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 $\left(b\right)$.

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 $a=e,b=3,f\left(a,b\right)=e^3,f_x\left(a,b\right)=3e^2$

Therefore, equation of tangent in $x$direction is

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

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

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

The equation of plane containing the given lines and point is

$f_x\left(a,b\right)\left(x-a\right)+f_y\left(a,b\right)\left(y-b\right)=z-f\left(a,b\right)$

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

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

$\Rightarrow 3e^{2}x+e^{3}y-z=5e^3$