Second Derivatives of Parametric Equations - AP Calculus BC

Let C be a curve defined by parametric equations $x(t)$ and $y(t)$ . At a certain time $t_0$ , it is known that $\frac{dx}{dt} = -2$ , $\frac{dy}{dt} = 4$ , and $\frac{d}{dt}\left(\frac{dy}{dx}\right) = 3$ . What is the value of $\frac{d^2y}{dx^2}$ at $t=t_0$ ?
Answer: $-\frac{3}{2}$ The formula for the second derivative of a parametric curve is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$ . We are given all the necessary values in the problem statement. Plugging them in, we get $\frac{d^2y}{dx^2} = \frac{3}{-2} = -\frac{3}{2}$ .
The position of a particle is given by the parametric equations $x(t) = t^3$ and $y(t) = t^2$ . Which of the following is the correct expression for $\frac{d^2y}{dx^2}$ ?
Answer: $\frac{-2}{9t^4}$ First, find the derivatives with respect to $t$ : $\frac{dx}{dt} = 3t^2$ and $\frac{dy}{dt} = 2t$ . Then, $\frac{dy}{dx} = \frac{2t}{3t^2} = \frac{2}{3t}$ . Next, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}\left(\frac{2}{3t}\right) = -\frac{2}{3t^2}$ . Finally, find the second derivative: $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{-2/(3t^2)}{3t^2} = -\frac{2}{9t^4}$ .
If $x(t)=\sqrt{t}$ and $y(t)=t^2$ with $t>0$ , what is $\dfrac{d^2y}{dx^2}$ ?
Answer: $12t$ This problem requires finding the second derivative using $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}$ . We have $\frac{dy}{dt} = 2t$ and $\frac{dx}{dt} = \frac{1}{2\sqrt{t}} = \frac{1}{2}t^{-1/2}$ , so $\frac{dy}{dx} = \frac{2t}{\frac{1}{2}t^{-1/2}} = 4t^{3/2}$ . Differentiating with respect to $t$ : $\frac{d}{dt}[4t^{3/2}] = 4 \cdot \frac{3}{2}t^{1/2} = 6t^{1/2}$ . Finally, dividing by $\frac{dx}{dt} = \frac{1}{2}t^{-1/2}$ gives $\frac{d^2y}{dx^2} = 6t^{1/2} \cdot 2t^{1/2} = 12t$ . Choice C incorrectly uses $6t$ instead of $12t$ , likely from forgetting to multiply by the reciprocal of $dx/dt$ . When $dx/dt$ involves fractional powers, be extra careful with the algebra in the final division step.
For $x=e^t$ and $y=te^t$ , what is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ?
Answer: $\dfrac{1}{e^t}$ This problem involves finding the second derivative for exponential parametric equations. Using the parametric second derivative formula, we start with $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = e^t + te^t = e^t(1+t)$ . The first derivative is $\frac{dy}{dx} = \frac{e^t(1+t)}{e^t} = $1 + t$ . Differentiating with respect to $t$ gives $\frac{d}{dt}(1+t) = 1$ . Finally, $\frac{d^2y}{dx^2} = \frac{1}{e^t}$ . Choice A incorrectly includes an extra $t$ term, likely from misapplying the product rule. Remember that when the first derivative simplifies nicely (here to $1+t$ ), the second derivative calculation becomes much simpler.
A particle moves in the xy-plane with position given by $x(t) = \cos(t^2)$ and $y(t) = \sin(t^2)$ for $t \ge 0$ . What is the value of $\frac{d^2y}{dx^2}$ at $t=\frac{\sqrt{\pi}}{2}$ ?
Answer: $-2\sqrt{2}$ First, find the derivatives with respect to $t$ : $\frac{dx}{dt} = -2t\sin(t^2)$ and $\frac{dy}{dt} = 2t\cos(t^2)$ . Then, $\frac{dy}{dx} = \frac{2t\cos(t^2)}{-2t\sin(t^2)} = -\cot(t^2)$ . Next, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}(-\cot(t^2)) = -(-\csc^2(t^2)) \cdot 2t = 2t\csc^2(t^2)$ . Finally, $\frac{d^2y}{dx^2} = \frac{2t\csc^2(t^2)}{-2t\sin(t^2)} = -\frac{\csc^2(t^2)}{\sin(t^2)} = -\frac{1}{\sin^3(t^2)}$ . At $t=\frac{\sqrt{\pi}}{2}$ , we have $t^2 = \frac{\pi}{4}$ . The value is $\frac{-1}{\sin^3(\pi/4)} = \frac{-1}{(\sqrt{2}/2)^3} = \frac{-1}{2\sqrt{2}/8} = -\frac{8}{2\sqrt{2}} = -2\sqrt{2}$ .
