AP Precalculus Quiz: Parametrization Of Implicitly Defined Functions
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Parametrization Of Implicitly Defined FunctionsQuestion 1 of 20

Based on the problem described, determine the parametrization for x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 using tt.

r⃗(t)=⟨16cos⁡t, 9sin⁡t⟩\vec r(t)=\langle 16\cos t,\,9\sin t\rangle
r⃗(t)=⟨4cos⁡t, 3sin⁡t⟩\vec r(t)=\langle 4\cos t,\,3\sin t\rangle
r⃗(t)=⟨4sin⁡t, 3sin⁡t⟩\vec r(t)=\langle 4\sin t,\,3\sin t\rangle
r⃗(t)=⟨4cos⁡t, 9sin⁡t⟩\vec r(t)=\langle 4\cos t,\,9\sin t\rangle
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AP Precalculus Quiz

AP Precalculus Quiz: Parametrization Of Implicitly Defined Functions

Practice Parametrization Of Implicitly Defined Functions in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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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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Question 1

Based on the problem described, determine the parametrization for x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 using tt.

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

A curve is parametrized by x(t)=−2+sec⁡(t)x(t) = -2 + \sec(t) and y(t)=4+3tan⁡(t)y(t) = 4 + 3\tan(t). Which of the following equations represents the curve?

  1. (x+2)2−(y−4)29=1(x+2)^2 - \frac{(y-4)^2}{9} = 1 (correct answer)
  2. (y−4)29−(x+2)2=1\frac{(y-4)^2}{9} - (x+2)^2 = 1
  3. (x−2)2−(y+4)29=1(x-2)^2 - \frac{(y+4)^2}{9} = 1
  4. (x+2)2+(y−4)29=1(x+2)^2 + \frac{(y-4)^2}{9} = 1
Explanation: To find the rectangular equation, we eliminate the parameter tt. Isolate the trigonometric functions: sec⁡(t)=x+2\sec(t) = x+2 and tan⁡(t)=y−43\tan(t) = \frac{y-4}{3}. Using the identity sec⁡2(t)−tan⁡2(t)=1\sec^2(t) - \tan^2(t) = 1, we get (x+2)2−(y−43)2=1(x+2)^2 - (\frac{y-4}{3})^2 = 1, which simplifies to (x+2)2−(y−4)29=1(x+2)^2 - \frac{(y-4)^2}{9} = 1.

Question 3

A curve is given by the parametric equations x(t)=t−1x(t) = t-1 and y(t)=t2+3y(t) = t^2+3. Which of the following is the rectangular equation for the curve?

  1. y=(x−1)2+3y = (x-1)^2 + 3
  2. x=(y+1)2+3x = (y+1)^2 + 3
  3. y=x2+2x+4y = x^2+2x+4
  4. y=(x+1)2+3y = (x+1)^2 + 3 (correct answer)
Explanation: To eliminate the parameter, we solve one equation for tt and substitute into the other. From x(t)=t−1x(t) = t-1, we find t=x+1t = x+1. Substituting this into the equation for y(t)y(t) gives y=(x+1)2+3y = (x+1)^2 + 3.

Question 4

Which of the following parametrizations represents the circle x2+y2=r2x^2+y^2=r^2 traced clockwise starting from the point (r,0)(r,0)?

  1. x(t)=rcos⁡(t),y(t)=rsin⁡(t)x(t) = r\cos(t), y(t) = r\sin(t) for 0≤t<2π0 \le t < 2\pi
  2. x(t)=rsin⁡(t),y(t)=rcos⁡(t)x(t) = r\sin(t), y(t) = r\cos(t) for 0≤t<2π0 \le t < 2\pi
  3. x(t)=rcos⁡(t),y(t)=−rsin⁡(t)x(t) = r\cos(t), y(t) = -r\sin(t) for 0≤t<2π0 \le t < 2\pi (correct answer)
  4. x(t)=−rcos⁡(t),y(t)=rsin⁡(t)x(t) = -r\cos(t), y(t) = r\sin(t) for 0≤t<2π0 \le t < 2\pi
Explanation: At t=0t=0, we need the point to be (r,0)(r,0). For choice C, x(0)=rcos⁡(0)=rx(0) = r\cos(0) = r and y(0)=−rsin⁡(0)=0y(0) = -r\sin(0) = 0, so it starts at the correct point. As tt increases from 0, x(t)x(t) decreases and y(t)y(t) becomes negative, which corresponds to clockwise motion from (r,0)(r,0).

