AP Calculus BC Quiz: Defining And Differentiating Parametric Equations
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Defining And Differentiating Parametric EquationsQuestion 1 of 20

A parametric path is given by x(t)=etx(t)=e^t and y(t)=tety(t)=t e^t; what is dydx\frac{dy}{dx} in terms of tt?

et+tetet\dfrac{e^t+te^t}{e^t}
etet+tet\dfrac{e^t}{e^t+te^t}
tetet\dfrac{te^t}{e^t}
et+tettet\dfrac{e^t+te^t}{t e^t}
(et)(et+tet)(e^t)(e^t+te^t)
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AP Calculus BC Quiz

AP Calculus BC Quiz: Defining And Differentiating Parametric Equations

Practice Defining And Differentiating Parametric Equations in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Defining And Differentiating Parametric Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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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.

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Question 1

A parametric path is given by x(t)=etx(t)=e^t and y(t)=tety(t)=t e^t; what is dydx\frac{dy}{dx} in terms of tt?

  1. et+tetet\dfrac{e^t+te^t}{e^t} (correct answer)
  2. etet+tet\dfrac{e^t}{e^t+te^t}
  3. tetet\dfrac{te^t}{e^t}
  4. et+tettet\dfrac{e^t+te^t}{t e^t}
  5. (et)(et+tet)(e^t)(e^t+te^t)
Explanation: This parametric curve requires applying dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} to find the derivative. For x(t)=etx(t) = e^t, we have dx/dt=etdx/dt = e^t. For y(t)=tety(t) = t e^t, we use the product rule: dy/dt=et+tetdy/dt = e^t + t e^t. The chain rule then gives dy/dx=(et+tet)/etdy/dx = (e^t + t e^t) / e^t. Choice B incorrectly inverts this ratio by placing dx/dtdx/dt in the numerator instead of the denominator. The essential parametric differentiation approach is to compute dy/dtdy/dt and dx/dtdx/dt separately, then divide the first by the second.

Question 2

A curve is x(t)=t2+12tx(t)=t^2+\frac{1}{2}t and y(t)=t2−12ty(t)=t^2-\frac{1}{2}t; find dydx\frac{dy}{dx}.

  1. 2t−1/22t+1/2\dfrac{2t-1/2}{2t+1/2} (correct answer)
  2. 2t+1/22t−1/2\dfrac{2t+1/2}{2t-1/2}
  3. t2−12t2t+1/2\dfrac{t^2-\tfrac12 t}{2t+1/2}
  4. (2t−1/2)(2t+1/2)(2t-1/2)(2t+1/2)
  5. 2t−1/2t2+12t\dfrac{2t-1/2}{t^2+\tfrac12 t}
Explanation: This parametric curve applies dy/dx = (dy/dt)/(dx/dt). For x(t) = t² + t/2, we calculate dx/dt = 2t + 1/2. For y(t) = t² - t/2, we get dy/dt = 2t - 1/2. Using the chain rule yields dy/dx = (2t - 1/2)/(2t + 1/2). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator. The essential parametric differentiation approach is to compute both component derivatives, then divide dy/dt by dx/dt.

Question 3

A particle moves with x(t)=12t2x(t)=\frac{1}{2}t^2 and y(t)=14t4y(t)=\frac{1}{4}t^4; compute dydx\frac{dy}{dx}.

  1. t3t\dfrac{t^3}{t} (correct answer)
  2. tt3\dfrac{t}{t^3}
  3. t4/4t\dfrac{t^4/4}{t}
  4. (t3)(t)(t^3)(t)
  5. t3t2/2\dfrac{t^3}{t^2/2}
Explanation: This parametric curve requires dy/dx=(dy/dt)/(dx/dt)dy/dx = (dy/dt)/(dx/dt). For x(t)=12t2x(t) = \frac{1}{2}t^2, we find dx/dt=tdx/dt = t. For y(t)=14t4y(t) = \frac{1}{4}t^4, we get dy/dt=t3dy/dt = t^3. Using the parametric derivative formula yields dy/dx=t3/t=t2dy/dx = t^3/t = t^2. Choice A shows this result correctly. The standard parametric differentiation method involves finding derivatives of both components, then dividing dy/dtdy/dt by dx/dtdx/dt.

