AP Calculus BC Quiz: Second Derivatives Of Parametric Equations
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Second Derivatives Of Parametric EquationsQuestion 1 of 20

If x(t)=t2x(t)=t^2 and y(t)=ty(t)=\sqrt{t} for t>0t>0, what is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

−316t7/2\dfrac{-3}{16t^{7/2}}
−116t7/2\dfrac{-1}{16t^{7/2}}
−38t5/2\dfrac{-3}{8t^{5/2}}
316t7/2\dfrac{3}{16t^{7/2}}
−316t5/2\dfrac{-3}{16t^{5/2}}
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AP Calculus BC Quiz

AP Calculus BC Quiz: Second Derivatives Of Parametric Equations

Practice Second Derivatives Of 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 Second Derivatives Of 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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Question 1

If x(t)=t2x(t)=t^2 and y(t)=ty(t)=\sqrt{t} for t>0t>0, what is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

  1. −316t7/2\dfrac{-3}{16t^{7/2}} (correct answer)
  2. −116t7/2\dfrac{-1}{16t^{7/2}}
  3. −38t5/2\dfrac{-3}{8t^{5/2}}
  4. 316t7/2\dfrac{3}{16t^{7/2}}
  5. −316t5/2\dfrac{-3}{16t^{5/2}}
Explanation: This problem tests the skill of finding second derivatives of parametric equations. The formula for the second derivative is d2ydx2=ddt(dy/dtdx/dt)dx/dt\frac{d^2 y}{dx^2} = \frac{ \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) }{ dx/dt }. Here, dx/dt=2tdx/dt = 2t and dy/dt=12t−1/2dy/dt = \frac{1}{2} t^{-1/2}, so dydx=12t−1/22t=14t3/2\frac{dy}{dx} = \frac{ \frac{1}{2} t^{-1/2} }{ 2t } = \frac{1}{4 t^{3/2}}. Differentiating this with respect to t gives ddt[14t−3/2]=14⋅(−32)t−5/2=−38t5/2\frac{d}{dt} [ \frac{1}{4} t^{-3/2} ] = \frac{1}{4} \cdot \left( -\frac{3}{2} \right) t^{-5/2} = -\frac{3}{8 t^{5/2}}. Dividing by dx/dt yields [−38t5/2]/2t=−316t7/2[-\frac{3}{8 t^{5/2}}] / 2t = -\frac{3}{16 t^{7/2}}. A tempting distractor is choice C, −3/(8t5/2)-3/(8 t^{5/2}), which forgets the final division by dx/dt. A transferable strategy for computing parametric second derivatives is to consistently apply the formula ddt(dydx)/dxdt\frac{d}{dt}\left( \frac{dy}{dx} \right) / \frac{dx}{dt} and verify with specific values if possible.

Question 2

A particle moves with x(t)=etx(t)=e^t and y(t)=tety(t)=t e^t; what is d2ydx2\dfrac{d^2y}{dx^2}?

  1. 1et\dfrac{1}{e^t} (correct answer)
  2. t+1et\dfrac{t+1}{e^t}
  3. tet\dfrac{t}{e^t}
  4. ete^t
  5. 1t+1\dfrac{1}{t+1}
Explanation: To find the second derivative of this parametric curve, we use d2ydx2=ddt(dy/dtdx/dt)⋅1dx/dt\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}. First, dydt=et+tet=et(1+t)\frac{dy}{dt} = e^t + te^t = e^t(1+t) and dxdt=et\frac{dx}{dt} = e^t, giving dydx=et(1+t)et=1+t\frac{dy}{dx} = \frac{e^t(1+t)}{e^t} = 1+t. Differentiating with respect to tt: ddt[1+t]=1\frac{d}{dt}[1+t] = 1. Finally, dividing by dxdt=et\frac{dx}{dt} = e^t yields d2ydx2=1et\frac{d^2y}{dx^2} = \frac{1}{e^t}. Choice B incorrectly includes (t+1)(t+1) in the numerator, confusing the first derivative expression with the second derivative. Remember that when dy/dxdy/dx simplifies to a function of tt alone, its derivative is straightforward before the final division by dx/dtdx/dt.