Question 5

The function ff is given by f(x)=e2x+1f(x) = e^{2x} + 1. Which of the following is a parametrization for the graph of the inverse function, f−1f^{-1}?

  1. x(t)=t,y(t)=e2t+1x(t) = t, y(t) = e^{2t}+1
  2. x(t)=e2t+1,y(t)=tx(t) = e^{2t}+1, y(t) = t (correct answer)
  3. x(t)=t,y(t)=ln⁡(t−1)2x(t) = t, y(t) = \frac{\ln(t-1)}{2}
  4. x(t)=t,y(t)=e−2t−1x(t) = t, y(t) = e^{-2t}-1
Explanation: If a function y=f(x)y=f(x) is invertible, a parametrization for its inverse f−1f^{-1} is given by x(t)=f(t)x(t) = f(t) and y(t)=ty(t) = t. In this case, f(t)=e2t+1f(t) = e^{2t}+1. So, the parametrization for f−1f^{-1} is x(t)=e2t+1,y(t)=tx(t) = e^{2t}+1, y(t) = t.

Question 6

Consider the implicitly defined curve x2/3+y2/3=4x^{2/3} + y^{2/3} = 4. Which of the following is a valid parametrization of this curve?

  1. x(t)=2cos⁡(t),y(t)=2sin⁡(t)x(t) = 2\cos(t), y(t) = 2\sin(t)
  2. x(t)=4cos⁡3(t),y(t)=4sin⁡3(t)x(t) = 4\cos^3(t), y(t) = 4\sin^3(t)
  3. x(t)=8cos⁡3(t),y(t)=8sin⁡3(t)x(t) = 8\cos^3(t), y(t) = 8\sin^3(t) (correct answer)
  4. x(t)=8cos⁡(t),y(t)=8sin⁡(t)x(t) = 8\cos(t), y(t) = 8\sin(t)
Explanation: To check the parametrization, substitute x(t)x(t) and y(t)y(t) into the equation. For choice C, we have (8cos⁡3(t))2/3+(8sin⁡3(t))2/3=82/3(cos⁡3(t))2/3+82/3(sin⁡3(t))2/3=(83)2cos⁡2(t)+(83)2sin⁡2(t)=22cos⁡2(t)+22sin⁡2(t)=4(cos⁡2(t)+sin⁡2(t))=4(1)=4(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

Which of the following is a parametrization for the entire parabola x=y2x=y^2?

  1. x(t)=t,y(t)=tx(t) = t, y(t) = \sqrt{t} for t≥0t \ge 0
  2. x(t)=t2,y(t)=tx(t) = t^2, y(t) = t for all real tt (correct answer)
  3. x(t)=cos⁡(t),y(t)=cos⁡(t)x(t) = \cos(t), y(t) = \sqrt{\cos(t)}
  4. x(t)=t2,y(t)=tx(t) = t^2, y(t) = t for t≥0t \ge 0
Explanation: Let y(t)=ty(t) = t. Since yy can take any real value, tt must range over all real numbers. Substituting y=ty=t into the equation x=y2x=y^2 gives x(t)=t2x(t) = t^2. This parametrization traces the entire parabola because as tt goes from −∞-\infty to +∞+\infty, yy covers all real numbers, and xx covers all non-negative numbers.

Question 8

Which parametrization represents the lower branch of the hyperbola y216−x29=1\frac{y^2}{16} - \frac{x^2}{9} = 1?

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

Question 9

The parametrization x(t)=acos⁡(t)+hx(t) = a\cos(t)+h and y(t)=bsin⁡(t)+ky(t) = b\sin(t)+k describes an ellipse. The conversion to the rectangular form (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 relies on which fundamental trigonometric identity?