Question 4

A parametric curve has x(t)=t2+tan⁡tx(t)=t^2+\tan t and y(t)=t2+sec⁡ty(t)=t^2+\sec t; find dydx\frac{dy}{dx}.

  1. 2t+sec⁡ttan⁡t2t+sec⁡2t\dfrac{2t+\sec t\tan t}{2t+\sec^2 t} (correct answer)
  2. 2t+sec⁡2t2t+sec⁡ttan⁡t\dfrac{2t+\sec^2 t}{2t+\sec t\tan t}
  3. t2+sec⁡t2t+sec⁡2t\dfrac{t^2+\sec t}{2t+\sec^2 t}
  4. (2t+sec⁡ttan⁡t)(2t+sec⁡2t)(2t+\sec t\tan t)(2t+\sec^2 t)
  5. 2t+sec⁡ttan⁡tt2+tan⁡t\dfrac{2t+\sec t\tan t}{t^2+\tan t}
Explanation: This parametric curve uses dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. For x(t)=t2+tan⁡tx(t) = t^2 + \tan t, we calculate dxdt=2t+sec⁡2t\frac{dx}{dt} = 2t + \sec^2 t. For y(t)=t2+sec⁡ty(t) = t^2 + \sec t, we get dydt=2t+sec⁡ttan⁡t\frac{dy}{dt} = 2t + \sec t \tan t. Using the parametric derivative gives dydx=2t+sec⁡ttan⁡t2t+sec⁡2t\frac{dy}{dx} = \frac{2t + \sec t \tan t}{2t + \sec^2 t}. Choice B incorrectly inverts the fraction. The fundamental parametric differentiation technique is to compute derivatives of both components, then divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}.

Question 5

Given x(t)=sin⁡(2t)x(t)=\sin(2t) and y(t)=cos⁡(3t)y(t)=\cos(3t), compute dydx\frac{dy}{dx} in terms of tt.

  1. −3sin⁡(3t)2cos⁡(2t)\dfrac{-3\sin(3t)}{2\cos(2t)} (correct answer)
  2. 2cos⁡(2t)−3sin⁡(3t)\dfrac{2\cos(2t)}{-3\sin(3t)}
  3. −sin⁡(3t)cos⁡(2t)\dfrac{-\sin(3t)}{\cos(2t)}
  4. (−3sin⁡(3t))(2cos⁡(2t))(-3\sin(3t))(2\cos(2t))
  5. cos⁡(3t)2cos⁡(2t)\dfrac{\cos(3t)}{2\cos(2t)}
Explanation: This parametric differentiation uses the chain rule dy/dx = (dy/dt)/(dx/dt). For x(t) = sin(2t), we find dx/dt = 2cos(2t). For y(t) = cos(3t), we get dy/dt = -3sin(3t). The parametric derivative formula gives dy/dx = (-3sin(3t))/(2cos(2t)). Choice B incorrectly inverts the fraction by placing dx/dt in the numerator. The key parametric differentiation strategy is to compute both component derivatives, then form the ratio dy/dt over dx/dt.

Question 6

If x(t)=sec⁡tx(t)=\sec t and y(t)=tan⁡ty(t)=\tan t, what is dydx\dfrac{dy}{dx} in terms of tt?

  1. sec⁡ttan⁡tsec⁡2t\dfrac{\sec t\tan t}{\sec^2 t}
  2. sec⁡2tsec⁡ttan⁡t\dfrac{\sec^2 t}{\sec t\tan t} (correct answer)
  3. sec⁡ttan⁡tsec⁡t\dfrac{\sec t\tan t}{\sec t}
  4. sec⁡2tsec⁡t\dfrac{\sec^2 t}{\sec t}
  5. sec⁡2ttan⁡t\dfrac{\sec^2 t}{\tan t}
Explanation: This problem involves parametric differentiation, where we find the derivative dy/dx for curves defined by parametric equations x(t) and y(t). To compute dy/dx using the chain rule, recognize that dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. First, differentiate x(t) = sec t to get dx/dt = sec t tan t, and y(t) = tan t to get dy/dt = sec² t. Then, divide these derivatives to obtain dy/dx = sec² t / (sec t tan t). A tempting distractor might be sec t tan t / sec² t, which fails because it incorrectly inverts the ratio of the derivatives. Always remember the general strategy for parametric derivatives: compute dy/dt and dx/dt separately, then form their quotient for dy/dx.

