V. Let \omega = f(x, y, z) = 3 square root of (4x^2 + 4y^2 + z^2) and P = (1, 2, 4). A. Find the...

Question:

V. Let {eq}\, \omega \, = \, f(x, \, y, \, z) \, = \, 3 \, \sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}} \, {/eq} and {eq}\, P \, = \, (1, \, 2, \, 4) {/eq}.

  • A. Find the gradient {eq}\, \nabla \, f(P) \, {/eq} of {eq}\, f \, {/eq} at {eq}\, P {/eq}, and then find an equation of the tangent plane to the ellipsoid {eq}\, f(x, \, y, \, z) \, = \, 18 \, {/eq} at {eq}\, P {/eq}.
  • B. Find the directional derivative {eq}\, f'_u (P) {/eq}, if {eq}\, \mathbf{u} \, {/eq} is the unit vector whose direction is the same as that of {eq}\, \mathbf{v} \, = \overrightarrow{PQ} {/eq}, where {eq}\, Q \, = \, (7, \, 5, \, 6) {/eq}.
  • C. In the above definition of {eq}\, \omega \, = \, f(x, \, y, \, z) \, {/eq} assume that {eq}\, x, \, y, \, {/eq} and {eq}\, z \, {/eq} are differentiable functions of {eq}\, t \, {/eq} such that when {eq}\, t \, = \, 1 {/eq}, then {eq}\, x \, = \, 1, \quad y \, = \, 2, \, {/eq} and {eq}\, z \, = \, 4, \, {/eq} and {eq}\, \dfrac{\mathrm{d} x}{\mathrm{d} t} \, = \, 3, \quad \dfrac{\mathrm{d} y}{\mathrm{d} t} \, = \, 5, \, {/eq} and {eq}\, \dfrac{\mathrm{d} z}{\mathrm{d} t} \, = \, 7 {/eq}. Find {eq}\dfrac{\mathrm{d} \omega}{\mathrm{d} t} \, {/eq} when {eq}\, t \, = \, 1 {/eq}.

Application of Gradient to find Directional Derivatives and Equation of Tangent Plane and Normal Line:

The directional derivative of f(x, y, z) at P in the direction of {eq}\vec v {/eq} is {eq}D_{\hat v} f=\nabla f(P) \cdot \hat v, {/eq} where {eq}\nabla f(P) {/eq} denotes the gradient of f at the point P and {eq}\hat v=\frac{\vec v}{|\vec v|} {/eq} is the unit vector.

The normal vector to the surface is found out using gradient. This normal being orthogonal to the plane will also be orthogonal to any vector in the required plane. So we find a vector lying in the required plane and substitute their dot product to be zero which gives the required equation of the tangent plane.

Answer and Explanation:

{eq}\displaystyle \, \omega \, = \, f(x, \, y, \, z) \, = \, 3 \, \sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}} \, {/eq}

Differentiate partially with respect to x, y and z, we get:

{eq}\displaystyle \frac{\partial \omega}{\partial x}=\frac{\partial f}{\partial x}=\frac{3(8x)}{2 \sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}=\frac{12x}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}\\ \displaystyle \frac{\partial \omega}{\partial y}=\frac{\partial f}{\partial y}=\frac{3(8y)}{2 \sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}=\frac{12y}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}\\ \displaystyle \frac{\partial \omega}{\partial z}=\frac{\partial f}{\partial z}=\frac{3(2z)}{2 \sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}=\frac{3z}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}} {/eq}

Evaluating the partial derivatives at the point P(1, 2, 4), we get:

{eq}\displaystyle \frac{\partial \omega}{\partial x}=\frac{\partial f}{\partial x}=\frac{12}{\sqrt{4 \ + \ 16 \ + \ 16}}=\frac{12}{6}=2\\ \displaystyle \frac{\partial \omega}{\partial y}=\frac{\partial f}{\partial y}=\frac{24}{\sqrt{4 \ + \ 16 \ + \ 16}}=\frac{24}{6}=4\\ \displaystyle \frac{\partial \omega}{\partial z}=\frac{\partial f}{\partial z}=\frac{12}{\sqrt{4 \ + \ 16 \ + \ 16}}=\frac{12}{6}=2 {/eq}

The gradient {eq}\, \nabla \, f(P) \, {/eq} of {eq}\, f \, {/eq} at {eq}\, P {/eq} is given as:

{eq}\begin{align} \displaystyle \, \nabla \, f \,&=\frac{\partial f}{\partial x} \vec i+\frac{\partial f}{\partial y} \vec j+\frac{\partial f}{\partial z} \vec k\\ \displaystyle &=\frac{12x}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}} \vec i+\frac{12y}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}} \vec j+\frac{3z}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}} \vec k\\ \displaystyle \, \nabla \, f(P) \,&=\color{blue}{2 \vec i + 4 \vec j + 2 \vec k} \end{align} {/eq}

Let {eq}(x,y,z) {/eq} be an arbitrary point on the required tangent plane. Also, {eq}P(1, \ 2, \ 4) {/eq} lies on the tangent plane. Therefore, the vector joining these points {eq}(x-1)i+(y-2)j+(z-4)k {/eq} lies on the required plane.

