Find the length of the indicated portion of the curve.
-$\mathbf { r } ( t ) = ( 4 \cos t ) \mathbf { i } + ( 4 \sin t ) \mathbf { j } + 5 t \mathbf { k } , 0 \leq t \leq \pi / 2$
A) $\frac { 3 t } { 2 } \pi$
B) $\frac { \sqrt { 41 } t } { 2 } \pi$
C) $\frac { \sqrt { 57 } \mathrm { t } } { 2 } \pi$
D) $\frac { \sqrt { 41 } t } { 4 } \pi$

Question

Quizplus · Accepted Answer

The length of a curve defined by a vector function $\mathbf{r}(t)$ from $t = a$ to $t = b$ is given by the integral $\int_{a}^{b} \|\mathbf{r}'(t)\| dt$. Differentiating $\mathbf{r}(t) = (4\cos t)\mathbf{i} + (4\sin t)\mathbf{j} + 5t\mathbf{k}$ with respect to $t$, we get $\mathbf{r}'(t) = (-4\sin t)\mathbf{i} + (4\cos t)\mathbf{j} + 5\mathbf{k}$. The magnitude of $\mathbf{r}'(t)$ is $\sqrt{(-4\sin t)^2 + (4\cos t)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$. Integrating from $0$ to $\pi/2$, the length is $\int_{0}^{\pi/2} \sqrt{41} dt = \sqrt{41}t |_{0}^{\pi/2} = \frac{\sqrt{41}\pi}{2}$.

Find the Length of the Indicated Portion of the Curve r(t)=(4cos⁡t)i+(4sin⁡t)j+5tk,0≤t≤π/2\mathbf { r } ( t ) = ( 4 \cos t ) \mathbf { i } + ( 4 \sin t ) \mathbf { j } + 5 t \mathbf { k } , 0 \leq t \leq \pi / 2r(t)=(4cost)i+(4sint)j+5tk,0≤t≤π/2

Find the Length of the Indicated Portion of the Curve $\mathbf { r } ( t ) = ( 4 \cos t ) \mathbf { i } + ( 4 \sin t ) \mathbf { j } + 5 t \mathbf { k } , 0 \leq t \leq \pi / 2$