Question 3

Given x(t)=ln⁡tx(t)=\ln t and y(t)=t2y(t)=t^2 for t>0t>0, compute d2ydx2\dfrac{d^2y}{dx^2}.

  1. 2t22t^2
  2. 4t24t^2 (correct answer)
  3. 2t2t
  4. 4t4t
  5. 4t\dfrac{4}{t}
Explanation: This problem tests the skill of finding the second derivative of parametric equations. The formula for d²y/dx² is (x' y'' - y' x'') / (x')³, where primes denote derivatives with respect to t. For x(t) = ln t and y(t) = t², compute x' = 1/t, x'' = -1/t², y' = 2t, and y'' = 2. Substituting yields [(1/t) · 2 - 2t · (-1/t²)] / (1/t)³ = (2/t + 2/t) / (1/t³) = (4/t) · t³ = 4t². A tempting distractor like 4t fails because it neglects the cubic power in the denominator. A transferable strategy for second derivatives is to handle inverse functions like logarithms by ensuring derivatives are correctly computed.

Question 4

For x(t)=t3x(t)=t^3 and y(t)=t2+1y(t)=t^2+1, what is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

  1. −29t4\dfrac{-2}{9t^4} (correct answer)
  2. 29t4\dfrac{2}{9t^4}
  3. −29t2\dfrac{-2}{9t^2}
  4. −23t2\dfrac{-2}{3t^2}
  5. 23t2\dfrac{2}{3t^2}
Explanation: This problem tests the skill of finding second derivatives of parametric equations. The formula for the second derivative is d2ydx2=ddt(dy/dtdx/dt)dx/dt\frac{d^2 y}{dx^2} = \frac{ \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) }{ dx/dt }. Here, dx/dt=3t2dx/dt = 3t^2 and dy/dt=2tdy/dt = 2t, so dydx=2t/3t2=2/(3t)\frac{dy}{dx} = 2t / 3t^2 = 2/(3t). Differentiating this with respect to t gives −2/(3t2)-2/(3t^2). Dividing by dx/dt yields −2/(3t2)/3t2=−2/(9t4)-2/(3t^2) / 3t^2 = -2/(9t^4). A tempting distractor is choice C, −2/(9t2)-2/(9t^2), which forgets the final division by dx/dt, resulting in a lower power in the denominator. A transferable strategy for computing parametric second derivatives is to consistently apply the formula ddt(dydx)/dxdt\frac{d}{dt}\left( \frac{dy}{dx} \right) / \frac{dx}{dt} and verify with specific values if possible.

Question 5

A curve is parametrized by x(t)=tan⁡tx(t)=\tan t and y(t)=sec⁡ty(t)=\sec t; what is d2ydx2\dfrac{d^2y}{dx^2}?

  1. sec⁡ttan⁡2t\dfrac{\sec t}{\tan^2 t}
  2. sec⁡2ttan⁡t\dfrac{\sec^2 t}{\tan t}
  3. 1sec⁡t\dfrac{1}{\sec t} (correct answer)
  4. sec⁡ttan⁡t\dfrac{\sec t}{\tan t}
  5. 1tan⁡t\dfrac{1}{\tan t}
Explanation: For this parametric curve, we need to find d2ydx2\frac{d^2y}{dx^2} using the formula ddt(dy/dtdx/dt)⋅1dx/dt\frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}. We have dydt=sec⁡ttan⁡t\frac{dy}{dt} = \sec t \tan t and dxdt=sec⁡2t\frac{dx}{dt} = \sec^2 t, giving dydx=sec⁡ttan⁡tsec⁡2t=tan⁡tsec⁡t=sin⁡tcos⁡t⋅cos⁡t=sin⁡t\frac{dy}{dx} = \frac{\sec t \tan t}{\sec^2 t} = \frac{\tan t}{\sec t} = \frac{\sin t}{\cos t} \cdot \cos t = \sin t. Differentiating with respect to tt: ddt[sin⁡t]=cos⁡t\frac{d}{dt}[\sin t] = \cos t. Finally, dividing by dxdt=sec⁡2t\frac{dx}{dt} = \sec^2 t yields d2ydx2=cos⁡tsec⁡2t=cos⁡t⋅cos⁡2t=cos⁡3t=1sec⁡t\frac{d^2y}{dx^2} = \frac{\cos t}{\sec^2 t} = \cos t \cdot \cos^2 t = \cos^3 t = \frac{1}{\sec t}. Choice A incorrectly applies the quotient rule to the original ratio instead of first simplifying dy/dxdy/dx. Simplifying dy/dxdy/dx before differentiating often makes parametric second derivative calculations much cleaner.

