Suppose we have the parametric surface given by \vec r(u, w) = \left \langle u \cos w, u \sin...

Question:

Suppose we have the parametric surface given by {eq}\vec r(u, w) = \left \langle u \cos w, u \sin w, w \right \rangle {/eq} for {eq}0 \leq u \leq 2 \pi {/eq} and {eq}0 \leq w \leq 5. {/eq} Find an equation of the tangent plane to this surface where u = 3 and {eq}w =\frac{\pi}{2}. {/eq}

Equation of the Tangent Plane:

Equation of the tangent plane for the parametric surface is {eq}\ \displaystyle \vec{n} \cdot \left \langle x-x_{0}, y-y_{0}, z-z_{0} \right \rangle =0 {/eq}. In this equation {eq}\ \displaystyle \vec{n} {/eq} is indicating the normal vector. The normal vector is obtained by evaluating the cross product of the tangent vectors, {eq}\ \displaystyle \vec r_{u} {/eq} and {eq}\ \displaystyle \vec r_{w} {/eq}. The point {eq}\ \displaystyle \left( x_{0}, y_{0}, z_{0} \right) {/eq} is obtained by evaluating the actual value of given parameteric surface.

Answer and Explanation:

The parametric surface is given by {eq}\displaystyle \vec r(u, w) = \left \langle u \cos w, u \sin w, w \right \rangle {/eq} where {eq}\displaystyle u = 3 {/eq} and {eq}\displaystyle w =\frac{\pi}{2} {/eq}.

Finding an equation of the tangent plane to the surface:

{eq}\begin{align*} \displaystyle \vec r(u, w) &= \left \langle u \cos w, u \sin w, w \right \rangle \\ \displaystyle \vec r\left( 3, \frac{\pi}{2} \right) &= \left \langle 3 \cos \left( \frac{\pi}{2} \right), 3 \sin \left( \frac{\pi}{2} \right), \frac{\pi}{2} \right \rangle \\ \displaystyle \vec r\left( 3, \frac{\pi}{2} \right) &= \left \langle 0, 3, \frac{\pi}{2} \right \rangle \\ \displaystyle \vec r_{u}(u, w) &= \left \langle \cos \left(w\right), \sin \left(w\right), 0 \right \rangle \\ \displaystyle \vec r_{u}\left( 3, \frac{\pi}{2} \right) &=\left \langle \cos \left(\frac{\pi}{2}\right), \sin \left(\frac{\pi}{2}\right), 0 \right \rangle \\ \displaystyle \vec r_{u}\left( 3, \frac{\pi}{2} \right) &=\left \langle 0, 1, 0 \right \rangle \\ \displaystyle \vec r_{w}(u, w) &= \left \langle -u\sin \left(w\right), u\cos \left(w\right), 1 \right \rangle \\ \displaystyle \vec r_{w}\left( 3, \frac{\pi}{2} \right) &=\left \langle -3\sin \left(\frac{\pi}{2}\right), 3\cos \left(\frac{\pi}{2}\right), 1 \right \rangle \\ \displaystyle \vec r_{w}\left( 3, \frac{\pi}{2} \right) &=\left \langle -3, 0, 1 \right \rangle \\ \displaystyle \vec r_{u}\left( 3, \frac{\pi}{2} \right) \times \vec r_{w}\left( 3, \frac{\pi}{2} \right) &=\begin{vmatrix} i & j & k\\ 0 & 1 & 0\\ -3 & 0 & 1 \end{vmatrix} \\ \displaystyle &=((1)(1)-(0)(0))\vec{i}-((1)(0)-(-3)(0))\vec{j}+((0)(0)-(-3)(1))\vec{k} \\ \displaystyle \vec r_{u}\left( 3, \frac{\pi}{2} \right) \times \vec r_{w}\left( 3, \frac{\pi}{2} \right) &=1 \vec{i} + 0\vec{j}+ 3 \vec{k} \\ \displaystyle \vec{n} \cdot \left \langle x-x_{0}, y-y_{0}, z-z_{0} \right \rangle &=0 \\ \displaystyle \left \langle 1, 0, 3 \right \rangle \cdot \left \langle x-0, y-3, z-\frac{\pi}{2} \right \rangle &=0 \\ \displaystyle (1)(x-0)+(0)(y-3)+(3)\left( z-\frac{\pi}{2} \right) &=0 \\ \displaystyle x+3z-\frac{3\pi }{2} &=0 \end{align*} {/eq}

Therefore, the equation of the tangent plane is {eq}\ \displaystyle \mathbf{\color{blue}{ x+3z-\frac{3\pi }{2} =0 }} {/eq}.


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Tangent Plane to the Surface

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 3
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