Calculate the arc length of the indicated portion of the curve r(t).
-$$\mathbf { r } ( \mathrm { t } ) = 4 \mathrm { ti } + \left( 6 \cos \frac { 1 } { 2 } \mathrm { t } \right) \mathbf { j } + \left( 6 \sin \frac { 1 } { 2 } \mathrm { t } \right) \mathbf { k } ; - 8 \leq \mathrm { t } \leq 1$$
A) -35
B) 45
C) -175
D) 225

Question

Quizplus · Accepted Answer

The length of a curve $\mathbf{r}(t)$ from $a$ to $b$ is given by:
$$\int_a^b ||\mathbf{r}'(t)|| dt$$
where $\mathbf{r}'(t)$ is the derivative of $\mathbf{r}(t)$ with respect to $t$.

Now, let's find $\mathbf{r}'(t)$:
$$\mathbf{r}'(t) = 4\mathbf{i} - 3\sin\frac{t}{2}\mathbf{j} + 3\cos\frac{t}{2}\mathbf{k}$$
So,
\begin{align*}
||\mathbf{r}'(t)|| &= ||4\mathbf{i} - 3\sin\frac{t}{2}\mathbf{j} + 3\cos\frac{t}{2}\mathbf{k}|| \
&= \sqrt{4^2 + \left(-3\sin\frac{t}{2}ight)^2 + \left(3\cos\frac{t}{2}ight)^2} \
&= \sqrt{16 + 9\sin^2\frac{t}{2} + 9\cos^2\frac{t}{2}} \
&= \sqrt{25} \
&= 5
\end{align*}

Therefore, the length of the curve from $-8$ to $1$ is:
$$\int_{-8}^1 ||\mathbf{r}'(t)|| dt = \int_{-8}^1 5 dt = 45$$

Hence, the answer is B.

Calculate the Arc Length of the Indicated Portion of the Curve