$\text { For the smooth curve } r ( t ) \text {, find the parametric equations for the line that is tangent to } r \text { at the given parameter value } t = t _ { 0 } \text {. }$
-$$r ( t ) = ( 10 \sin t ) i - ( 3 \cos 3 t ) j + e ^ { - 8 t _ { k } } \mathbf { k } ; t _ { 0 } = 0$$
A) x = 10, y = -3t, z = -8 + t
B) x = 10t, y = -3, z = 1 - t
C) x = 10t, y = 3, z = 1 + t
D) x = 10t, y = -3, z = 1 - 8t

Question

Quizplus · Accepted Answer

The tangent line at a point on a curve is given by the derivative of the curve at that point. For the curve $r(t) = (10\sin t)i - (3\cos 3t)j + e^{-8t}k$, we differentiate each component with respect to $t$ to find the tangent vector at any point $t$. At $t_0 = 0$, we have:- $x(t) = 10\sin t$, so $dx/dt = 10\cos t$. At $t = 0$, $dx/dt = 10$.- $y(t) = -3\cos 3t$, so $dy/dt = 9\sin 3t$. At $t = 0$, $dy/dt = 0$.- $z(t) = e^{-8t}$, so $dz/dt = -8e^{-8t}$. At $t = 0$, $dz/dt = -8$.The point on the curve at $t_0 = 0$ is $r(0) = (10\sin 0)i - (3\cos 0)j + e^{-8\cdot0}k = 10i - 3j + 1k$.The parametric equations for the tangent line at $t_0 = 0$ are given by the point on the curve plus $t$ times the direction of the tangent vector:- $x = 10 + 10t$- $y = -3 + 0t = -3$- $z = 1 - 8t$Therefore, the correct answer is $x = 10t, y = -3, z = 1 - 8t$, which corresponds to choice D.

- r(t)=(10sin⁡t)i−(3cos⁡3t)j+e−8tkk;t0=0r ( t ) = ( 10 \sin t ) i - ( 3 \cos 3 t ) j + e ^ { - 8 t _ { k } } \mathbf { k } ; t _ { 0 } = 0r(t)=(10sint)i−(3cos3t)j+e−8tk​k;t0​=0

- $r ( t ) = ( 10 \sin t ) i - ( 3 \cos 3 t ) j + e ^ { - 8 t _ { k } } \mathbf { k } ; t _ { 0 } = 0$