Question 6

A curve is given by x(t)=etx(t)=e^t and y(t)=t2y(t)=t^2. What is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

  1. 2−2te2t\dfrac{2-2t}{e^{2t}} (correct answer)
  2. 2−2tet\dfrac{2-2t}{e^{t}}
  3. 2tet\dfrac{2t}{e^{t}}
  4. 2et\dfrac{2}{e^{t}}
  5. 2−2tt2\dfrac{2-2t}{t^2}
Explanation: This problem requires finding the second derivative of a parametric curve using d2ydx2=ddt(dy/dtdx/dt)dx/dt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt}. We have dy/dt=2t\frac{dy/dt} = 2t and dx/dt=et\frac{dx/dt} = e^t, so dy/dx=2tet\frac{dy/dx} = \frac{2t}{e^t}. Differentiating using the quotient rule: ddt(2tet)=2et−2tete2t=2et(1−t)e2t=2(1−t)et\frac{d}{dt}\left(\frac{2t}{e^t}\right) = \frac{2e^t - 2te^t}{e^{2t}} = \frac{2e^t(1-t)}{e^{2t}} = \frac{2(1-t)}{e^t}. Then d2ydx2=2(1−t)/etet=2(1−t)e2t=2−2te2t\frac{d^2y}{dx^2} = \frac{2(1-t)/e^t}{e^t} = \frac{2(1-t)}{e^{2t}} = \frac{2-2t}{e^{2t}}. Choice D shows 2et\frac{2}{e^t}, which omits the crucial (1−t)(1-t) factor from the quotient rule differentiation. When finding parametric second derivatives, carefully apply the quotient rule to \frac{dy/dx} before dividing by dxdt\frac{dx}{dt}.

Question 7

If x=etx=e^t and y=t2y=t^2, what is d2ydx2\dfrac{d^2y}{dx^2} expressed in terms of tt?

  1. 2e2t\dfrac{2}{e^{2t}}
  2. 2(1−t)e2t\dfrac{2(1-t)}{e^{2t}} (correct answer)
  3. 2tet\dfrac{2t}{e^{t}}
  4. 2(t−1)e2t\dfrac{2(t-1)}{e^{2t}}
  5. 2et\dfrac{2}{e^{t}}
Explanation: This problem requires finding d²y/dx² for x = eᵗ and y = t². Using the parametric second derivative formula d²y/dx² = [d/dt(dy/dx)]/(dx/dt), we first find dy/dx = (dy/dt)/(dx/dt) = 2t/eᵗ. Next, we differentiate dy/dx with respect to t: d/dt[2t/eᵗ] = [2·eᵗ - 2t·eᵗ]/(e²ᵗ) = 2eᵗ(1 - t)/(e²ᵗ) = 2(1 - t)/eᵗ. Finally, dividing by dx/dt = eᵗ gives d²y/dx² = [2(1 - t)/eᵗ]/eᵗ = 2(1 - t)/e²ᵗ. Choice D incorrectly has (t - 1) instead of (1 - t) in the numerator, which is a common sign error. Remember to carefully apply the quotient rule and maintain proper sign conventions throughout the calculation.

Question 8

For what interval of tt is the curve defined by the parametric equations x(t)=t3−3tx(t) = t^3 - 3t and y(t)=t2y(t) = t^2 concave up?