If $x(t)=e^t$ and $y(t)=t e^t$ , what is $\dfrac{d^2y}{dx^2}$ expressed in terms of $t$ ?
Answer: $\dfrac{1}{e^t}$ 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( $\frac{dy/dt}{dx/dt} \right$ ) }{ 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( $\frac{dy}{dx} \right$ ) / \frac{dx}{dt} and verify with specific values if possible.
A particle is moving in the xy-plane with position given by parametric equations $(x(t), y(t))$ . At time $t=2$ , it is known that $\frac{dx}{dt} = 3$ , $\frac{dy}{dt} = -1$ , $\frac{d^2x}{dt^2} = 0$ , and $\frac{d^2y}{dt^2} = 4$ . What is the value of $\frac{d^2y}{dx^2}$ at $t=2$ ?
Answer: $\frac{4}{9}$ The formula for the second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$ . First, we find the derivative of $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ with respect to $t$ using the quotient rule: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{\frac{d^2y}{dt^2}\frac{dx}{dt} - \frac{dy}{dt}\frac{d^2x}{dt^2}}{(\frac{dx}{dt})^2}$ . At $t=2$ , this is $\frac{(4)(3) - (-1)(0)}{(3)^2} = \frac{12}{9} = \frac{4}{3}$ . Now, we can find the second derivative: $\frac{d^2y}{dx^2} = \frac{4/3}{dx/dt} = \frac{4/3}{3} = \frac{4}{9}$ .
A curve is given by $x(t)=e^t$ and $y(t)=t^2$ . What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ?
Answer: $\dfrac{2-2t}{e^{2t}}$ This problem requires finding the second derivative of a parametric curve using $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt}$ . We have $\frac{dy/dt} = 2t$ and $\frac{dx/dt} = e^t$ , so $\frac{dy/dx} = \frac{2t}{e^t}$ . Differentiating using the quotient rule: $\frac{d}{dt}\left(\frac{2t}{e^t}\right) = \frac{2e^t - 2te^t}{e^{2t}} = \frac{2e^t(1-t)}{e^{2t}} = \frac{2(1-t)}{e^t}$ . Then $\frac{d^2y}{dx^2} = \frac{2(1-t)/e^t}{e^t} = \frac{2(1-t)}{e^{2t}} = \frac{2-2t}{e^{2t}}$ . Choice D shows $\frac{2}{e^t}$ , which omits the crucial $(1-t)$ factor from the quotient rule differentiation. When finding parametric second derivatives, carefully apply the quotient rule to $\frac{dy/dx}$ before dividing by $\frac{dx}{dt}$ .
For $x(t)=t^3$ and $y(t)=t^2+1$ , what is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ?
Answer: $\dfrac{-2}{9t^4}$ This problem tests the skill of finding second derivatives of parametric equations. The formula for the second derivative is $\frac{d^2 y}{dx^2} = \frac{ \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) }{ dx/dt }$ . Here, $ dx/dt = $3t^2$ $ and $ dy/dt = 2t $, so $\frac{dy}{dx} = 2t / 3t^2 = 2/(3t)$ . Differentiating this with respect to t gives $ -2/( $3t^2$ ) $. Dividing by dx/dt yields $ -2/( $3t^2$ ) / $3t^2$ = -2/( $9t^4$ ) $. A tempting distractor is choice C, $ -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 $\frac{d}{dt}\left( \frac{dy}{dx} \right) / \frac{dx}{dt}$ and verify with specific values if possible.
A curve is defined by the parametric equations $x = e^{2t}$ and $y = t^2$ . Which of the following is an expression for $\frac{d^2y}{dx^2}$ in terms of $x$ ?
