Answer to Question #340924 in Calculus for ser

Question

: Let 𝑓(𝑥, 𝑦) = 𝑥^2 + 𝑦^3 . Find the slope of the line tangent to this surface at the point (-1, 1) and lying in the plane x = -1

Answer 1

$f\left(x,y\right)=x^2+y^3$ , $M=\left(-1,1\right)$ ,

$df\left(x,y\right)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=2xdx+{3y}^2dy$ ,

The equation of a plane tangent to a point $M=\left(x_0,y_0\right)$ of the surface  $z=f\left(x,y\right)$  has the form:

$z-z_0={f\prime}_x\left(x_0,y_0\right)\left(x-x_0\right)+{f\prime}_y\left(x_0,y_0\right)\left(y-y_0\right),\ z_0=f\left(x_0,y_0\right)$ ,

$z_0=f\left(x_0,y_0\right)=2, {f\prime}_x\left(x_0,y_0\right)=-2, {f\prime}_y\left(x_0,y_0\right)=3$ ,

Then

$z-2=-2\left(x+1\right)+3\left(y-1\right)$ ,

By $x=-1$ given:

$z-2=3\left(y-1\right)=k\left(y-1\right),\ k=3$.

Answer: the slope of the line tangent to this surface at the point (-1, 1) and lying in the plane x=–1 is k=3