Question 7

A particle moves with x(t)=t2−3tx(t)=t^2-3t and y(t)=4t+sin⁡ty(t)=4t+\sin t; find dydx\dfrac{dy}{dx}.

  1. 4+cos⁡t2t−3\dfrac{4+\cos t}{2t-3} (correct answer)
  2. 2t−34+cos⁡t\dfrac{2t-3}{4+\cos t}
  3. 4+cos⁡t2t+3\dfrac{4+\cos t}{2t+3}
  4. 4−cos⁡t2t−3\dfrac{4-\cos t}{2t-3}
  5. 2t−34−cos⁡t\dfrac{2t-3}{4-\cos t}
Explanation: This problem involves parametric differentiation, where we find the derivative dy/dx for curves defined by parametric equations x(t) and y(t). To compute dy/dx using the chain rule, recognize that dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. First, differentiate x(t) = t² - 3t to get dx/dt = 2t - 3, and y(t) = 4t + sin t to get dy/dt = 4 + cos t. Then, divide these derivatives to obtain dy/dx = (4 + cos t) / (2t - 3). A tempting distractor might be (2t - 3) / (4 + cos t), which fails because it incorrectly inverts the ratio of the derivatives. Always remember the general strategy for parametric derivatives: compute dy/dt and dx/dt separately, then form their quotient for dy/dx.

Question 8

Given x(t)=t2+tx(t)=t^2+t and y(t)=t22+ty(t)=\frac{t^2}{2}+t, compute dydx\frac{dy}{dx} in terms of tt.

  1. t+12t+1\dfrac{t+1}{2t+1} (correct answer)
  2. 2t+1t+1\dfrac{2t+1}{t+1}
  3. t2/2+t2t+1\dfrac{t^2/2+t}{2t+1}
  4. (t+1)(2t+1)(t+1)(2t+1)
  5. t+1t2+t\dfrac{t+1}{t^2+t}
Explanation: This parametric differentiation uses dy/dx = (dy/dt)/(dx/dt). For x(t) = t² + t, we calculate dx/dt = 2t + 1. For y(t) = t²/2 + t, we get dy/dt = t + 1. Using the chain rule gives dy/dx = (t + 1)/(2t + 1). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator. The key parametric derivative technique is to differentiate each component separately, then form the ratio with dy/dt on top.

Question 9

For x(t)=t2−4x(t)=t^2-4 and y(t)=1ty(t)=\frac{1}{t} with t≠0t\ne0, determine dydx\frac{dy}{dx}.

  1. −1/t22t\dfrac{-1/t^2}{2t} (correct answer)
  2. 2t−1/t2\dfrac{2t}{-1/t^2}
  3. −1/t2t2−4\dfrac{-1/t^2}{t^2-4}
  4. (−1/t2)(2t)(-1/t^2)(2t)
  5. 1/t2t\dfrac{1/t}{2t}
Explanation: This parametric differentiation problem uses dy/dx = (dy/dt)/(dx/dt). For x(t) = t² - 4, we calculate dx/dt = 2t. For y(t) = 1/t, we get dy/dt = -1/t². The chain rule gives dy/dx = (-1/t²)/(2t) = -1/(2t³). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator. The fundamental parametric derivative approach is to differentiate each component separately, then divide dy/dt by dx/dt.

Question 10

Given x(t)=t3x(t)=t^3 and y(t)=ln⁡(t)y(t)=\ln(t) for t>0t>0, determine dydx\frac{dy}{dx} as a function of tt.

  1. 1/t3t2\dfrac{1/t}{3t^2} (correct answer)
  2. 3t21/t\dfrac{3t^2}{1/t}
  3. ln⁡t3t2\dfrac{\ln t}{3t^2}
  4. (1/t)(3t2)(1/t)(3t^2)
  5. 1/tt3\dfrac{1/t}{t^3}
Explanation: This parametric differentiation requires dy/dx = (dy/dt)/(dx/dt). For x(t) = t³, we calculate dx/dt = 3t². For y(t) = ln(t), we get dy/dt = 1/t. The chain rule gives dy/dx = (1/t)/(3t²) = 1/(3t³). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator. The reliable parametric derivative strategy is to find derivatives of both components, then form their proper quotient.

