What is the slope of the line tangent to the surface at that point $(2,-1)$ lying in the plane $y=-1$?

Question

For the function $f(x,y)=x(x+y^5)$,What is the slope of the line tangent to the surface at that point $(2,-1)$ lying in the plane $y=-1$?

I know how to find the linear approximation of any function,but i'm not getting any approach of finding the slope.

Please suggest any method/formula....

Answer 1

Since the plane is parallel to the $z$-axis, this question is basically asking you about a directional derivative of $f$ at the point $(2,-1)$. For this plane, $y$ is constant, so the particular directional derivative required is just the partial $x$-derivative of $f$ at this point ($\nabla f\cdot (1,0)=f_x$). I expect that you’ll be able to take it from here.

Answer 2

Let's take $\frac{\partial f}{\partial x}=2x+y^5$ if y = -1 (is constant), $\frac{\partial f}{\partial x}=2x-1$, so the slope of the intersection curve generated by f(x, y) and the plane y = -1 at (2, -1) is $\frac{\partial f}{\partial x}\Big|_{x=2}=3$. If you keep y = -1, the function is only going to change along the X axis, that gives the intersection curve with the plane y = -1, so use $\frac{\partial f}{\partial x}=2x+y_0^5$ if y keeps constant.