This vector and the normal vector to the surface will be perpendicular to each other. Hence, their dot product will be zero, which gives us the equation of the required tangent plane.

{eq}\displaystyle \left ((x-1) \vec i+(y-2) \vec j+(z-4) \vec k \right ).\left (2 \vec i + 4 \vec j + 2 \vec k \right )=0\\ \displaystyle \Rightarrow 2(x-1)+4(y-2)+2(z-4)=0\\ \displaystyle \Rightarrow 2x+4y+2z-2-8-8=0\\ \displaystyle \Rightarrow 2x+4y+2z-18=0\\ \displaystyle \Rightarrow x+2y+z=9 {/eq}

Hence, an equation of the tangent plane to the ellipsoid {eq}\, f(x, \, y, \, z) \, = \, 18 \, {/eq} at {eq}\, P {/eq} is {eq}\color{blue}{x+2y+z=9}. {/eq}


B)

{eq}\displaystyle \vec u=(7-1) \hat i+(5-2) \hat j+(6-4) \hat k=6 \hat i+3 \hat j+2 \hat k\\ \displaystyle |\vec u|=\sqrt{6^2+3^2+2^2}=\sqrt{36+9+4}=7\\ \displaystyle \hat u=\frac{\vec u}{|\vec u|}=\frac{6 \hat i+3 \hat j+2 \hat k}{7} {/eq}

The directional derivative of {eq}f {/eq} at P in the direction of {eq}\vec u {/eq} is:

{eq}\begin{align} \displaystyle D_{\vec u}f &=\nabla f(P) \cdot \hat u\\ \displaystyle &= \left ( 2 \vec i + 4 \vec j + 2 \vec k \right ) \cdot \left ( \frac{6 \hat i+3 \hat j+2 \hat k}{7} \right )\\ \displaystyle &= \frac{2(6)+4(3)+2(2)}{7} \\ \displaystyle &= \frac{28}{7} \\ \displaystyle &= 4 \end{align} {/eq}

Therefore, the directional derivative {eq}\, f'_u (P)=\color{blue}{7}. {/eq}


C) When {eq}\, t \, = \, 1 {/eq}, then {eq}\, x \, = \, 1, \quad y \, = \, 2, \, {/eq} and {eq}\, z \, = \, 4, \, {/eq} and {eq}\, \dfrac{\mathrm{d} x}{\mathrm{d} t} \, = \, 3, \quad \dfrac{\mathrm{d} y}{\mathrm{d} t} \, = \, 5, \, {/eq} and {eq}\, \dfrac{\mathrm{d} z}{\mathrm{d} t} \, = \, 7 {/eq}.

We have {eq}\displaystyle \frac{\partial \omega}{\partial x}=\frac{12x}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}\\ \displaystyle \frac{\partial \omega}{\partial y}=\frac{12y}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}}\\ \displaystyle \frac{\partial \omega}{\partial z}=\frac{3z}{\sqrt{4x^{2} \, + \, 4y^{2} \, + \, z^{2}}} {/eq}

Finding the value of the derivatives at t=1, we get

We have {eq}\displaystyle \frac{\partial \omega}{\partial x}|_{t=1}=2\\ \displaystyle \frac{\partial \omega}{\partial y}|_{t=1}=4\\ \displaystyle \frac{\partial \omega}{\partial z}|_{t=1}=2 {/eq}

Using Chain Rule, we have

{eq}\begin{align} \displaystyle \frac{d \omega}{d t} &=\frac{\partial \omega}{\partial x}\frac{dx}{dt}+\frac{\partial \omega}{\partial y}\frac{dy}{dt}+\frac{\partial \omega}{\partial z}\frac{dz}{dt}\\ \displaystyle \frac{d \omega}{d t}|_{t=1} &=\frac{\partial \omega}{\partial x}|_{t=1} \frac{dx}{dt}|_{t=1}+\frac{\partial \omega}{\partial y}|_{t=1} \frac{dy}{dt}|_{t=1}+\frac{\partial \omega}{\partial z}|_{t=1} \frac{dz}{dt}|_{t=1}\\ \displaystyle &=2 (3)+4 (5)+2 (7)\\ \displaystyle &=6+20+14\\ &=40 \end{align} {/eq}

Therefore, we get {eq}\frac{d \omega}{d t} =\color{blue}{40}. {/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
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