You need to review the notion of a directional derivative and, in particular, the partial derivative. For the plane perpendicular to the -axis, the slope of the curve will be . That's it. I have no idea where you're getting -1/0 from.
Hi
I need help with the followoing question:
Find the slopes of the curves of intersection of the surface z= f(x,y) with the planes perpendicular to the x axis and y axis respectively at a given point.
z= sin (4x+y) at (0, pi/2)
I did fx(x,y)= 4 cos(4x+y) and at this point i get 0
To find the perpendicular planes i think you do -1/0 but that is not defined.
is this the correct method?
kind regards
This can be done without partial derivatives:
The "plane perpendicular to the x-axis" at is the y-z plane, x= 0. Set x= 0 in the formula: z= sin(y) and the derivative of that is cos(y) which is at .
The "plane perpendicular to the y-axis" at is the plane : and the derivative of that is which is at .
HallsofIvy: If your goal is to get through your course on multivariable calculus without ever learning what a partial derivative is, you're not going to make it. By the way, I think you're second answer is incorrect (you didn't differentiate).
One more thing, the derivatives you're doing ARE partial derivatives.