Question 11

A moving point has x(t)=sec⁡tx(t)=\sec t and y(t)=tan⁡ty(t)=\tan t; find dydx\frac{dy}{dx} in terms of tt.

  1. sec⁡2tsec⁡ttan⁡t\dfrac{\sec^2 t}{\sec t\tan t} (correct answer)
  2. sec⁡ttan⁡tsec⁡2t\dfrac{\sec t\tan t}{\sec^2 t}
  3. tan⁡tsec⁡t\dfrac{\tan t}{\sec t}
  4. (sec⁡2t)(sec⁡ttan⁡t)(\sec^2 t)(\sec t\tan t)
  5. tan⁡tsec⁡ttan⁡t\dfrac{\tan t}{\sec t\tan t}
Explanation: This parametric curve uses the chain rule dy/dx = (dy/dt)/(dx/dt). For x(t) = sec t, we find dx/dt = sec t tan t. For y(t) = tan t, we get dy/dt = sec² t. Applying the parametric derivative formula yields dy/dx = (sec² t)/(sec t tan t) = sec t/tan t = 1/sin t. Choice B incorrectly inverts the fraction. The standard parametric differentiation technique involves finding both derivatives with respect to t, then forming their proper ratio.

Question 12

For the parametric curve x(t)=sin⁡tx(t)=\sin t and y(t)=cos⁡t+ty(t)=\cos t+t, find dydx\dfrac{dy}{dx} in terms of tt.

  1. −sin⁡t+1cos⁡t\dfrac{-\sin t+1}{\cos t} (correct answer)
  2. cos⁡t−sin⁡t+1\dfrac{\cos t}{-\sin t+1}
  3. −sin⁡t+1-\sin t+1
  4. cos⁡t\cos t
  5. −sin⁡tcos⁡t+t\dfrac{-\sin t}{\cos t}+t
Explanation: This problem involves parametric differentiation to find dy/dx for the given functions x(t) and y(t). To find dy/dx, we apply the chain rule, which states that dy/dx = (dy/dt) / (dx/dt), assuming dx/dt is not zero. First, compute dx/dt = cos t from x(t) = sin t, and dy/dt = -sin t + 1 from y(t) = cos t + t. Therefore, dy/dx = (-sin t + 1) / cos t, which matches choice A. A tempting distractor is choice B, which is the reciprocal (cos t)/(-sin t + 1), but that would be dx/dy instead of dy/dx. In general, for parametric equations, always compute the ratio of the y-derivative to the x-derivative with respect to the parameter to obtain the correct slope.

Question 13

Given x(t)=t3−tx(t)=t^3- t and y(t)=t3+ty(t)=t^3+t, find dydx\frac{dy}{dx} in terms of tt.

  1. 3t2+13t2−1\dfrac{3t^2+1}{3t^2-1} (correct answer)
  2. 3t2−13t2+1\dfrac{3t^2-1}{3t^2+1}
  3. t3+t3t2−1\dfrac{t^3+t}{3t^2-1}
  4. (3t2+1)(3t2−1)(3t^2+1)(3t^2-1)
  5. 3t2+1t3−t\dfrac{3t^2+1}{t^3-t}
Explanation: This parametric differentiation applies dy/dx = (dy/dt)/(dx/dt). For x(t) = t³ - t, we calculate dx/dt = 3t² - 1. For y(t) = t³ + t, we get dy/dt = 3t² + 1. The parametric derivative gives dy/dx = (3t² + 1)/(3t² - 1). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator. The fundamental parametric differentiation technique is to compute derivatives of both components, then divide dy/dt by dx/dt.

Question 14

If x(t)=t−cos⁡tx(t)=t-\cos t and y(t)=1+sin⁡ty(t)=1+\sin t, what is dydx\dfrac{dy}{dx} at t=π2t=\dfrac{\pi}{2}?

