a) Find equations of the tangent plane and the normal line to z = h(x, y) = ln(\sqrt{x^{2} +...


a) Find equations of the tangent plane and the normal line to {eq}z = h(x, y) = ln(\sqrt{x^{2} + y^{2}} {/eq} at {eq}(3, 4, \ln(5)) {/eq}.

b) Find {eq}\frac{dw}{dt} {/eq} of {eq}w = \ln(x^{2} + y); \, x = 2t; \, y = 4 - t {/eq} by two different ways. Compare the results and provide a conclusion.

Tangent Plane and Normal Line to the Surface:

The normal vector to the surface is found out using a 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:

a) The given surface is {eq}\displaystyle S: \ z = \ln(\sqrt{x^{2} + y^{2}}) {/eq}

Let {eq}\displaystyle f(x,y,z)= \ln(\sqrt{x^{2} + y^{2}})-z. {/eq}

The normal vector to the surface is the gradient of {eq}f(x,y,z), {/eq}

{eq}\begin{align} \displaystyle \nabla f(x,y,z) &=\frac{\partial f}{\partial x}\vec i+\frac{\partial f}{\partial y} \vec j+\frac{\partial f}{\partial z} \vec k\\ \displaystyle &=\frac{1}{\sqrt{x^{2} + y^{2}}} \cdot \frac{x}{\sqrt{x^{2} + y^{2}}} \vec i +\frac{1}{\sqrt{x^{2} + y^{2}}} \cdot \frac{y}{\sqrt{x^{2} + y^{2}}} \vec j - \vec k\\ \displaystyle &= \frac{x}{x^{2} + y^{2}} \vec i +\frac{y}{x^{2} + y^{2}} \vec j - \vec k\\ \displaystyle \nabla f(3, \ 4 \ \ln 5) &= \frac{3}{3^{2} + 4^{2}} \vec i +\frac{4}{3^{2} + 4^{2}} \vec j - \vec k\\ \displaystyle &= \frac{3}{25} \vec i +\frac{4}{25} \vec j - \vec k\\ \end{align} {/eq}

Let {eq}(x, \ y, \ z) {/eq} be an arbitrary point on the required tangent plane. Also, {eq}P(3, \ 4 \ \ln 5) {/eq} lies on the tangent plane. Therefore, the vector joining these points {eq}(x-3) \vec i+(y-4) \vec j+(z- \ln 5) \vec k {/eq} lies on the required plane.

This vector is perpendicular to the normal vector to the surface, resulting in their dot product to be zero, which gives us the equation of the required tangent plane.

{eq}\displaystyle \left ( (x-3) \vec i+(y-4) \vec j+(z- \ln 5) \vec k \right ) \cdot \left ( \frac{3}{25} \vec i +\frac{4}{25} \vec j - \vec k \right )=0\\ \displaystyle \Rightarrow \frac{3}{25}(x-3)+\frac{4}{25}(y-4)-(z- \ln 5)=0 \\ \displaystyle \Rightarrow 3(x-3)+4(y-4)-25(z- \ln 5)=0 \\ \displaystyle \Rightarrow 3x-9+4y-16-25z+25 \ln 5=0 \\ \displaystyle \Rightarrow 3x+4y-25z=25(1- \ln 5) \\ {/eq}

So, the equation of the required tangent plane is {eq}\color{blue}{ 3x+4y-25z=25(1- \ln 5)}. {/eq}

Now, the equation of the normal line to the given surface at {eq}P(3, \ 4, \ \ln 5) {/eq} will be {eq}\displaystyle \frac{x-3}{ \frac{3}{25}}=\frac{y-4}{ \frac{4}{25}}=\frac{z-\ln 5}{-1}\\ \displaystyle \frac{25(x-3)}{ 3}=\frac{25(y-4)}{4}=-z+\ln 5\\ {/eq} or in parametric form as {eq}(x,y,z)=(3, \ 4, \ \ln 5) + t \ (\frac{3}{25}, \ \frac{4}{25}, \ -1), \ t \ \in \ R. {/eq}

The required equation of the normal line is {eq}\color{blue}{(3, \ 4, \ \ln 5) + t \ (\frac{3}{25}, \ \frac{4}{25}, \ -1)}, \ t \ \in \ R. {/eq}

b) Given {eq}\displaystyle w = \ln(x^{2} + y); \, x = 2t; \, y = 4 - t {/eq}

Now, we find the partial derivatives of w with respect to x and y.

{eq}\begin{align} \displaystyle \frac{\partial w}{\partial x}&=\frac{2x}{x^2+y}\\ \displaystyle \frac{\partial w}{\partial y}&=\frac{1}{x^2+y}\\ \end{align} {/eq}

Writing the derivatives in terms of t, we get:

{eq}\begin{align} \displaystyle \frac{\partial w}{\partial x}&=\frac{4t}{4t^2-t+4}\\ \displaystyle \frac{\partial w}{\partial y}&=\frac{1}{4t^2-t+4}\\ \end{align} {/eq}

We then find the derivatives of x and y with respect to t.

{eq}\begin{align} \displaystyle \frac{\mathrm{d} x}{\mathrm{d} t}&=2\\ \displaystyle \frac{\mathrm{d} y}{\mathrm{d} t}&=-1\\ \end{align} {/eq}

Now using Chain Rule,

{eq}\begin{align} \displaystyle \frac{\mathrm{d} w}{\mathrm{d} t}&=\frac{\partial w}{\partial x}\frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\partial w}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} t}\\ \displaystyle &=2 \left ( \frac{4t}{4t^2-t+4} \right )-\frac{1}{4t^2-t+4}\\ \displaystyle \frac{\mathrm{d} w}{\mathrm{d} t} &= \color{blue}{ \frac{8t-1}{4t^2-t+4}} \ ...(1)\\ \end{align} {/eq}

In an alternate way, we substitute the values of x and y directly in the function w and differentiate with respect to t. So, w becomes

{eq}\displaystyle w = \ln(4t^{2} -t+ 4). {/eq}


{eq}\begin{align} \displaystyle \frac{\mathrm{d} w}{\mathrm{d} t}&=\frac{1}{4t^2-t+4} \cdot \left ( 8t-1 \right )\\ \displaystyle \frac{\mathrm{d} w}{\mathrm{d} t} &= \color{blue}{ \frac{8t-1}{4t^2-t+4}} \ ...(2)\\ \end{align} {/eq}

From (1) and (2), we can conclude that both are equal and we can find the derivative using anyway.

Learn more about this topic:

Tangent Plane to the Surface

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 3

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