Parametrization of Implicitly Defined Functions - AP Precalculus

1. Which of the following is a parametrization for the parabola given by $x = -3(y+1)^2 + 2$?
   Answer: $x(t) = -3(t+1)^2+2, y(t) = t$ For a relation where $x$ is given as a function of $y$, the simplest parametrization is to set $y=t$. Substituting $t$ for $y$ in the equation gives $x = -3(t+1)^2 + 2$. Thus, a valid parametrization is $x(t) = -3(t+1)^2+2, y(t) = t$.
2. Based on the problem described, determine the parametrization after $\vec r(t)=\mathbf{A}\langle \cos t,\sin t\rangle$, $\mathbf{A}=\begin{bmatrix}2&0\\0&5\end{bmatrix}$.
   Answer: $\vec r(t)=\langle 2\cos t,\,5\sin t\rangle$ 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.
3. Which of the following is a parametrization for the entire parabola $x=y^2$?
   Answer: $x(t) = t^2, y(t) = t$ for all real $t$ 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.
4. Which of the following is a parametrization of the parabola defined by $y = (x+4)^2 - 1$?
   Answer: $x(t) = t, y(t) = (t+4)^2 - 1$ The most direct way to parametrize a function $y=f(x)$ is to let $x=t$. Substituting $t$ for $x$ in the equation gives $y = (t+4)^2 - 1$. Thus, a valid parametrization is $x(t) = t, y(t) = (t+4)^2 - 1$. Choice D represents the same parabola but is the expanded form, which is also correct, but choice A directly reflects the substitution.
5. Based on the problem described, determine the parametrization for $\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ applied to $\langle 2\cos t,2\sin t\rangle$.
   Answer: $\vec r(t)=\langle -2\sin t,\,2\cos t\rangle$ This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on matrix transformations that rotate parametric curves. Parametrization combined with matrix operations allows systematic transformation of curves, where the given matrix performs a 90° counterclockwise rotation. In this scenario, the matrix A = [[0, -1], [1, 0]] is the standard 90° counterclockwise rotation matrix applied to ⟨2cos(t), 2sin(t)⟩. Choice B is correct because matrix multiplication gives: [[0, -1], [1, 0]] × [2cos(t), 2sin(t)]ᵀ = [0·2cos(t) + (-1)·2sin(t), 1·2cos(t) + 0·2sin(t)]ᵀ = [-2sin(t), 2cos(t)]ᵀ. Choice C is incorrect because it has the wrong signs, giving ⟨2sin(t), -2cos(t)⟩, which would represent a 90° clockwise rotation instead. To help students: Recognize special rotation matrices - [[0, -1], [1, 0]] rotates 90° counterclockwise, while [[0, 1], [-1, 0]] rotates 90° clockwise. Practice matrix multiplication carefully to avoid sign errors.
6. Using the scenario above, determine the parametrization for $x^2+y^2=25$ using $t$.
   Answer: $\vec r(t)=\langle 5\cos t,\,5\sin t\rangle$ This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on converting a circle equation to parametric form. Parametrization involves expressing an implicitly defined function in terms of one or more parameters, allowing us to trace all points on the curve systematically. In this scenario, the function is implicitly defined by x² + y² = 25 (a circle with radius 5) and requires parameter t to represent the angle from the positive x-axis. Choice D is correct because x = 5cos(t) and y = 5sin(t) satisfy the circle equation: (5cos(t))² + (5sin(t))² = 25cos²(t) + 25sin²(t) = 25(cos²(t) + sin²(t)) = 25(1) = 25. Choice A is incorrect because it reverses sine and cosine, which would trace the circle starting from the positive y-axis instead of the positive x-axis. To help students: Emphasize that for a circle x² + y² = r², the standard parametrization is x = r·cos(t), y = r·sin(t), where r is the radius. Practice verifying parametrizations by substituting back into the original equation.
