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 = 2t $ and $ dy/dt = \frac{1}{2} t^{-1/2} $, so $ \frac{dy}{dx} = \frac{ \frac{1}{2} t^{-1/2} }{ 2t } = \frac{1}{4 t^{3/2}} $. Differentiating this with respect to t gives $ \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 $ [-\frac{3}{8 t^{5/2}}] / 2t = -\frac{3}{16 t^{7/2}} $. A tempting distractor is choice C, $ -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 $ \frac{d}{dt}\left( \frac{dy}{dx} \right) / \frac{dx}{dt} $ and verify with specific values if possible.

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.

For this parametric curve, we need to find $\frac{d^2y}{dx^2}$ using the formula $\frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}$. We have $\frac{dy}{dt} = \sec t \tan t$ and $\frac{dx}{dt} = \sec^2 t$, giving $\frac{dy}{dx} = \frac{\sec t \tan t}{\sec^2 t} = \frac{\tan t}{\sec t} = \frac{\sin t}{\cos t} \cdot \cos t = \sin t$. Differentiating with respect to $t$: $\frac{d}{dt}[\sin t] = \cos t$. Finally, dividing by $\frac{dx}{dt} = \sec^2 t$ yields $\frac{d^2y}{dx^2} = \frac{\cos t}{\sec^2 t} = \cos t \cdot \cos^2 t = \cos^3 t = \frac{1}{\sec t}$. Choice A incorrectly applies the quotient rule to the original ratio instead of first simplifying $dy/dx$. Simplifying $dy/dx$ before differentiating often makes parametric second derivative calculations much cleaner.

To find the second derivative of this parametric curve, we use $\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy/dt}{dx/dt}) \cdot \frac{1}{dx/dt}$. 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 $t$: $\frac{d}{dt}[1+t] = 1$. Finally, dividing by $\frac{dx}{dt} = e^t$ yields $\frac{d^2y}{dx^2} = \frac{1}{e^t}$. Choice B incorrectly includes $(t+1)$ in the numerator, confusing the first derivative expression with the second derivative. Remember that when $dy/dx$ simplifies to a function of $t$ alone, its derivative is straightforward before the final division by $dx/dt$.

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 $\frac{dx}{dt}$.

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.

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.

Finding the second derivative of a parametric curve requires $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt}$. We have $\frac{dy/dt} = 3t^2$ and $\frac{dx/dt} = \frac{1}{t}$, giving $\frac{dy/dx} = \frac{3t^2}{1/t} = 3t^3$. Differentiating: $\frac{d}{dt}(3t^3) = 9t^2$. Therefore, $\frac{d^2y}{dx^2} = \frac{9t^2}{1/t} = 9t^2 \cdot t = 9t^3$. Choice A shows $9t^2$, 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 $t$-differentiation to $x$-differentiation.

This problem involves finding the second derivative for a logarithmic-polynomial parametric curve. Given $x(t) = \ln t$ and $y(t) = t^2$ with $t > 0$, we have $\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = 2t$. The first derivative is $\frac{dy}{dx} = \frac{2t}{1/t} = 2t^2$. To find the second derivative: $\frac{d}{dt}(2t^2) = 4t$, so $\frac{d^2y}{dx^2} = \frac{4t}{1/t} = 4t^2$. Choice B ($\frac{2}{t^2}$) incorrectly inverts the relationship between $t$ and the derivative. The key insight for logarithmic parametric curves is that $\frac{dx}{dt} = \frac{1}{t}$ often leads to powers of $t$ in the second derivative.

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.