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Q 38.

Expert-verifiedFound in: Page 944

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

$z=0$

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

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

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

The equation of plane containing the given lines and point is

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

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

$\Rightarrow z=0$

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