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Answers without the blur. Sign up and see all textbooks for free! Q 39.

Expert-verified Found in: Page 944 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In Exercises $37-42$, use the partial derivatives of role="math" localid="1650186824938" $f\left(x,y\right)={x}^{y}$ and the point $\left(e,3\right)$ specified to $\left(a\right)$ find the equation of the line tangent to the surface defined by the function in the $x$ direction, $\left(b\right)$ find the equation of the line tangent to the surface defined by the function in the $y$ direction, and $\left(c\right)$ find the equation of the plane containing the lines you found in parts $\left(a\right)$ and $\left(b\right)$.

Part $\left(a\right)$, The equation of the line tangent to the surface defined by the function in the $x$ direction is

$x=e+t,y=3,z={e}^{3}+3{e}^{2}t$

Part $\left(b\right)$, The equation of the line tangent to the surface defined by the function in the $y$ direction is

$x=e,y=3+t,z={e}^{3}+{e}^{3}t$

Part $\left(c\right)$, The equation of the plane containing the lines you found in parts $\left(a\right)$ and $\left(b\right)$ is

localid="1650216415404" $3{e}^{2}x+{e}^{3}y-z=5{e}^{3}$

See the step by step solution

## Part (a) Step 1. Explanation of part a, Finding equation of tangent in x direction

The line tangent to the surface at $\left(a,b,f\left(a,b\right)\right)$ in the $x$ direction is given by the parametric equation

$x=a+t,y=b,z=f\left(a,b\right)+{f}_{x}\left(a,b\right)t$

Now we have function $f\left(x,y\right)={x}^{y}$ and point role="math" localid="1650215016730" $\left(e,3\right)$

So $a=e,b=3,f\left(a,b\right)={e}^{3},{f}_{x}\left(a,b\right)=3{e}^{2}$

Therefore, equation of tangent in $x$ direction is

$x=e+t,y=3,z={e}^{3}+3{e}^{2}t$

## Part (b)  Step 1. Explanation of part b, Finding equation of tangent in y direction

The line tangent to the surface at $\left(a,b,f\left(a,b\right)\right)$ in the $y$ direction is given by the parametric equation

$x=a,y=b+t,z=f\left(a,b\right)+{f}_{y}\left(a,b\right)t$

Now we have function $f\left(x,y\right)={x}^{y}$ and point $\left(e,3\right)$

So role="math" localid="1650215656272" $a=e,b=3,f\left(a,b\right)={e}^{3},{f}_{y}\left(a,b\right)={e}^{3}$

Therefore, equation of tangent in $y$ direction is

$x=e,y=3+t,z={e}^{3}+{e}^{3}t$

## Part (c)  Step 1. Explanation of part c, Finding equation of plane

The equation of plane containing the given lines and point is

${f}_{x}\left(a,b\right)\left(x-a\right)+{f}_{y}\left(a,b\right)\left(y-b\right)=z-f\left(a,b\right)$

$⇒3{e}^{2}\left(x-e\right)+{e}^{3}\left(y-3\right)=z-{e}^{3}$

$⇒3{e}^{2}x-3{e}^{3}+{e}^{3}y-3{e}^{3}=z-{e}^{3}$

$⇒3{e}^{2}x+{e}^{3}y-z=5{e}^{3}$ ### Want to see more solutions like these? 