$\text { Find } a _ { \mathbf { T } } \text { at time } t = 0 \text { for the plane curve } \mathbf { r } ( t ) = e ^ { 2 t } \mathbf { i } + e ^ { - 7 t } \mathbf { j } \text {. Round your answer to }$ three decimal places.
A)-40.625
B)-39.209
C)-111.667
D)-46.016
E)-43.988

Question

Quizplus · Accepted Answer

To find the acceleration vector $a_T$ at time $t = 0$ for the given plane curve $\mathbf{r}(t) = e^{2t}\mathbf{i} + e^{-7t}\mathbf{j}$, we first need to find the velocity vector $\mathbf{v}(t)$ by differentiating $\mathbf{r}(t)$ with respect to $t$, and then find the acceleration vector $\mathbf{a}(t)$ by differentiating $\mathbf{v}(t)$ with respect to $t$. Finally, we find the tangential component of the acceleration, $a_T$, at $t = 0$.1. Find the velocity vector $\mathbf{v}(t)$:$$\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt}(e^{2t}\mathbf{i} + e^{-7t}\mathbf{j}) = 2e^{2t}\mathbf{i} - 7e^{-7t}\mathbf{j}$$2. Find the acceleration vector $\mathbf{a}(t)$:$$\mathbf{a}(t) = \frac{d}{dt}\mathbf{v}(t) = \frac{d}{dt}(2e^{2t}\mathbf{i} - 7e^{-7t}\mathbf{j}) = 4e^{2t}\mathbf{i} + 49e^{-7t}\mathbf{j}$$3. Evaluate $\mathbf{a}(t)$ at $t = 0$:$$\mathbf{a}(0) = 4e^{2(0)}\mathbf{i} + 49e^{-7(0)}\mathbf{j} = 4\mathbf{i} + 49\mathbf{j}$$4. Find the magnitude of $\mathbf{a}(0)$ to get $a_T$ at $t = 0$:$$a_T = |\mathbf{a}(0)| = \sqrt{4^2 + 49^2} = \sqrt{16 + 2401} = \sqrt{2417} \approx 49.162$$It appears there has been a mistake in the calculation or interpretation of the question as none of the provided options match the calculated result. The correct approach involves finding the velocity and acceleration vectors, and then calculating the magnitude of the acceleration vector at $t = 0$. However, the calculation provided does not align with the options given, suggesting a possible misunderstanding of the question's requirements or a mistake in the calculation process.

Three Decimal Places