  1. 1+sin⁡tcos⁡t\dfrac{1+\sin t}{\cos t}
  2. cos⁡t1+sin⁡t\dfrac{\cos t}{1+\sin t}
  3. cos⁡t1+sin⁡t∣t=π/2=0\dfrac{\cos t}{1+\sin t}\Big|_{t=\pi/2}=0 (correct answer)
  4. cos⁡t1+sin⁡t∣t=π/2=12\dfrac{\cos t}{1+\sin t}\Big|_{t=\pi/2}=\dfrac{1}{2}
  5. cos⁡t1+sin⁡t∣t=π/2=1\dfrac{\cos t}{1+\sin t}\Big|_{t=\pi/2}=1
Explanation: This problem requires finding dy/dx for parametric equations using the chain rule relationship dy/dx = (dy/dt)/(dx/dt). With x(t) = t - cos t and y(t) = 1 + sin t, we compute dx/dt = 1 + sin t and dy/dt = cos t. Therefore, dy/dx = cos t/(1 + sin t). At t = π/2, we have cos(π/2) = 0 and sin(π/2) = 1, giving dy/dx = 0/(1 + 1) = 0/2 = 0. Choice A shows the general formula without evaluation, which is incomplete. For parametric problems asking for the derivative at a specific point, always substitute the parameter value after finding the general derivative formula.

Question 15

Given x(t)=e2tx(t)=e^{2t} and y(t)=et+1y(t)=e^t+1, compute dydx\frac{dy}{dx} in terms of tt.

  1. et2e2t\dfrac{e^t}{2e^{2t}} (correct answer)
  2. 2e2tet\dfrac{2e^{2t}}{e^t}
  3. et+12e2t\dfrac{e^t+1}{2e^{2t}}
  4. (et)(2e2t)(e^t)(2e^{2t})
  5. ete2t\dfrac{e^t}{e^{2t}}
Explanation: This parametric differentiation requires dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. For x(t)=e2tx(t) = e^{2t}, we find dxdt=2e2t\frac{dx}{dt} = 2e^{2t}. For y(t)=et+1y(t) = e^t + 1, we get dydt=et\frac{dy}{dt} = e^t. The parametric derivative formula gives dydx=et2e2t=12et\frac{dy}{dx} = \frac{e^t}{2e^{2t}} = \frac{1}{2e^t}. Choice A shows the unreduced form. The reliable parametric derivative strategy is to differentiate each component separately, then form their proper quotient.

Question 16

For x(t)=t2+1tx(t)=\frac{t^2+1}{t} and y(t)=t2−1ty(t)=\frac{t^2-1}{t}, t≠0t\ne0, determine dydx\frac{dy}{dx}.

  1. 1+1/t21−1/t2\dfrac{1+1/t^2}{1-1/t^2} (correct answer)
  2. 1−1/t21+1/t2\dfrac{1-1/t^2}{1+1/t^2}
  3. (t2−1)/t1−1/t2\dfrac{(t^2-1)/t}{1-1/t^2}
  4. (1−1/t2)(1+1/t2)(1-1/t^2)(1+1/t^2)
  5. 1−1/t2(t2+1)/t\dfrac{1-1/t^2}{(t^2+1)/t}
Explanation: This parametric differentiation applies dy/dx = (dy/dt)/(dx/dt). For x(t) = (t² + 1)/t, we can rewrite as t + 1/t, so dx/dt = 1 - 1/t². For y(t) = (t² - 1)/t = t - 1/t, we get dy/dt = 1 + 1/t². The parametric derivative formula yields dy/dx = (1 + 1/t²)/(1 - 1/t²). Choice B incorrectly inverts the fraction. The standard parametric differentiation method involves finding derivatives of both components, then dividing dy/dt by dx/dt.

Question 17

For x(t)=tcos⁡tx(t)=t\cos t and y(t)=tsin⁡ty(t)=t\sin t, determine dydx\frac{dy}{dx} in terms of tt.

  1. sin⁡t+tcos⁡tcos⁡t−tsin⁡t\dfrac{\sin t+t\cos t}{\cos t-t\sin t} (correct answer)
  2. cos⁡t−tsin⁡tsin⁡t+tcos⁡t\dfrac{\cos t-t\sin t}{\sin t+t\cos t}
  3. tsin⁡tcos⁡t−tsin⁡t\dfrac{t\sin t}{\cos t-t\sin t}
  4. (sin⁡t+tcos⁡t)(cos⁡t−tsin⁡t)(\sin t+t\cos t)(\cos t-t\sin t)
  5. sin⁡t+tcos⁡ttcos⁡t\dfrac{\sin t+t\cos t}{t\cos t}
Explanation: This parametric differentiation uses dy/dx = (dy/dt)/(dx/dt). For x(t) = t cos t, we apply the product rule: dx/dt = cos t - t sin t. For y(t) = t sin t, we get dy/dt = sin t + t cos t. The parametric derivative gives dy/dx = (sin t + t cos t)/(cos t - t sin t). Choice B incorrectly inverts the fraction. The key parametric derivative technique is to differentiate each component separately, then form the ratio with dy/dt in the numerator.