  1. t<−1t < -1 or t>1t > 1
  2. −1<t<1-1 < t < 1 (correct answer)
  3. t>0t > 0
  4. t<0t < 0
Explanation: First, find the derivatives: dxdt=3t2−3\frac{dx}{dt} = 3t^2 - 3 and dydt=2t\frac{dy}{dt} = 2t. Then, dydx=2t3t2−3\frac{dy}{dx} = \frac{2t}{3t^2-3}. Next, differentiate dydx\frac{dy}{dx} with respect to tt: ddt(dydx)=2(3t2−3)−2t(6t)(3t2−3)2=6t2−6−12t2(3(t2−1))2=−6t2−69(t2−1)2=−6(t2+1)9(t2−1)2\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2(3t^2-3) - 2t(6t)}{(3t^2-3)^2} = \frac{6t^2-6-12t^2}{(3(t^2-1))^2} = \frac{-6t^2-6}{9(t^2-1)^2} = \frac{-6(t^2+1)}{9(t^2-1)^2}. Finally, d2ydx2=d/dt(dy/dx)dx/dt=−6(t2+1)/[9(t2−1)2]3(t2−1)=−6(t2+1)27(t2−1)3\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{-6(t^2+1)/[9(t^2-1)^2]}{3(t^2-1)} = \frac{-6(t^2+1)}{27(t^2-1)^3}. For the curve to be concave up, d2ydx2>0\frac{d^2y}{dx^2} > 0. Since the numerator −6(t2+1)-6(t^2+1) is always negative, the denominator 27(t2−1)327(t^2-1)^3 must be negative. This occurs when t2−1<0t^2-1 < 0, which means t2<1t^2 < 1, so −1<t<1-1 < t < 1.

Question 9

A particle is moving in the xy-plane with position given by parametric equations (x(t),y(t))(x(t), y(t)). At time t=2t=2, it is known that dxdt=3\frac{dx}{dt} = 3, dydt=−1\frac{dy}{dt} = -1, d2xdt2=0\frac{d^2x}{dt^2} = 0, and d2ydt2=4\frac{d^2y}{dt^2} = 4. What is the value of d2ydx2\frac{d^2y}{dx^2} at t=2t=2?

  1. 40\frac{4}{0}, which is undefined
  2. 49\frac{4}{9} (correct answer)
  3. 43\frac{4}{3}
  4. 44
Explanation: The formula for the second derivative is d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}. First, we find the derivative of dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} with respect to tt using the quotient rule: ddt(dydx)=d2ydt2dxdt−dydtd2xdt2(dxdt)2\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{\frac{d^2y}{dt^2}\frac{dx}{dt} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^2}. At t=2t=2, this is (4)(3)−(−1)(0)(3)2=129=43\frac{(4)(3) - (-1)(0)}{(3)^2} = \frac{12}{9} = \frac{4}{3}. Now, we can find the second derivative: d2ydx2=4/3dx/dt=4/33=49\frac{d^2y}{dx^2} = \frac{4/3}{dx/dt} = \frac{4/3}{3} = \frac{4}{9}.

Question 10

A curve is defined by parametric equations x=f(t)x = f(t) and y=g(t)y = g(t), where ff and gg are twice-differentiable functions. Which of the following gives an expression for d2ydx2\frac{d^2y}{dx^2}?

  1. g′′(t)f′′(t)\frac{g''(t)}{f''(t)}
  2. g′′(t)f′(t)−g′(t)f′′(t)(f′(t))2\frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^2}
  3. g′′(t)f′(t)\frac{g''(t)}{f'(t)}
  4. g′′(t)f′(t)−g′(t)f′′(t)(f′(t))3\frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^3} (correct answer)
Explanation: The first derivative is dydx=g′(t)f′(t)\frac{dy}{dx} = \frac{g'(t)}{f'(t)}. To find the second derivative, we must differentiate dydx\frac{dy}{dx} with respect to tt and divide by dxdt=f′(t)\frac{dx}{dt} = f'(t). Using the quotient rule, ddt(g′(t)f′(t))=g′′(t)f′(t)−g′(t)f′′(t)(f′(t))2\frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) = \frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^2}. Dividing this by f′(t)f'(t) gives d2ydx2=g′′(t)f′(t)−g′(t)f′′(t)(f′(t))3\frac{d^2y}{dx^2} = \frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^3}.

