First, find the derivatives: $$\frac{dx}{dt} = e^t$$ and $$\frac{dy}{dt} = 3t^2-3$$. Then $$\frac{dy}{dx} = \frac{3t^2-3}{e^t}$$. Differentiating with respect to $$t$$: $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{6t e^t - (3t^2-3)e^t}{(e^t)^2} = \frac{-3t^2+6t+3}{e^t}$$. The second derivative is $$\frac{d^2y}{dx^2} = \frac{(-3t^2+6t+3)/e^t}{e^t} = \frac{-3(t^2-2t-1)}{e^{2t}}$$. For the curve to be concave up, we need $$\frac{d^2y}{dx^2} > 0$$. Since $$e^{2t}$$ is always positive, we need $$-3(t^2-2t-1) > 0$$, which simplifies to $$t^2-2t-1 < 0$$. The roots of $$t^2-2t-1=0$$ are $$t = 1 \pm \sqrt{2}$$. Since the parabola $$z=t^2-2t-1$$ opens upward, it is negative between its roots. Therefore, the curve is concave up for $$1-\sqrt{2} < t < 1+\sqrt{2}$$.

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 $$\frac{-(1+2)}{e^{2(1)}(1+1)^2} = \frac{-3}{4e^2}$$.

First, find the derivatives: $$\frac{dx}{dt} = 3t^2 - 3$$ and $$\frac{dy}{dt} = 2t$$. Then, $$\frac{dy}{dx} = \frac{2t}{3t^2-3}$$. Next, differentiate $$\frac{dy}{dx}$$ with respect to $$t$$: $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2(3t^2-3) - 2t(6t)}{(3t^2-3)^2} = \frac{6t^2-6-12t^2}{(3(t^2-1))^2} = \frac{-6t^2-6}{9(t^2-1)^2} = \frac{-6(t^2+1)}{9(t^2-1)^2}$$. Finally, $$\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} = \frac{-6(t^2+1)/[9(t^2-1)^2]}{3(t^2-1)} = \frac{-6(t^2+1)}{27(t^2-1)^3}$$. For the curve to be concave up, $$\frac{d^2y}{dx^2} > 0$$. Since the numerator $$-6(t^2+1)$$ is always negative, the denominator $$27(t^2-1)^3$$ must be negative. This occurs when $$t^2-1 < 0$$, which means $$t^2 < 1$$, so $$-1 < t < 1$$.

First, find derivatives: $$\frac{dx}{dt} = 1 - \cos(t)$$ and $$\frac{dy}{dt} = \sin(t)$$. Then, $$\frac{dy}{dx} = \frac{\sin(t)}{1-\cos(t)}$$. Using trigonometric identities, this simplifies to $$\cot(t/2)$$. Now, $$\frac{d}{dt}(\cot(t/2)) = -\frac{1}{2}\csc^2(t/2)$$. The second derivative is $$\frac{d^2y}{dx^2} = \frac{-\frac{1}{2}\csc^2(t/2)}{1-\cos(t)} = \frac{-1}{2\sin^2(t/2)(1-\cos t)} = \frac{-1}{2\sin^2(t/2)(2\sin^2(t/2))} = \frac{-1}{4\sin^4(t/2)}$$. For $$0 < t < 2\pi$$, $$t/2$$ is in $$(0, \pi)$$, so $$\sin(t/2) \neq 0$$. Thus, $$\sin^4(t/2)$$ is always positive. The entire expression for $$\frac{d^2y}{dx^2}$$ is therefore always negative. This means the path is always concave down.

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.

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 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.

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.

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 $\frac{dx}{dt}$.