  1. sec⁡2(t)−tan⁡2(t)=1\sec^2(t) - \tan^2(t) = 1
  2. sin⁡(2t)=2sin⁡(t)cos⁡(t)\sin(2t) = 2\sin(t)\cos(t)
  3. tan⁡(t)=sin⁡(t)cos⁡(t)\tan(t) = \frac{\sin(t)}{\cos(t)}
  4. sin⁡2(t)+cos⁡2(t)=1\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)\cos(t) and sin⁡(t)\sin(t) to get cos⁡(t)=x−ha\cos(t) = \frac{x-h}{a} and sin⁡(t)=y−kb\sin(t) = \frac{y-k}{b}. Then we substitute these into the Pythagorean identity sin⁡2(t)+cos⁡2(t)=1\sin^2(t) + \cos^2(t) = 1, which results in the standard rectangular equation for an ellipse.

Question 10

Which of the following is a parametrization for the ellipse given by the equation (x+1)29+(y−4)225=1\frac{(x+1)^2}{9} + \frac{(y-4)^2}{25} = 1?

  1. x(t)=−1+3cos⁡(t),y(t)=4+5sin⁡(t)x(t) = -1 + 3\cos(t), y(t) = 4 + 5\sin(t) (correct answer)
  2. x(t)=1+3cos⁡(t),y(t)=−4+5sin⁡(t)x(t) = 1 + 3\cos(t), y(t) = -4 + 5\sin(t)
  3. x(t)=−1+9cos⁡(t),y(t)=4+25sin⁡(t)x(t) = -1 + 9\cos(t), y(t) = 4 + 25\sin(t)
  4. x(t)=−1+5cos⁡(t),y(t)=4+3sin⁡(t)x(t) = -1 + 5\cos(t), y(t) = 4 + 3\sin(t)
Explanation: The standard parametrization for an ellipse (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 is x(t)=h+acos⁡(t)x(t) = h + a\cos(t) and y(t)=k+bsin⁡(t)y(t) = k + b\sin(t). Here, the center is (h,k)=(−1,4)(h,k) = (-1, 4). The horizontal semi-axis is a=9=3a = \sqrt{9} = 3, and the vertical semi-axis is b=25=5b = \sqrt{25} = 5. So, x(t)=−1+3cos⁡(t)x(t) = -1 + 3\cos(t) and y(t)=4+5sin⁡(t)y(t) = 4 + 5\sin(t).

Question 11

A curve is given by x(t)=etx(t) = e^t and y(t)=2e2t−1y(t) = 2e^{2t} - 1. Which of the following rectangular equations represents the curve?

  1. y=2x2−1y = 2x^2 - 1 for all real xx
  2. y=2x−1y = 2x - 1 for x>0x>0
  3. y=x2−1y = x^2 - 1 for all real xx
  4. y=2x2−1y = 2x^2 - 1 for x>0x > 0 (correct answer)
Explanation: From x(t)=etx(t) = e^t, we can see that xx must be positive. We can also write e2t=(et)2=x2e^{2t} = (e^t)^2 = x^2. Substituting this into the equation for y(t)y(t) gives y=2x2−1y = 2x^2 - 1. Since x=etx=e^t, the domain of the rectangular equation is restricted to x>0x>0.

Question 12

A curve is defined by the parametric equations x(t)=5+2cos⁡(t)x(t) = 5 + 2\cos(t) and y(t)=−1+6sin⁡(t)y(t) = -1 + 6\sin(t). Which of the following is the equation of the curve in rectangular coordinates?