Question 18

A curve is defined by x(t)=t3−2tx(t)=t^3-2t and y(t)=4t2+ty(t)=4t^2+t; what is dydx\frac{dy}{dx}?

  1. 8t+13t2−2\dfrac{8t+1}{3t^2-2} (correct answer)
  2. 3t2−28t+1\dfrac{3t^2-2}{8t+1}
  3. 4t2+t3t2−2\dfrac{4t^2+t}{3t^2-2}
  4. (8t+1)(3t2−2)(8t+1)(3t^2-2)
  5. 8t+1t3−2t\dfrac{8t+1}{t^3-2t}
Explanation: This parametric curve requires the chain rule formula dy/dx = (dy/dt)/(dx/dt). For x(t) = t³ - 2t, we calculate dx/dt = 3t² - 2. For y(t) = 4t² + t, we get dy/dt = 8t + 1. Applying the parametric derivative rule yields dy/dx = (8t + 1)/(3t² - 2). Choice B incorrectly inverts the fraction by placing dx/dt in the numerator. The reliable parametric differentiation method is to compute both component derivatives, then divide dy/dt by dx/dt.

Question 19

A curve is given by x(t)=t3x(t)=t^3 and y(t)=1ty(t)=\dfrac{1}{t} for t≠0t\ne0; what is dydx\dfrac{dy}{dx}?

  1. −1/t23t2\dfrac{-1/t^2}{3t^2}
  2. 3t2−1/t2\dfrac{3t^2}{-1/t^2}
  3. 3t23t^2
  4. −1t2-\dfrac{1}{t^2}
  5. −13t4-\dfrac{1}{3t^4} (correct answer)
Explanation: This problem involves parametric differentiation to find dy/dxdy/dx for the given functions x(t)x(t) and y(t)y(t). To find dy/dxdy/dx, we apply the chain rule, which states that dy/dx=(dy/dt)/(dx/dt)dy/dx = (dy/dt) / (dx/dt), assuming dx/dtdx/dt is not zero. First, compute dx/dt=3t2dx/dt = 3t^2 from x(t)=t3x(t) = t^3, and dy/dt=−1/t2dy/dt = -1/t^2 from y(t)=1/ty(t) = 1/t. Therefore, dy/dx=(−1/t2)/(3t2)=−1/(3t4)dy/dx = (-1/t^2) / (3t^2) = -1/(3t^4), which matches choice E. A tempting distractor is choice A, which is the unsimplified form (−1/t2)/(3t2)(-1/t^2)/(3t^2) but is actually equal to E, though the question likely expects the simplified version. In general, for parametric equations, always compute the ratio of the y-derivative to the x-derivative with respect to the parameter to obtain the correct slope.

Question 20

A curve is described by x(t)=t4x(t)=t^4 and y(t)=t2+3ty(t)=t^2+3t; find dydx\frac{dy}{dx} as a function of tt.

  1. 2t+34t3\dfrac{2t+3}{4t^3} (correct answer)
  2. 4t32t+3\dfrac{4t^3}{2t+3}
  3. t2+3t4t3\dfrac{t^2+3t}{4t^3}
  4. (2t+3)(4t3)(2t+3)(4t^3)
  5. 2t+3t4\dfrac{2t+3}{t^4}
Explanation: This parametric curve applies dy/dx = (dy/dt)/(dx/dt). For x(t) = t⁴, we calculate dx/dt = 4t³. For y(t) = t² + 3t, we get dy/dt = 2t + 3. Using the chain rule yields dy/dx = (2t + 3)/(4t³). Choice B incorrectly inverts the fraction by putting dx/dt in the numerator instead of dy/dt. The standard parametric differentiation method involves finding derivatives of both components, then dividing dy/dt by dx/dt.