Answer to Question #270563 in Calculus for Hemanth

Question

Find the slopes of the tangents to the curves obtained by slicing the surface

x^{2}e^{y}-yze^{x}=0 with planes x=1 and y=1 at the point (1, 1), using implicit

differentiation.

Answer 1

$x^2e^y-yze^x=0$

$x^2e^y=yze^x$

Differentiate Z with respect to x to get the first slope tangent$\frac{dz}{dx}$

$\frac{dz}{dx}=\frac{2xe^{y-x}-zy}{y}$

Replace (1, 1,0) to get the slope since at z plane z=0

$\frac{dz}{dx}=2$

The first slope is $2$

The second slope

Differentiate z with respect to y

$\frac{dz}{dy}\\$

$\frac{dz}{dy}=\frac{x^2e^{y-x}-z}{y}$

Replace (1, 1,0) in the equation

$\frac{dz}{dy}=1$

The second slope is $1$