# Curvature Question

shards5

## 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.

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)

shards5
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.

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)