$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$
-$$\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } + 1 ) \mathbf { i } + ( \ln ( \sec t ) - 9 ) \mathbf { j } + 9 \mathbf { k } , - \pi / 2

Question

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$
-$$\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } + 1 ) \mathbf { i } + ( \ln ( \sec t ) - 9 ) \mathbf { j } + 9 \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi / 2$$
A) $\mathbf { a } = ( \cos t ) \mathbf { T } + ( \cos t ) \mathbf { N }$
B) $a = \left( \sec ^ { 2 } t \right) T + ( \cos t ) \mathbf { N }$
C) $\mathbf { a } = ( \csc t ) \mathbf { T } + ( \sec t ) \mathbf { N }$
D) $\mathbf { a } = ( \sec t \tan t ) \mathbf { T } + ( \sec t ) \mathbf { N }$

Quizplus · Accepted Answer

We start by finding the velocity and acceleration vectors:

$$\mathbf{r}(t)=(t+1)\mathbf{i}+(\ln(\sec t)-9)\mathbf{j}+9\mathbf{k}$$

$$\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=\mathbf{i}+\frac{\sec t \tan t}{\sec t}\mathbf{j}+0\mathbf{k}=\mathbf{i}+\tan t\mathbf{j}$$

$$\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=\frac{d}{dt}\left(\mathbf{i}+\tan t\mathbf{j}\right)=0\mathbf{i}+\sec^2t\mathbf{j}+\tan t\sec t\mathbf{k}$$

Now we express the acceleration vector in terms of $\mathbf{T}$ and $\mathbf{N}$. We know that $\mathbf{T}=\frac{\mathbf{v}}{\left|\mathbf{v}\right|}$, so we need to find $\left|\mathbf{v}\right|$:

$$\left|\mathbf{v}\right|=\left|\mathbf{i}+\tan t\mathbf{j}\right|=\sqrt{1+\tan^2t}=\sec t$$

Now we can find $\mathbf{T}$:

$$\mathbf{T}=\frac{\mathbf{v}}{\left|\mathbf{v}\right|}=\frac{\mathbf{i}+\tan t\mathbf{j}}{\sec t}=\cos t\mathbf{i}+\sin t\mathbf{j}$$

To find $\mathbf{N}$, we take the derivative of $\mathbf{T}$ with respect to $t$:

$$\frac{d\mathbf{T}}{dt}=-\sin t\mathbf{i}+\cos t\mathbf{j}$$

And then we normalize it:

$$\mathbf{N}=\frac{\frac{d\mathbf{T}}{dt}}{\left|\frac{d\mathbf{T}}{dt}\right|}=-\sin t\mathbf{i}+\cos t\mathbf{j}$$

So the acceleration vector can be expressed as:

$$\mathbf{a}=\left(\sec t\tan t\cos t\right)\mathbf{T}+\left(\sec^2t\sin t\right)\mathbf{N}$$

Comparing with the answer choices, we see that the only option that matches is D.

For the curve r(t), write the acceleration in the form aTT + anN. \text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. } For the curve r(t), write the acceleration in the form aTT + anN.

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$