  1. (x+5)24+(y−1)236=1\frac{(x+5)^2}{4} + \frac{(y-1)^2}{36} = 1
  2. (x−5)22+(y+1)26=1\frac{(x-5)^2}{2} + \frac{(y+1)^2}{6} = 1
  3. (x−5)24+(y+1)236=1\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1 (correct answer)
  4. (x−5)2+(y+1)2=1(x-5)^2 + (y+1)^2 = 1
Explanation: To eliminate the parameter tt, we isolate the trigonometric functions. From x=5+2cos⁡(t)x = 5 + 2\cos(t), we get cos⁡(t)=x−52\cos(t) = \frac{x-5}{2}. From y=−1+6sin⁡(t)y = -1 + 6\sin(t), we get sin⁡(t)=y+16\sin(t) = \frac{y+1}{6}. Using the identity cos⁡2(t)+sin⁡2(t)=1\cos^2(t) + \sin^2(t) = 1, we substitute to get (x−52)2+(y+16)2=1(\frac{x-5}{2})^2 + (\frac{y+1}{6})^2 = 1, which simplifies to (x−5)24+(y+1)236=1\frac{(x-5)^2}{4} + \frac{(y+1)^2}{36} = 1.

Question 13

Which of the following is a parametrization of the hyperbola given by (y+3)24−(x−1)236=1\frac{(y+3)^2}{4} - \frac{(x-1)^2}{36} = 1?

  1. x(t)=1+2tan⁡(t),y(t)=−3+6sec⁡(t)x(t) = 1 + 2\tan(t), y(t) = -3 + 6\sec(t)
  2. x(t)=1+6sec⁡(t),y(t)=−3+2tan⁡(t)x(t) = 1 + 6\sec(t), y(t) = -3 + 2\tan(t)
  3. x(t)=−1+36tan⁡(t),y(t)=3+4sec⁡(t)x(t) = -1 + 36\tan(t), y(t) = 3 + 4\sec(t)
  4. x(t)=1+6tan⁡(t),y(t)=−3+2sec⁡(t)x(t) = 1 + 6\tan(t), y(t) = -3 + 2\sec(t) (correct answer)
Explanation: The standard parametrization for a vertical hyperbola (y−k)2b2−(x−h)2a2=1\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 is x(t)=h+atan⁡(t)x(t) = h + a\tan(t) and y(t)=k+bsec⁡(t)y(t) = k + b\sec(t). Here, the center is (h,k)=(1,−3)(h,k) = (1, -3), a=36=6a = \sqrt{36} = 6, and b=4=2b = \sqrt{4} = 2. So, x(t)=1+6tan⁡(t)x(t) = 1 + 6\tan(t) and y(t)=−3+2sec⁡(t)y(t) = -3 + 2\sec(t).

Question 14

A curve is defined parametrically by x(t)=3tan⁡(t)−2x(t) = 3\tan(t) - 2 and y(t)=5sec⁡(t)+1y(t) = 5\sec(t) + 1. What type of conic section does this parametrization represent?

  1. A circle
  2. A parabola
  3. An ellipse
  4. A hyperbola (correct answer)
Explanation: To identify the conic, we eliminate the parameter tt. From the given equations, we have tan⁡(t)=x+23\tan(t) = \frac{x+2}{3} and sec⁡(t)=y−15\sec(t) = \frac{y-1}{5}. Using the Pythagorean identity sec⁡2(t)−tan⁡2(t)=1\sec^2(t) - \tan^2(t) = 1, we get (y−15)2−(x+23)2=1\left(\frac{y-1}{5}\right)^2 - \left(\frac{x+2}{3}\right)^2 = 1. This equation, (y−1)225−(x+2)29=1\frac{(y-1)^2}{25} - \frac{(x+2)^2}{9} = 1, is the standard form of a hyperbola.

Question 15

A circle is defined by the equation (x−3)2+(y+2)2=16(x-3)^2 + (y+2)^2 = 16. Which of the following is a valid parametrization of this circle?

