For the curve r(t), find an equation for the indicated plane at the given value of t.
-$\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } - 9 ) \mathbf { i } + ( \ln ( \sec \mathrm { t } ) - 1 ) \mathbf { j } + 7 \mathbf { k } , - \pi / 2

Question

For the curve r(t), find an equation for the indicated plane at the given value of t.
-$\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } - 9 ) \mathbf { i } + ( \ln ( \sec \mathrm { t } ) - 1 ) \mathbf { j } + 7 \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi / 2$; normal plane at $t = 10 \pi$.
A) $x = - 10 \pi - 9$
B) $x = 9 - 10 \pi$
C) $x = 10 \pi + 9$
D) $x = 10 \pi - 9$

Quizplus · Accepted Answer

The normal plane to the curve at a given point is perpendicular to the tangent vector at that point. The tangent vector is the derivative of the position vector r(t). For the given curve, the x-component of r(t) is (t - 9), so its derivative with respect to t is 1. This means the tangent vector at any point has a component in the x-direction of 1, indicating that the normal plane's equation will be of the form x = constant. At $t = 10\pi$, the x-component of r(t) is $10\pi - 9$, so the equation of the normal plane is $x = 10\pi - 9$.

For the Curve R(t), Find an Equation for the Indicated