# Thread: Differential calculus- first order partial derviatives

1. ## Differential calculus- first order partial derviatives

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

2. You need to review the notion of a directional derivative and, in particular, the partial derivative. For the plane perpendicular to the $\displaystyle y$-axis, the slope of the curve will be $\displaystyle f_x(0,\pi/2)=0$. That's it. I have no idea where you're getting -1/0 from.

3. This can be done without partial derivatives:

The "plane perpendicular to the x-axis" at $\displaystyle (0, \pi/2)$ 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 $\displaystyle cos(\pi/2)= 0$ at $\displaystyle (0, \pi/2)$.

The "plane perpendicular to the y-axis" at $\displaystyle (0, \pi/2)$ is the plane $\displaystyle y= \pi/2$: $\displaystyle z= sin(4x+\pi/2)$ and the derivative of that is $\displaystyle 4 cos(4x+ \pi/2)$ which is $\displaystyle 4 sin(\pi/2)= 4$ at $\displaystyle (0, \pi/2)$.

4. 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.

5. thanks halls of ivy and ojones! but ojones- there really isn't any need to get so worked up over hall of ivy's post- any help is good and we all make mistakes