# Curvature Question

## Homework Statement

Find the curvature K(t) of the curve r(t) = (-4sin(t)) i + (-4sin(t)) j + (5cos(t)) k.

## Homework Equations

K(t) = |r'(t) x r"(t)| / |r'(t)|3

## The Attempt at a Solution

r'(t) = (-4cos(t))i + (-4cos(t))j + (-5sin(t))k
r"(t) = (4sin(t))i + 4sin(t))j + (-5cos(t))k
|r'(t)| = sqrt(16cos2(t) + 16cos2(t) + 25sin2(t))
r'(t) x r"(t) = [20cos(t)+20sin(t)]i - [20cos(t) + 20sin(t)]k +0j
|r'(t) x r"(t)| = sqrt([20cos(t)+20sin(t)]^2 + [-20cos(t) - 20sin(t)]^2)
sqrt([20cos(t)+20sin(t)]^2 + [-20cos(t) - 20sin(t)]^2)/[sqrt(16cos2(t) + 16cos2(t) + 25sin2(t))]3
But it isn't, so I am confused as to what I am doing wrong.

lanedance
Homework Helper
think you added a minus... though i didn't check the original cross product
r'(t) x r"(t) = [20cos(t)+20sin(t)]i - [20cos(t) + 20sin(t)]k +0j
so shouldn't it become
|r'(t) x r"(t)| = sqrt([20cos(t)+20sin(t)]^2 + [20cos(t) + 20sin(t)]^2)
|r'(t) x r"(t)| = sqrt(2[20cos(t)+20sin(t)]^2)

I don't think that would be it since the cross product formula says that for the middle number it is subtraction. Besides, it is being squared, even if it is negative the answer which results will be positive.

lanedance
Homework Helper
r(t) = (-4sin(t)) i + (-4sin(t)) j + (5cos(t)) k.
r'(t) = (-4cos(t)) i + (-4cos(t)) j + (-5sin(t)) k.
r''(t) = (4sin(t)) i + (4sin(t)) j + (-5cos(t)) k.

ok redoing the cross product you get
|r'(t) x r"(t)| = i(20c^2 + 20s^2) + j(-20s^2 - 20c^2) + k(-16cs +16cs)

which becomes
|r'(t) x r"(t)| = i(1) + j(-1) + k(0)