  1. x(t)=3+16cos⁡(t),y(t)=−2+16sin⁡(t)x(t) = 3 + 16\cos(t), y(t) = -2 + 16\sin(t)
  2. x(t)=−3+4cos⁡(t),y(t)=2+4sin⁡(t)x(t) = -3 + 4\cos(t), y(t) = 2 + 4\sin(t)
  3. x(t)=3+4cos⁡(t),y(t)=−2+4sin⁡(t)x(t) = 3 + 4\cos(t), y(t) = -2 + 4\sin(t) (correct answer)
  4. x(t)=3+4cos⁡(t),y(t)=−2+4sin⁡(t)x(t) = 3 + 4\cos(t), y(t) = -2 + 4\sin(t) for 0≤t<π0 \le t < \pi
Explanation: The standard parametrization for a circle with center (h,k)(h, k) and radius rr is x(t)=h+rcos⁡(t)x(t) = h + r\cos(t) and y(t)=k+rsin⁡(t)y(t) = k + r\sin(t). The given equation has center (3,−2)(3, -2) and radius r=16=4r = \sqrt{16} = 4. Therefore, the correct parametrization is x(t)=3+4cos⁡(t)x(t) = 3 + 4\cos(t), y(t)=−2+4sin⁡(t)y(t) = -2 + 4\sin(t) for a full cycle, 0≤t<2π0 \le t < 2\pi.

Question 16

Which of the following parametrizations represents the top half of the circle x2+y2=9x^2 + y^2 = 9, traced counter-clockwise?

  1. x(t)=3cos⁡(t),y(t)=3sin⁡(t)x(t) = 3\cos(t), y(t) = 3\sin(t) for 0≤t≤2π0 \le t \le 2\pi
  2. x(t)=3cos⁡(t),y(t)=3sin⁡(t)x(t) = 3\cos(t), y(t) = 3\sin(t) for −π/2≤t≤π/2-\pi/2 \le t \le \pi/2
  3. x(t)=3sin⁡(t),y(t)=3cos⁡(t)x(t) = 3\sin(t), y(t) = 3\cos(t) for 0≤t≤π0 \le t \le \pi
  4. x(t)=3cos⁡(t),y(t)=3sin⁡(t)x(t) = 3\cos(t), y(t) = 3\sin(t) for 0≤t≤π0 \le t \le \pi (correct answer)
Explanation: The standard counter-clockwise parametrization for a circle of radius 3 is x(t)=3cos⁡(t),y(t)=3sin⁡(t)x(t) = 3\cos(t), y(t) = 3\sin(t). The top half of the circle corresponds to y≥0y \ge 0. The function y(t)=3sin⁡(t)y(t) = 3\sin(t) is non-negative for 0≤t≤π0 \le t \le \pi. Therefore, this domain for tt traces the top half of the circle.

Question 17

Using the scenario above, determine the parametrization for x2−y2=1x^2-y^2=1 using tt.

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

Based on the problem described, determine the parametrization after r⃗(t)=A⟨cos⁡t,sin⁡t⟩\vec r(t)=\mathbf{A}\langle \cos t,\sin t\rangle, A=[2005]\mathbf{A}=\begin{bmatrix}2&0\\0&5\end{bmatrix}.

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

Based on the problem described, what is the matrix representation for translating ⟨3cos⁡t,3sin⁡t⟩\langle 3\cos t,3\sin t\rangle by b⃗=⟨1,−2⟩\vec b=\langle 1,-2\rangle?

  1. r⃗(t)=[3cos⁡t3sin⁡t]+[1−2]\vec r(t)=\begin{bmatrix}3\cos t\\3\sin t\end{bmatrix}+\begin{bmatrix}1\\-2\end{bmatrix} (correct answer)
  2. r⃗(t)=[1−201][3cos⁡t3sin⁡t]\vec r(t)=\begin{bmatrix}1&-2\\0&1\end{bmatrix}\begin{bmatrix}3\cos t\\3\sin t\end{bmatrix}
  3. r⃗(t)=[3cos⁡(t+1)3sin⁡(t−2)]\vec r(t)=\begin{bmatrix}3\cos(t+1)\\3\sin(t-2)\end{bmatrix}
  4. r⃗(t)=[3cos⁡t3sin⁡t]+[−21]\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

Based on the problem described, what is the role of tt in r⃗(t)=⟨3cos⁡t,3sin⁡t⟩\vec r(t)=\langle 3\cos t,3\sin t\rangle?

  1. tt fixes one point, so the curve is a single constant location.
  2. tt is the radius, so changing tt changes the circle size.
  3. tt selects an angle, tracing all points on x2+y2=9x^2+y^2=9. (correct answer)
  4. tt replaces yy, so xx becomes an implicit function of tt 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.