Question 11

Let C be a curve defined by parametric equations x(t)x(t) and y(t)y(t). At a certain time t0t_0, it is known that dxdt=−2\frac{dx}{dt} = -2, dydt=4\frac{dy}{dt} = 4, and ddt(dydx)=3\frac{d}{dt}\left(\frac{dy}{dx}\right) = 3. What is the value of d2ydx2\frac{d^2y}{dx^2} at t=t0t=t_0?

  1. 33
  2. 43\frac{4}{3}
  3. −32-\frac{3}{2} (correct answer)
  4. −23-\frac{2}{3}
Explanation: The formula for the second derivative of a parametric curve is d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}. We are given all the necessary values in the problem statement. Plugging them in, we get d2ydx2=3−2=−32\frac{d^2y}{dx^2} = \frac{3}{-2} = -\frac{3}{2}.

Question 12

A curve is defined by the parametric equations x(t)=t2+tx(t) = t^2 + t and y(t)=t4+t2y(t) = t^4 + t^2. What is the value of d2ydx2\frac{d^2y}{dx^2} at the point where t=1t=1?

  1. 103\frac{10}{3}
  2. 77
  3. 109\frac{10}{9} (correct answer)
  4. 59\frac{5}{9}
Explanation: First, find derivatives with respect to tt: dxdt=2t+1\frac{dx}{dt} = 2t+1 and dydt=4t3+2t\frac{dy}{dt} = 4t^3+2t. Then, dydx=4t3+2t2t+1\frac{dy}{dx} = \frac{4t^3+2t}{2t+1}. Now, differentiate dydx\frac{dy}{dx} with respect to tt: ddt(dydx)=(12t2+2)(2t+1)−(4t3+2t)(2)(2t+1)2\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(12t^2+2)(2t+1) - (4t^3+2t)(2)}{(2t+1)^2}. At t=1t=1, dxdt=3\frac{dx}{dt} = 3 and ddt(dydx)=(14)(3)−(6)(2)32=42−129=309=103\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(14)(3) - (6)(2)}{3^2} = \frac{42-12}{9} = \frac{30}{9} = \frac{10}{3}. Finally, d2ydx2=d/dt(dy/dx)dx/dt=10/33=109\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{10/3}{3} = \frac{10}{9}.

Question 13

For the curve defined by the parametric equations x(t)=12t2x(t) = \frac{1}{2}t^2 and y(t)=14t4−43t3y(t) = \frac{1}{4}t^4 - \frac{4}{3}t^3, what is the value of tt for which d2ydx2=0\frac{d^2y}{dx^2} = 0?

  1. t=0t=0
  2. t=4t=4
  3. t=2t=2 (correct answer)
  4. There is no such value of tt.
Explanation: First, find the derivatives with respect to tt: dxdt=t\frac{dx}{dt} = t and dydt=t3−4t2\frac{dy}{dt} = t^3 - 4t^2. Then, dydx=t3−4t2t=t2−4t\frac{dy}{dx} = \frac{t^3-4t^2}{t} = t^2 - 4t for t≠0t \neq 0. Next, differentiate dydx\frac{dy}{dx} with respect to tt: ddt(t2−4t)=2t−4\frac{d}{dt}(t^2 - 4t) = 2t - 4. Finally, d2ydx2=2t−4t\frac{d^2y}{dx^2} = \frac{2t-4}{t}. Setting this equal to zero gives 2t−4=02t-4=0, so t=2t=2.

Question 14

For x(t)=ln⁡tx(t)=\ln t and y(t)=t3y(t)=t^3, what is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

  1. 9t29t^2
  2. 3t23t^2
  3. 6t36t^3
  4. 9t39t^3 (correct answer)
  5. 6t26t^2
Explanation: Finding the second derivative of a parametric curve requires d2ydx2=ddt(dy/dtdx/dt)dx/dt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt}. We have dy/dt=3t2\frac{dy/dt} = 3t^2 and dx/dt=1t\frac{dx/dt} = \frac{1}{t}, giving dy/dx=3t21/t=3t3\frac{dy/dx} = \frac{3t^2}{1/t} = 3t^3. Differentiating: ddt(3t3)=9t2\frac{d}{dt}(3t^3) = 9t^2. Therefore, d2ydx2=9t21/t=9t2⋅t=9t3\frac{d^2y}{dx^2} = \frac{9t^2}{1/t} = 9t^2 \cdot t = 9t^3. Choice A shows 9t29t^2, which is the derivative of \frac{dy/dx} but forgets to divide by \frac{dx/dt}. Remember that the parametric second derivative formula requires dividing by \frac{dx/dt} at the end to convert from tt-differentiation to xx-differentiation.