Answer: $\frac{1 - \ln x}{2x^2}$ First, find derivatives with respect to $t$ : $\frac{dx}{dt} = 2e^{2t}$ and $\frac{dy}{dt} = 2t$ . Then $\frac{dy}{dx} = \frac{2t}{2e^{2t}} = te^{-2t}$ . Next, $\frac{d}{dt}(te^{-2t}) = 1 \cdot e^{-2t} + t(-2e^{-2t}) = e^{-2t}(1-2t)$ . Then $\frac{d^2y}{dx^2} = \frac{e^{-2t}(1-2t)}{2e^{2t}} = \frac{1-2t}{2e^{4t}}$ . We are given $x=e^{2t}$ , so $\ln x = 2t$ , which implies $t = \frac{1}{2}\ln x$ . Also, $e^{4t} = (e^{2t})^2 = x^2$ . Substituting these into the expression for the second derivative gives $\frac{d^2y}{dx^2} = \frac{1-2(\frac{1}{2}\ln x)}{2x^2} = \frac{1-\ln x}{2x^2}$ .
For the curve defined by the parametric equations $x(t) = \frac{1}{2}t^2$ and $y(t) = \frac{1}{4}t^4 - \frac{4}{3}t^3$ , what is the value of $t$ for which $\frac{d^2y}{dx^2} = 0$ ?
Answer: $t=2$ First, find the derivatives with respect to $t$ : $\frac{dx}{dt} = t$ and $\frac{dy}{dt} = t^3 - 4t^2$ . Then, $\frac{dy}{dx} = \frac{t^3-4t^2}{t} = t^2 - 4t$ for $t \neq 0$ . Next, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}(t^2 - 4t) = 2t - 4$ . Finally, $\frac{d^2y}{dx^2} = \frac{2t-4}{t}$ . Setting this equal to zero gives $2t-4=0$ , so $t=2$ .
A curve in the xy-plane is defined by the parametric equations $x(t) = \cos(t)$ and $y(t) = \sin(2t)$ . What is the value of $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{4}$ ?
Answer: $-8$ First, find the derivatives with respect to $t$ : $\frac{dx}{dt} = -\sin(t)$ and $\frac{dy}{dt} = 2\cos(2t)$ . Then, $\frac{dy}{dx} = \frac{2\cos(2t)}{-\sin(t)}$ . To find $\frac{d^2y}{dx^2}$ , we compute $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(-4\sin(2t))(-\sin(t)) - (2\cos(2t))(-\cos(t))}{(-\sin(t))^2} = \frac{4\sin(2t)\sin(t) + 2\cos(2t)\cos(t)}{\sin^2(t)}$ . At $t=\frac{\pi}{4}$ , $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{4\sin(\pi/2)\sin(\pi/4) + 2\cos(\pi/2)\cos(\pi/4)}{\sin^2(\pi/4)} = \frac{4(1)(\sqrt{2}/2) + 2(0)(\sqrt{2}/2)}{(\sqrt{2}/2)^2} = \frac{2\sqrt{2}}{1/2} = 4\sqrt{2}$ . Also at $t=\frac{\pi}{4}$ , $\frac{dx}{dt} = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$ . Therefore, $\frac{d^2y}{dx^2} = \frac{4\sqrt{2}}{-\sqrt{2}/2} = -8$ .
With $x(t)=e^t$ and $y(t)=t e^t$ , what is $\dfrac{d^2 y}{dx^2}$ as a function of $t$ ?
Answer: $\dfrac{1}{e^t}$ This problem tests the skill of finding the second derivative of parametric equations. The formula for $d^2 y / dx^2$ is $(x' y'' - y' x'') / (x')^3$ , where primes denote derivatives with respect to t. For $x(t) = e^t$ and $y(t) = t e^t$ , compute $x' = e^t$ , $x'' = e^t$ , $y' = e^t + t e^t = e^t(t + 1)$ , and $y'' = e^t(t + 1) + e^t = e^t(t + 2)$ . Substituting gives $[e^t \cdot e^t(t + 2) - e^t(t + 1) \cdot e^t] / (e^t)^3 = [e^{2t}(t + 2) - e^{2t}(t + 1)] / e^{3t} = e^{2t} / e^{3t} = 1/e^t$ . A tempting distractor like $e^t$ 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.
A curve is given by $x(t)=e^t$ and $y(t)=t e^t$ . What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ?
Answer: $\dfrac{1}{e^t}$ This problem involves finding the second derivative for exponential parametric equations. The parametric second derivative formula is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy/dt}{dx/dt})}{dx/dt}$ . First, $\frac{dy}{dt} = e^t + te^t = e^t(1+t)$ and $\frac{dx}{dt} = e^t$ , giving $\frac{dy}{dx} = \frac{e^t(1+t)}{e^t} = 1+t$ . Differentiating with respect to $t$ : $\frac{d}{dt}(1+t) = 1$ . Thus, $\frac{d^2y}{dx^2} = \frac{1}{e^t}$ . Choice C shows $\frac{t+1}{e^t}$ , which incorrectly includes the first derivative expression in the numerator instead of its derivative. When computing parametric second derivatives, remember to differentiate the simplified first derivative expression, not just substitute it into a formula.
