$\text { Find the arc length parameter along the curve from the point where } t = 0 \text { by evaluating } s = \int _ { 0 } ^ { t } | v ( \tau ) | d \tau \text {. }$
-r(t) = (5cos t)i + (5sin t)j + 2tk
A) $\frac { \sqrt { 29 } } { 2 } t$
B) $\sqrt { 29 } t$
C) $3 \sqrt { 6 } t$
D) $\sqrt { 33 } \mathrm { t }$

Question

$\text { Find the arc length parameter along the curve from the point where } t = 0 \text { by evaluating } s = \int _ { 0 } ^ { t } | v ( \tau ) | d \tau \text {. }$
-r(t) = (5cos t)i + (5sin t)j + 2tk 
A) $\frac { \sqrt { 29 } } { 2 } t$
B) $\sqrt { 29 } t$
C) $3 \sqrt { 6 } t$
D) $\sqrt { 33 } \mathrm { t }$

Quizplus · Accepted Answer

To find the arc length parameter $s$, we first need to find the velocity vector $v(t)$ by differentiating the position vector $r(t) = (5\cos t)i + (5\sin t)j + 2tk$. Differentiating each component with respect to $t$, we get $v(t) = (-5\sin t)i + (5\cos t)j + 2k$. The magnitude of $v(t)$ is $|v(t)| = \sqrt{(-5\sin t)^2 + (5\cos t)^2 + 2^2} = \sqrt{25\sin^2 t + 25\cos^2 t + 4} = \sqrt{25 + 4} = \sqrt{29}$. Therefore, the arc length parameter $s$ is given by $s = \int_{0}^{t} |v(\tau)| d\tau = \int_{0}^{t} \sqrt{29} d\tau = \sqrt{29}t$.

-R(t) = (5cos T)i + (5sin T)j + 2tk 292t\frac { \sqrt { 29 } } { 2 } t229​​t

-R(t) = (5cos T)i + (5sin T)j + 2tk $\frac { \sqrt { 29 } } { 2 } t$