$\text { For the smooth curve } r ( t ) \text {, find the parametric equations for the line that is tangent to } r \text { at the given parameter value } t = t _ { 0 } \text {. }$
-$$\mathbf { r } ( \mathrm { t } ) = ( 9 \tan \mathrm { t } ) \mathbf { i } - ( 3 \sin \mathrm { t } ) \mathbf { j } + \left( 10 \cos ^ { 2 } \mathrm { t } \right) \mathbf { k } ; \mathrm { t } _ { 0 } = \frac { \pi } { 4 }$$
A) $x = 9 + 18 t , y = - \frac { 3 \sqrt { 2 } } { 2 } + \frac { 3 \sqrt { 2 } } { 2 } t , z = 5 + 10 t$
B) $x = 9 + 18 t , y = - \frac { 3 \sqrt { 2 } } { 2 } - \frac { 3 \sqrt { 2 } } { 2 } t , z = 5 - 10 t$
C) $x = 9 + 18 t , y = - \frac { 3 } { 2 } - \frac { 3 } { 2 } t , z = 10 - 10 t$
D) $x = 18 + 9 t , y = - \frac { 3 \sqrt { 2 } } { 2 } - \frac { 3 \sqrt { 2 } } { 2 } t , z = - 10 + 5 t$

Question

Quizplus · Accepted Answer

We need to find the derivative of the curve, $r(t)$, and evaluate it at $t_0 = \frac{\pi}{4}$ to get the direction vector of the tangent line.

$$r(t) = (9	an t)i - (3\sin t)j + (10\cos^2 t)k$$

$$r'(t) = 9\sec^2 t\;i - 3\cos t \;j -20\cos t\sin t\;k$$

evaluated at $t_0= \frac{\pi}{4}$

$$r'\left(\frac{\pi}{4}ight) = 9\sqrt{2}i - \frac{3}{\sqrt{2}}j - 10k$$

So the parametric equation of the tangent line passing through the point $r(\frac{\pi}{4})$ is given by

\begin{align*} x &= 9 + 18t \ y &= -\frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}t \ z &= 5 - 10t \end{align*}

which is choice $\boxed{	extbf{(B)}}$.

- A) x=9+18t,y=−322+322t,z=5+10tx = 9 + 18 t , y = - \frac { 3 \sqrt { 2 } } { 2 } + \frac { 3 \sqrt { 2 } } { 2 } t , z = 5 + 10 tx=9+18t,y=−232​​+232​​t,z=5+10t

- A) $x = 9 + 18 t , y = - \frac { 3 \sqrt { 2 } } { 2 } + \frac { 3 \sqrt { 2 } } { 2 } t , z = 5 + 10 t$