Thread: Calculus problem

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

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

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 . So what you need to do is find where

Can you take it from here?

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

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

-O