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Find an equation of the tangent plane to the parametric surface x=-2rcos\theta, y=-3rsin\theta,...

Question:

Find an equation of the tangent plane to the parametric surface {eq}x=-2rcos\theta{/eq}, {eq}y=-3rsin\theta{/eq}, {eq}z=r{/eq} at the point {eq}(-5\sqrt{2},-3\sqrt{2}){/eq} when {eq}r=2{/eq}, {eq}\theta=\frac{\pi}{4}{/eq}

Finding Equation of the Tangent Plane:

For the given parametric surface, we have to find the equation of the tangent plane to the surface. The suitable equation is {eq}\displaystyle \vec{n} \cdot \left \langle x-x_{0},y-y_{0}, z-z_{0} \right \rangle =0 {/eq}.

The cross product of the partial derivatives of the given parametric equations will provide the normal vector.

Answer and Explanation:

Let us consider the given parametric surface {eq}\displaystyle x=-2r \cos \theta {/eq}, {eq}\displaystyle y=-3r \sin \theta{/eq}, {eq}\displaystyle z=r{/eq} when {eq}\displaystyle r=2{/eq}, {eq}\displaystyle \theta=\frac{\pi}{4}{/eq}.

Finding the point:

{eq}\begin{align*} \displaystyle x &=-2r \cos \theta \\ \displaystyle x_{0} &=-2 (2) \cos \left( \frac{\pi}{4} \right) \\ \displaystyle x_{0} &=-2\sqrt{2} \\ \displaystyle y &=-3r \sin \theta \\ \displaystyle y_{0} &=-3 (2) \sin \left( \frac{\pi}{4} \right) \\ \displaystyle y_{0} &=-3\sqrt{2} \\ \displaystyle z &=r \\ \displaystyle z_{0} &=2 \\ \displaystyle \left( x_{0}, y_{0}, z_{0} \right) &=\left( -2\sqrt{2}, -3\sqrt{2}, 2 \right) \end{align*} {/eq}


Finding the normal vector at the point:

{eq}\begin{align*} \displaystyle u(r, \theta) &=\left \langle -2r \cos \theta, -3r \sin \theta, r \right \rangle \\ \displaystyle \vec{n} &=u_{r} \times u_{\theta} \\ \displaystyle u_{r} &=\left \langle -2\cos \left(\theta \right), -3\sin \left(\theta \right), 1 \right \rangle \\ \displaystyle u_{\theta} &=\left \langle 2r\sin \left(\theta \right), -3r\cos \left(\theta \right), 0 \right \rangle \\ \displaystyle u_{r} \times u_{\theta} &= \begin{vmatrix} i & j & k\\ -2\cos \left(\theta \right) & -3\sin \left(\theta \right) & 1\\ 2r\sin \left(\theta \right) & -3r\cos \left(\theta \right) & 0 \end{vmatrix} \\ \displaystyle u_{r} \times u_{\theta} &=((0)(-3\sin \left(\theta \right))-(-3r\cos \left(\theta \right))(1))\vec{i}-((0)(-2\cos \left(\theta \right))-(2r\sin \left(\theta \right))(1))\vec{j}+((-3r\cos \left(\theta \right))(-2\cos \left(\theta \right))-(2r\sin \left(\theta \right))(-3\sin \left(\theta \right)))\vec{k} \\ \displaystyle u_{r} \times u_{\theta} &=3r\cos \left(\theta \right) \vec{i} +2r\sin \left(\theta \right)\vec{j}+ 6r \vec{k} \\ \displaystyle \vec{n}(r, \theta) &=3(2)\cos \left(\frac{\pi}{4}\right) \vec{i} +2(2)\sin \left(\frac{\pi}{4}\right)\vec{j}+ 6(2) \vec{k} \\ \displaystyle \vec{n}(r, \theta) &=3\sqrt{2} \vec{i} + 2\sqrt{2}\vec{j}+ 12 \vec{k} \end{align*} {/eq}


Finding an equation of the tangent plane to the parametric surface:

{eq}\begin{align*} \displaystyle \vec{n} \cdot \left \langle x-x_{0},y-y_{0}, z-z_{0} \right \rangle &=0 \\ \displaystyle \left \langle 3\sqrt{2}, 2\sqrt{2}, 12 \right \rangle \cdot \left \langle x-\left( -2\sqrt{2} \right), y-\left( -3\sqrt{2} \right), z-2 \right \rangle &=0 \\ \displaystyle \left \langle 3\sqrt{2}, 2\sqrt{2}, 12 \right \rangle \cdot \left \langle x+2\sqrt{2}, y+3\sqrt{2}, z-2 \right \rangle &=0 \\ \displaystyle \left( 3\sqrt{2} \right)\left( x+2\sqrt{2} \right)+\left( 2\sqrt{2} \right)\left( y+3\sqrt{2} \right)+\left( 12 \right)\left( z-2 \right) &=0 \\ \displaystyle 3\sqrt{2}x+12+ 2\sqrt{2} y+12+12z-24 &=0 \\ \displaystyle 3\sqrt{2}x+ 2\sqrt{2} y+12z+12+12-24 &=0 \\ \displaystyle 3\sqrt{2}x+ 2\sqrt{2} y+12z &=0 \end{align*} {/eq}

The equation of the tangent plane to the parametric surface is {eq}\ \displaystyle \mathbf{\color{blue}{ 3\sqrt{2}x+ 2\sqrt{2} y+12z =0 }} {/eq}.


Learn more about this topic:

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Tangent Plane to the Surface

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 3
621

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