Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Select the correct answer.
$$x = \cos t , y = 2 e ^ { 6 t } , z = 2 e ^ { - 6 t } ; ( 1,2,2 )$$
A) $x = t , y = 2 - 24 t , z = 2 + 24 t$
B) $x = 1 , y = 2 - 24 t , z = 4 - 24 t$
C) $x = 1 , y = 2 + 12 t , z = 2 - 12 t$
D) $x = 0 , y = 2 - 24 t , z = 2 + 24 t$
E) $x = t , y = 2 + 12 t , z = 2 - 12 t$

Question

Quizplus · Accepted Answer

The derivatives at the point (1,2,2) are $x'(t) = -\sin t$, $y'(t) = 12e^{6t}$, $z'(t) = -12e^{-6t}$. At $t=0$, these give the direction vector (0,12,-12). The tangent line is $x=1$, $y=2+12t$, $z=2-12t$.

Find Parametric Equations for the Tangent Line to the Curve x=cos⁡t,y=2e6t,z=2e−6t;(1,2,2)x = \cos t , y = 2 e ^ { 6 t } , z = 2 e ^ { - 6 t } ; ( 1,2,2 )x=cost,y=2e6t,z=2e−6t;(1,2,2)

Find Parametric Equations for the Tangent Line to the Curve $x = \cos t , y = 2 e ^ { 6 t } , z = 2 e ^ { - 6 t } ; ( 1,2,2 )$