Question 15

For x(t)=ln⁡tx(t)=\ln t and y(t)=t2y(t)=t^2 with t>0t>0, what is d2ydx2\dfrac{d^2y}{dx^2}?​

  1. 2t22t^2
  2. 2t2\dfrac{2}{t^2}
  3. 4t24t^2 (correct answer)
  4. 2t2t
  5. 4t4t
Explanation: This problem involves finding the second derivative for a logarithmic-polynomial parametric curve. Given x(t)=ln⁡tx(t) = \ln t and y(t)=t2y(t) = t^2 with t>0t > 0, we have dxdt=1t\frac{dx}{dt} = \frac{1}{t} and dydt=2t\frac{dy}{dt} = 2t. The first derivative is dydx=2t1/t=2t2\frac{dy}{dx} = \frac{2t}{1/t} = 2t^2. To find the second derivative: ddt(2t2)=4t\frac{d}{dt}(2t^2) = 4t, so d2ydx2=4t1/t=4t2\frac{d^2y}{dx^2} = \frac{4t}{1/t} = 4t^2. Choice B (2t2\frac{2}{t^2}) incorrectly inverts the relationship between tt and the derivative. The key insight for logarithmic parametric curves is that dxdt=1t\frac{dx}{dt} = \frac{1}{t} often leads to powers of tt in the second derivative.

Question 16

If x(t)=etx(t)=e^t and y(t)=tety(t)=t e^t, what is d2ydx2\dfrac{d^2y}{dx^2} expressed in terms of tt?

  1. 1et\dfrac{1}{e^t} (correct answer)
  2. tet\dfrac{t}{e^t}
  3. t+1et\dfrac{t+1}{e^t}
  4. 1tet\dfrac{1}{t e^t}
  5. t+1e2t\dfrac{t+1}{e^{2t}}
Explanation: This problem tests the skill of finding second derivatives of parametric equations. The formula for the second derivative is \frac{d^2 y}{dx^2} = \frac{ \frac{d}{dt} \left( dy/dtdx/dt\frac{dy/dt}{dx/dt} \right) }{ dx/dt }. Here, dx/dt = e^t and dy/dt = e^t + t e^t = e^t (t+1), so \frac{dy}{dx} = t+1. Differentiating this with respect to t gives 1. Dividing by dx/dt yields 1 / e^t. A tempting distractor is choice C, (t+1)/e^t, which is the first derivative dy/dx, but we need the second. A transferable strategy for computing parametric second derivatives is to consistently apply the formula \frac{d}{dt}\left( dydx\frac{dy}{dx} \right) / \frac{dx}{dt} and verify with specific values if possible.

Question 17

For x(t)=t2+1x(t)=t^2+1 and y(t)=1ty(t)=\dfrac{1}{t}, t≠0t\neq 0, what is d2ydx2\dfrac{d^2y}{dx^2}?

  1. 34t5\dfrac{3}{4t^5} (correct answer)
  2. 32t4\dfrac{3}{2t^4}
  3. 3t5\dfrac{3}{t^5}
  4. −34t5-\dfrac{3}{4t^5}
  5. −34t4-\dfrac{3}{4t^4}
Explanation: The skill here is finding the second derivative of parametric equations. The formula for d²y/dx² is the derivative with respect to t of dy/dx divided by dx/dt. Compute dy/dt = -1/t² and dx/dt = 2t, so dy/dx = -1/(2t³). Differentiate dy/dx to get 3/(2t⁴). Divide by 2t to obtain 3/(4t⁵). A tempting distractor is -3/(4t⁵), which keeps a negative sign from the first derivative without accounting for the positive from differentiation. A transferable strategy for second derivatives is to track signs meticulously through each step to determine concavity accurately.