7. A hyperbola is defined by the equation $\frac{(x-5)^2}{16} - \frac{(y-2)^2}{9} = 1$. Which of the following is a parametrization for this hyperbola?
   Answer: $x(t) = 5 + 4\sec(t), y(t) = 2 + 3\tan(t)$ The standard parametrization for a horizontal hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ is $x(t) = h + a\sec(t)$ and $y(t) = k + b\tan(t)$, based on the identity $\sec^2(t) - \tan^2(t) = 1$. Here, the center is $(h,k) = (5, 2)$, $a = \sqrt{16} = 4$, and $b = \sqrt{9} = 3$. So, $x(t) = 5 + 4\sec(t)$ and $y(t) = 2 + 3\tan(t)$.
8. Which parametrization represents the lower branch of the hyperbola $\frac{y^2}{16} - \frac{x^2}{9} = 1$?
   Answer: $x(t) = 3\tan(t), y(t) = -4\sec(t)$ for $-\pi/2 < t < \pi/2$ 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.
9. A curve is given by the parametric equations $x(t) = t-1$ and $y(t) = t^2+3$. Which of the following is the rectangular equation for the curve?
   Answer: $y = (x+1)^2 + 3$ 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 $y = (x+1)^2 + 3$.
10. Consider the implicitly defined curve $x^{2/3} + y^{2/3} = 4$. Which of the following is a valid parametrization of this curve?
    Answer: $x(t) = 8\cos^3(t), y(t) = 8\sin^3(t)$ 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.
11. Using the scenario above, determine the parametrization for $x^2-y^2=1$ using $t$.
    Answer: $\vec r(t)=\langle \sec t,\,\tan t\rangle$ 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.
12. A curve is defined parametrically by $x(t) = 3\tan(t) - 2$ and $y(t) = 5\sec(t) + 1$. What type of conic section does this parametrization represent?
    Answer: A hyperbola 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.
13. A curve is defined by the parametric equations $x(t) = 5 + 2\cos(t)$ and $y(t) = -1 + 6\sin(t)$. Which of the following is the equation of the curve in rectangular coordinates?
    Answer: $\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1$ 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 $\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1$.
14. Which of the following is a parametrization of the hyperbola given by $\frac{(y+3)^2}{4} - \frac{(x-1)^2}{36} = 1$?
    Answer: $x(t) = 1 + 6\tan(t), y(t) = -3 + 2\sec(t)$ 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 $y(t) = -3 + 2\sec(t)$.
15. Which of the following parametrizations represents the top half of the circle $x^2 + y^2 = 9$, traced counter-clockwise?
    Answer: $x(t) = 3\cos(t), y(t) = 3\sin(t)$ for $0 \le t \le \pi$ 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.
16. A curve is given by $x(t) = e^t$ and $y(t) = 2e^{2t} - 1$. Which of the following rectangular equations represents the curve?
    Answer: $y = 2x^2 - 1$ for $x > 0$ 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 $x>0$.
17. The function $f$ is given by $f(x) = e^{2x} + 1$. Which of the following is a parametrization for the graph of the inverse function, $f^{-1}$?
    Answer: $x(t) = e^{2t}+1, y(t) = t$ 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 $x(t) = e^{2t}+1, y(t) = t$.
18. A circle is defined by the equation $(x-3)^2 + (y+2)^2 = 16$. Which of the following is a valid parametrization of this circle?
    Answer: $x(t) = 3 + 4\cos(t), y(t) = -2 + 4\sin(t)$ 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, $0 \le t < 2\pi$.
19. Based on the problem described, what is the role of $t$ in $\vec r(t)=\langle 3\cos t,3\sin t\rangle$?
    Answer: $t$ selects an angle, tracing all points on $x^2+y^2=9$. 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.
20. Based on the problem described, determine the parametrization for $\frac{x^2}{16}+\frac{y^2}{9}=1$ using $t$.
    Answer: $\vec r(t)=\langle 4\cos t,\,3\sin t\rangle$ 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.

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