Suggested languages for you:

Americas

Europe

Q 38.

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="1650186853142" $g\left(x,y\right)={\mathrm{tan}}^{-1}\left(x{y}^{2}\right)$ and the point role="math" localid="1650186870407" $\left(1,0\right)$ 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

$x=1+t,y=0,z=0$

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

$x=1,y=t,z=0$

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

$z=0$

See the step by step solution

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 $g\left(x,y\right)={\mathrm{tan}}^{-1}\left(x{y}^{2}\right)$ and point $\left(1,0\right)$

So localid="1650290477299" $a=1,b=0,f\left(a,b\right)=0,{f}_{x}\left(a,b\right)=0$

Therefore, equation of tangent in $x$ direction is

$x=1+t,y=0,z=0$

Step 2. 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 $g\left(x,y\right)={\mathrm{tan}}^{-1}\left(x{y}^{2}\right)$ and point $\left(1,0\right)$

So $a=1,b=0,f\left(a,b\right)=0,{f}_{y}\left(a,b\right)=0$

Therefore, equation of tangent in $y$ direction is

role="math" localid="1650290595556" $x=1,y=t,z=0$

Step 3. Explanation of part c, Finding equation of plane

The equation of plane containing the given lines and point is

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

$⇒0\left(x-1\right)+0\left(y-0\right)=z-0$

$⇒z=0$