Parametrization of Implicitly Defined Functions
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AP Precalculus › Parametrization of Implicitly Defined Functions
Using the scenario above, determine the parametrization for $x^2-y^2=1$ using $t$.
$\vec r(t)=\langle \sec t,,\tan t\rangle$
$\vec r(t)=\langle \csc t,,\cot t\rangle$
$\vec r(t)=\langle \cos t,,\sin t\rangle$
$\vec r(t)=\langle \tan t,,\sec t\rangle$
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.
Based on the problem described, determine the parametrization for $\frac{x^2}{16}+\frac{y^2}{9}=1$ using $t$.
$\vec r(t)=\langle 4\cos t,,9\sin t\rangle$
$\vec r(t)=\langle 16\cos t,,9\sin t\rangle$
$\vec r(t)=\langle 4\cos t,,3\sin t\rangle$
$\vec r(t)=\langle 4\sin t,,3\sin t\rangle$
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.
Based on the problem described, determine the parametrization after $\vec r(t)=\mathbf{A}\langle \cos t,\sin t\rangle$, $\mathbf{A}=\begin{bmatrix}2&0\0&5\end{bmatrix}$.
$\vec r(t)=\langle 2\sin t,,5\cos t\rangle$
$\vec r(t)=\langle 2\cos t,,5\cos t\rangle$
$\vec r(t)=\langle 2\cos t,,5\sin t\rangle$
$\vec r(t)=\langle 10\cos t,,\sin t\rangle$
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.
Based on the problem described, what is the matrix representation for translating $\langle 3\cos t,3\sin t\rangle$ by $\vec b=\langle 1,-2\rangle$?
$\vec r(t)=\begin{bmatrix}3\cos t\3\sin t\end{bmatrix}+\begin{bmatrix}1\-2\end{bmatrix}$
$\vec r(t)=\begin{bmatrix}3\cos(t+1)\3\sin(t-2)\end{bmatrix}$
$\vec r(t)=\begin{bmatrix}1&-2\0&1\end{bmatrix}\begin{bmatrix}3\cos t\3\sin t\end{bmatrix}$
$\vec r(t)=\begin{bmatrix}3\cos t\3\sin t\end{bmatrix}+\begin{bmatrix}-2\1\end{bmatrix}$
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.
Based on the problem described, what is the role of $t$ in $\vec r(t)=\langle 3\cos t,3\sin t\rangle$?
$t$ fixes one point, so the curve is a single constant location.
$t$ selects an angle, tracing all points on $x^2+y^2=9$.
$t$ is the radius, so changing $t$ changes the circle size.
$t$ 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.
Based on the problem described, determine the parametrization for $\mathbf{A}=\begin{bmatrix}0&-1\1&0\end{bmatrix}$ applied to $\langle 2\cos t,2\sin t\rangle$.
$\vec r(t)=\langle 2\sin t,,-2\cos t\rangle$
$\vec r(t)=\langle 2\cos t,,2\sin t\rangle$
$\vec r(t)=\langle -2\cos t,,-2\sin t\rangle$
$\vec r(t)=\langle -2\sin t,,2\cos t\rangle$
Explanation
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.
Using the scenario above, determine the parametrization for $x^2+y^2=25$ using $t$.
$\vec r(t)=\langle 25\cos t,,25\sin t\rangle$
$\vec r(t)=\langle 5\sin t,,5\cos t\rangle$
$\vec r(t)=\langle 5\cos t,,5\sin t\rangle$
$\vec r(t)=\langle 5t,,\sqrt{25-t^2}\rangle$
Explanation
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.
Using the scenario above, what is the matrix representation for rotating $\langle 4\cos t,2\sin t\rangle$ by $30^\circ$?
$\vec r(t)=\begin{bmatrix}\cos30&-\sin30\sin30&\cos30\end{bmatrix}!\begin{bmatrix}4\cos t\2\sin t\end{bmatrix}$
$\vec r(t)=\begin{bmatrix}\cos30&-\sin30&0\sin30&\cos30&0\end{bmatrix}!\begin{bmatrix}4\cos t\2\sin t\end{bmatrix}$
$\vec r(t)=\begin{bmatrix}\cos30&\sin30\-\sin30&\cos30\end{bmatrix}!\begin{bmatrix}4\cos t\2\sin t\end{bmatrix}$
$\vec r(t)=\begin{bmatrix}4\cos t&0\0&2\sin t\end{bmatrix}!\begin{bmatrix}\cos30\sin30\end{bmatrix}$
Explanation
This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on matrix transformations for rotating parametric curves. Parametrization combined with matrix operations allows us to transform curves systematically, where rotation matrices apply angular transformations to vector functions. In this scenario, we need to rotate the ellipse parametrization ⟨4cos(t), 2sin(t)⟩ by 30° using the standard 2D rotation matrix. Choice A is correct because the rotation matrix for angle θ is [[cos(θ), -sin(θ)], [sin(θ), cos(θ)]], so for 30°: [[cos(30°), -sin(30°)], [sin(30°), cos(30°)]] applied to the column vector [4cos(t), 2sin(t)]ᵀ. Choice B is incorrect because it uses the inverse rotation matrix (rotating by -30° instead of +30°), which has sine terms with opposite signs. To help students: Memorize the standard rotation matrix and remember that positive angles rotate counterclockwise. Practice matrix-vector multiplication to understand how each component transforms under rotation.
Using the scenario above, determine the parametrization for $(x-2)^2+(y+1)^2=9$ using $t$.
$\vec r(t)=\langle 3\cos t,,3\sin t\rangle$
$\vec r(t)=\langle 2+9\cos t,,-1+9\sin t\rangle$
$\vec r(t)=\langle -2+3\cos t,,1+3\sin t\rangle$
$\vec r(t)=\langle 2+3\cos t,,-1+3\sin t\rangle$
Explanation
This question tests AP-level precalculus skills in parametrizing implicitly defined functions, focusing on parametrizing translated circles from standard form. Parametrization involves expressing implicitly defined functions through parameters, where translated circles require shifting the standard circle parametrization. In this scenario, the equation (x-2)² + (y+1)² = 9 represents a circle with center (2, -1) and radius 3. Choice A is correct because it parametrizes a circle of radius 3 centered at (2, -1): x = 2 + 3cos(t) gives (x-2) = 3cos(t), and y = -1 + 3sin(t) gives (y+1) = 3sin(t), so (x-2)² + (y+1)² = 9cos²(t) + 9sin²(t) = 9. Choice D is incorrect because it uses center (-2, 1) instead of (2, -1), which would parametrize the circle (x+2)² + (y-1)² = 9. To help students: For a circle (x-h)² + (y-k)² = r², the parametrization is x = h + r·cos(t), y = k + r·sin(t). Always identify the center (h, k) and radius r from the standard form before parametrizing.