# Thread: Calculus problem

1. ## Calculus problem

At what point on the curve y = x^4 does the normal have a slope of 16?

2. Does "normal" mean perpendicular to the tangent?
If that's the case:
slope_normal(x) * slope_tangent(x) = -1
16 * slope_tangent(x) = -1

where slope_tangent(x) = y'(x)

16 * y'(x) = -1

Solve for x.

-O

3. Originally Posted by mathamatics112
At what point on the curve y = x^4 does the normal have a slope of 16?
If the slope of the normal line at that point is 16, then the slope of the tangent line at that point is $\displaystyle -\frac{1}{16}$. So what you need to do is find where $\displaystyle \frac{\,d}{\,dx}\left[x^4\right]=-\frac{1}{16}$

Can you take it from here?

4. ## I am stuck again :(

Ok so I am at this point where dx/dy (x^4) = -1/16.

from here I got 4x^3 = -1/16

i multipled by 4 * 16 to get -1/64 and got the third root of this to get me -1/4 which is my x. How do i get the y vaule? i tried subbing it in into x^4 and i got the wrong answer

5. (-1/4)^4=1/256 should be the answer. No?

-O