(a) Find the directional derivative of f at the given point in the indicated direction. f ( x...

Question:

(a) Find the directional derivative of {eq}f {/eq} at the given point in the indicated direction.

{eq}f(x,y) = x^2e^{-y}, \ (-2,0) {/eq} in the direction toward the point {eq}(2,-3) {/eq}

(b) Find the equations of the tangent plane and the normal line to

{eq}x^2 + 2y^2-3z^2 = 3, (2, -1, 1) {/eq}.

Application Of Partial Derivatives:

We have to solve the two problems by using the partial derivatives with respect to the given variables at the given point. The first problem is solving the directional derivative of the function {eq}f(x, y) {/eq} at the point and in the unit direction vector. The second problem is solving the equations of tangent plane and the normal line to the surface {eq}F(x, y, z)=0 {/eq} at the point.

Answer and Explanation:

(a) Given function is {eq}f(x,y) = x^2e^{-y}, {/eq} at the given point {eq}\ (-2,0) {/eq} in the direction toward the point {eq}(2,-3) {/eq}:

Finding the directional derivative:

{eq}\begin{align*} \displaystyle f(x,y) &= x^2e^{-y} \\ \displaystyle \frac{\partial f}{\partial x}&=\frac{\partial }{\partial x}\left(x^2e^{-y}\:\right) \\ \displaystyle f_{x}(x, y)&=2e^{-y}x \\ \displaystyle f_{x}(-2,0)&=2e^{-0}(-2)=-4 \\ \displaystyle \frac{\partial f}{\partial y}&=\frac{\partial }{\partial y}\left(x^2e^{-y}\:\right) \\ \displaystyle f_{y}(x, y)&=-e^{-y}x^2 \\ \displaystyle f_{y}(-2,0)&=-e^{-0}(-2)^2=-4 \\ \displaystyle \bigtriangledown f(x, y)&=\left \langle -4, -4 \right \rangle \\ \displaystyle \vec{u}&=\frac{1}{\sqrt{(2)^{2}+(-3)^{2}}}\left \langle 2,-3 \right \rangle \\ \displaystyle &=\frac{1}{\sqrt{13}}\left \langle 2,-3 \right \rangle \\ \displaystyle \vec{u}&=\left \langle \frac{2}{\sqrt{13}},\frac{-3}{\sqrt{13}} \right \rangle \\ \displaystyle D_{\vec{u}}f(P)&=\bigtriangledown f(P) \cdot \vec{u} \\ \displaystyle &=\left \langle -4, -4 \right \rangle \cdot \left \langle \frac{2}{\sqrt{13}}, \frac{-3}{\sqrt{13}} \right \rangle \\ \displaystyle &=(-4)\left(\frac{2}{\sqrt{13}}\right)+(-4)\left(\frac{-3}{\sqrt{13}}\right) \\ \displaystyle &=-\frac{8}{\sqrt{13}}+\frac{12}{\sqrt{13}} \\ \displaystyle D_{\vec{u}}f(P)&=\frac{4}{\sqrt{13}} \end{align*} {/eq}

Hence, the directional derivative is {eq}\ \displaystyle \mathbf{\color{blue}{ D_{\vec{u}}f(P)=\frac{4}{\sqrt{13}} }} {/eq}.


(b) Consider the given surface at point {eq}\displaystyle x^2+2y^2-3z^2 = 3, (2, -1, 1) {/eq}:

Finding the equations of the tangent plane:

{eq}\begin{align*} \displaystyle x^2+2y^2-3z^2-3&=0 \\ \displaystyle \frac{\partial F}{\partial x}&=\frac{\partial }{\partial x}\left(x^2+2y^2-3z^2-3\right) \\ \displaystyle F_{x}(x, y, z)&=2x \\ \displaystyle F_{x}(2, -1, 1)&=2(2)=4 \\ \displaystyle \frac{\partial F}{\partial y}&=\frac{\partial }{\partial y}\left(x^2+2y^2-3z^2-3\right) \\ \displaystyle F_{y}(x, y, z)&=4y \\ \displaystyle F_{y}(2, -1, 1)&=4(-1)=-4 \\ \displaystyle \frac{\partial F}{\partial z}&=\frac{\partial }{\partial z}\left(x^2+2y^2-3z^2-3\right) \\ \displaystyle F_{z}(x, y, z)&=-6z \\ \displaystyle F_{z}(2, -1, 1)&=-6(1)=-6 \\ \displaystyle F_{x}(P)(x-x_{0})+F_{y}(P)(y-y_{0})+F_{z}(P)(z-z_{0})&=0 \\ \displaystyle 4(x-2)+(-4)(y-(-1))+(-6)(z-1)&=0 \\ \displaystyle 4x-8-4y-4-6z+6&=0 \\ \displaystyle 4x-4y-6z-6&=0 \end{align*} {/eq}

Finding the equation of normal line:

{eq}\begin{align*} \displaystyle x&=x_{0}+F_{x}(P)t, \ y=y_{0}+F_{y}(P)t, \ z=z_{0}+F_{z}(P)t \\ \displaystyle x&=2+4t, \ y=-1+(-4)t, \ z=1+(-6)t \\ \displaystyle x&=2+4t, \ y=-1-4t, \ z=1-6t \end{align*} {/eq}

Hence, equation of tangent plane is {eq}\ \displaystyle \mathbf{\color{blue}{ 4x-4y-6z-6=0 }} {/eq} and the equation of normal line is {eq}\ \displaystyle \mathbf{\color{blue}{x=2+4t, \ y=-1-4t, \ z=1-6t }} {/eq}.


Learn more about this topic:

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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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