$
4. $x(t) = 3\cos(t), y(t) = 3\sin(t)$ for $0 \le t \le \pi$ (correct answer)

Explanation: The standard counter-clockwise parametrization for a circle of radius 3 is $x(t) = 3\cos(t), y(t) = 3\sin(t)$. The top half of the circle corresponds to $y \ge 0$. The function $y(t) = 3\sin(t)$ is non-negative for $0 \le t \le \pi$. Therefore, this domain for $t$ traces the top half of the circle.

Question 17

1. $\vec r(t)=\langle \cos t,\,\sin t\rangle$
2. $\vec r(t)=\langle \sec t,\,\tan t\rangle$ (correct answer)
3. $\vec r(t)=\langle \tan t,\,\sec t\rangle$
4. $\vec r(t)=\langle \csc t,\,\cot t\rangle$

Explanation: This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on hyperbola parametrization using trigonometric identities. Parametrization involves expressing implicitly defined functions through parameters, with hyperbolas requiring the hyperbolic identity sec²(t) - tan²(t) = 1. In this scenario, the equation x² - y² = 1 represents a hyperbola, which cannot use the circular identity cos²(t) + sin²(t) = 1. Choice B is correct because x = sec(t) and y = tan(t) satisfy the hyperbola equation: sec²(t) - tan²(t) = 1, which is a fundamental trigonometric identity. Choice A is incorrect because cos²(t) - sin²(t) = cos(2t), not 1, so this parametrization doesn't trace the given hyperbola. To help students: Distinguish between circle/ellipse parametrizations (using sin and cos) and hyperbola parametrizations (using sec and tan or hyperbolic functions). Emphasize the identity sec²(t) - tan²(t) = 1 as the hyperbolic analogue to cos²(t) + sin²(t) = 1.

Question 18

1. $\vec r(t)=\langle 2\cos t,\,5\sin t\rangle$ (correct answer)
2. $\vec r(t)=\langle 2\sin t,\,5\cos t\rangle$
3. $\vec r(t)=\langle 10\cos t,\,\sin t\rangle$
4. $\vec r(t)=\langle 2\cos t,\,5\cos t\rangle$

Explanation: This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on matrix transformations of parametric curves. Parametrization combined with matrix multiplication allows systematic transformation of curves, where diagonal matrices perform scaling operations on each component independently. In this scenario, the diagonal matrix A = [[2, 0], [0, 5]] scales the unit circle parametrization ⟨cos(t), sin(t)⟩. Choice A is correct because matrix multiplication gives: [[2, 0], [0, 5]] × [cos(t), sin(t)]ᵀ = [2cos(t), 5sin(t)]ᵀ, which represents an ellipse with horizontal semi-axis 2 and vertical semi-axis 5. Choice D is incorrect because it replaces sin(t) with cos(t) in the y-component, giving ⟨2cos(t), 5cos(t)⟩, which would trace a line segment rather than an ellipse. To help students: Emphasize that diagonal matrices scale each component independently - the (1,1) entry scales x and the (2,2) entry scales y. Practice matrix-vector multiplication step by step to avoid confusion.

Question 19

1. $\vec r(t)=\begin{bmatrix}3\cos t\\3\sin t\end{bmatrix}+\begin{bmatrix}1\\-2\end{bmatrix}$ (correct answer)
2. $\vec r(t)=\begin{bmatrix}1&-2\\0&1\end{bmatrix}\begin{bmatrix}3\cos t\\3\sin t\end{bmatrix}$
3. $\vec r(t)=\begin{bmatrix}3\cos(t+1)\\3\sin(t-2)\end{bmatrix}$
4. $\vec r(t)=\begin{bmatrix}3\cos t\\3\sin t\end{bmatrix}+\begin{bmatrix}-2\\1\end{bmatrix}$

Explanation: This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on translating parametric curves using vector addition. Parametrization combined with translations allows us to shift curves in the plane, where translation by vector b is achieved by adding b to each point on the curve. In this scenario, we translate the circle ⟨3cos(t), 3sin(t)⟩ by the vector b = ⟨1, -2⟩. Choice A is correct because translation is performed by vector addition: ⟨3cos(t), 3sin(t)⟩ + ⟨1, -2⟩ = ⟨3cos(t) + 1, 3sin(t) + (-2)⟩ = ⟨3cos(t) + 1, 3sin(t) - 2⟩, which can be written in matrix form as shown. Choice B is incorrect because it attempts to use matrix multiplication for translation, but translation requires addition, not multiplication by a non-square matrix. To help students: Emphasize that translation is always addition of vectors, while rotation and scaling use matrix multiplication. Practice both component-wise addition and the augmented matrix notation for affine transformations.

Question 20

1. $t$ fixes one point, so the curve is a single constant location.
2. $t$ is the radius, so changing $t$ changes the circle size.
3. $t$ selects an angle, tracing all points on $x^2+y^2=9$. (correct answer)
4. $t$ replaces $y$, so $x$ becomes an implicit function of $t$ only.

Explanation: This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on understanding the role of the parameter in circular motion. Parametrization involves expressing an implicitly defined function in terms of one or more parameters, with t typically representing an angle or time variable that traces the curve. In this scenario, the vector function ⟨3cos(t), 3sin(t)⟩ parametrizes the circle x² + y² = 9, and t acts as the angle parameter measured from the positive x-axis. Choice C is correct because as t varies from 0 to 2π, the angle sweeps around the circle, with each value of t corresponding to exactly one point on the circle x² + y² = 9. Choice B is incorrect because t is not the radius (which is fixed at 3) but rather the angular parameter that determines position on the circle. To help students: Use unit circle knowledge to connect angle t with coordinates (cos(t), sin(t)), then scale by the radius. Demonstrate with specific t values like 0, π/2, π, and 3π/2 to show how the point moves around the circle.