A curve is defined by the parametric equations $x(t) = e^t$ and $y(t) = \ln(t+1)$ for $t > -1$ . What is the value of $\frac{d^2y}{dx^2}$ at $t=1$ ?
Answer: $\frac{-3}{4e^2}$ First, find the derivatives with respect to $t$ : $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = \frac{1}{t+1}$ . Then, $\frac{dy}{dx} = \frac{1/(t+1)}{e^t} = \frac{e^{-t}}{t+1}$ . Next, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{-e^{-t}(t+1) - e^{-t}(1)}{(t+1)^2} = \frac{-e^{-t}(t+2)}{(t+1)^2}$ . Then, $\frac{d^2y}{dx^2} = \frac{-e^{-t}(t+2)/(t+1)^2}{e^t} = \frac{-(t+2)}{e^{2t}(t+1)^2}$ . At $t=1$ , the value is $\frac{-(1+2)}{e^{2(1)}(1+1)^2} = \frac{-3}{4e^2}$ .
Given $x(t)=\ln t$ and $y(t)=t^2$ for $t>0$ , compute $\dfrac{d^2y}{dx^2}$ .
Answer: $4t^2$ 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.
A curve is defined by parametric equations $x = f(t)$ and $y = g(t)$ , where $f$ and $g$ are twice-differentiable functions. Which of the following gives an expression for $\frac{d^2y}{dx^2}$ ?
Answer: $\frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^3}$ The first derivative is $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ . To find the second derivative, we must differentiate $\frac{dy}{dx}$ with respect to $t$ and divide by $\frac{dx}{dt} = f'(t)$ . Using the quotient rule, $\frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) = \frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^2}$ . Dividing this by $f'(t)$ gives $\frac{d^2y}{dx^2} = \frac{g''(t)f'(t) - g'(t)f''(t)}{(f'(t))^3}$ .
For $x(t)=t^2+1$ and $y(t)=\dfrac{1}{t}$ , $t\neq 0$ , what is $\dfrac{d^2y}{dx^2}$ ?
Answer: $\dfrac{3}{4t^5}$ 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.
A curve is defined by the parametric equations $x(t) = t^2 + t$ and $y(t) = t^4 + t^2$ . What is the value of $\frac{d^2y}{dx^2}$ at the point where $t=1$ ?
Answer: $\frac{10}{9}$ First, find derivatives with respect to $t$ : $\frac{dx}{dt} = 2t+1$ and $\frac{dy}{dt} = 4t^3+2t$ . Then, $\frac{dy}{dx} = \frac{4t^3+2t}{2t+1}$ . Now, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(12t^2+2)(2t+1) - (4t^3+2t)(2)}{(2t+1)^2}$ . At $t=1$ , $\frac{dx}{dt} = 3$ and $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(14)(3) - (6)(2)}{3^2} = \frac{42-12}{9} = \frac{30}{9} = \frac{10}{3}$ . Finally, $\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{10/3}{3} = \frac{10}{9}$ .
A curve is defined by $x(t)=\ln t$ and $y(t)=t^2$ for $t>0$ . What is $\dfrac{d^2y}{dx^2}$ ?
Answer: $4t^2$ This problem involves finding the second derivative when one component is logarithmic. The parametric second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy/dt}{dx/dt})}{dx/dt}$ . We have $\frac{dy}{dt} = 2t$ and $\frac{dx}{dt} = \frac{1}{t}$ , giving $\frac{dy}{dx} = \frac{2t}{\frac{1}{t}} = 2t^2$ . Differentiating with respect to $t$ : $\frac{d}{dt}(2t^2) = 4t$ . Thus, $\frac{d^2y}{dx^2} = \frac{4t}{\frac{1}{t}} = 4t \cdot t = 4t^2$ . Choice D shows $2t$ , which appears to be the result of forgetting to divide by $\frac{dx}{dt}$ in the final step. When finding parametric second derivatives, always complete the division by $\frac{dx}{dt}$ to get the correct expression.

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