Question 18

A curve is parameterized by x=etx=e^t and y=e−ty=e^{-t}. What is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

  1. 2et\dfrac{2}{e^t}
  2. 2e3t\dfrac{2}{e^{3t}} (correct answer)
  3. −2e3t-\dfrac{2}{e^{3t}}
  4. 1e2t\dfrac{1}{e^{2t}}
  5. −1e2t-\dfrac{1}{e^{2t}}
Explanation: The skill here is finding the second derivative of parametric equations. The formula for d²y/dx² is the derivative with respect to t of dy/dx, divided by dx/dt. First, compute dy/dx = -e^{-2t}. Then, differentiate this to get 2e^{-2t}. Finally, divide by dx/dt = e^t to obtain 2/e^{3t}. A tempting distractor like -2/e^{3t} fails because it incorrectly retains a negative sign from the first derivative without accounting for the double negative in differentiation. When computing second derivatives for parametric equations, always ensure you apply the chain rule correctly by dividing the derivative of the first derivative by dx/dt.

Question 19

With x(t)=etx(t)=e^t and y(t)=tety(t)=t e^t, what is d2ydx2\dfrac{d^2 y}{dx^2} as a function of tt?

  1. 1et\dfrac{1}{e^t} (correct answer)
  2. ete^t
  3. 1e2t\dfrac{1}{e^{2t}}
  4. tet\dfrac{t}{e^t}
  5. t+1et\dfrac{t+1}{e^t}
Explanation: This problem tests the skill of finding the second derivative of parametric equations. The formula for d2y/dx2d^2 y / dx^2 is (x′y′′−y′x′′)/(x′)3(x' y'' - y' x'') / (x')^3, where primes denote derivatives with respect to t. For x(t)=etx(t) = e^t and y(t)=tety(t) = t e^t, compute x′=etx' = e^t, x′′=etx'' = e^t, y′=et+tet=et(t+1)y' = e^t + t e^t = e^t(t + 1), and y′′=et(t+1)+et=et(t+2)y'' = e^t(t + 1) + e^t = e^t(t + 2). Substituting gives [et⋅et(t+2)−et(t+1)⋅et]/(et)3=[e2t(t+2)−e2t(t+1)]/e3t=e2t/e3t=1/et[e^t \cdot e^t(t + 2) - e^t(t + 1) \cdot e^t] / (e^t)^3 = [e^{2t}(t + 2) - e^{2t}(t + 1)] / e^{3t} = e^{2t} / e^{3t} = 1/e^t. A tempting distractor like ete^t fails because it might result from incorrectly inverting the denominator. A transferable strategy for second derivatives is to use the chain rule carefully when dealing with exponential functions.

Question 20

For x=etx=e^t and y=tety=te^t, what is d2ydx2\dfrac{d^2y}{dx^2} in terms of tt?

  1. t+1et\dfrac{t+1}{e^t}
  2. 1et\dfrac{1}{e^t} (correct answer)
  3. tet\dfrac{t}{e^t}
  4. t+2et\dfrac{t+2}{e^t}
  5. 1e2t\dfrac{1}{e^{2t}}
Explanation: This problem involves finding the second derivative for exponential parametric equations. Using the parametric second derivative formula, we start with dxdt=et\frac{dx}{dt} = e^t and dydt=et+tet=et(1+t)\frac{dy}{dt} = e^t + te^t = e^t(1+t). The first derivative is dydx=et(1+t)et=\frac{dy}{dx} = \frac{e^t(1+t)}{e^t} = 1 + t.Differentiatingwithrespectto. Differentiating with respect to tgivesgives\frac{d}{dt}(1+t) = 1.Finally,. Finally, \frac{d^2y}{dx^2} = \frac{1}{e^t}.ChoiceAincorrectlyincludesanextra. Choice A incorrectly includes an extra tterm,likelyfrommisapplyingtheproductrule.Rememberthatwhenthefirstderivativesimplifiesnicely(heretoterm, likely from misapplying the product rule. Remember that when the first derivative simplifies nicely (here to1+t$), the second derivative calculation becomes much simpler.