Finding the curvature of a space curve

[dlacombe13]

Homework Statement

Find the curvature of the car's path, K(t)
Car's Path: $r(t) = \Big< 40cos( \frac {2 \pi}{16}t ) , 40sin( \frac {2 \pi}{16}t ), \frac{20}{16}t \Big>$

Homework Equations

$K(t) = \frac { |r'(t)\:X \:r''(t)|}{|r'(t)|^3 }$

The Attempt at a Solution

This is part of a massive 6 part question, a,b,c,d,e,f. This is part e. I already have r'(t) and r''(t) as well as r'(t) x r''(t). I'm just getting totally lost in the algebra, and I don't know if I am on the right track:
$r'(t) = \Big< -5 \pi sin( \frac{2 \pi}{16}t) , 5 \pi cos( \frac{2 \pi}{16}t) , \frac{20}{16} \Big>$
$r''(t) = \Big< \frac{ -5 \pi ^2 cos( \frac{2 \pi}{16}t )}{8} , \frac{ -5 \pi ^2 sin( \frac{2 \pi}{16}t )}{8} , 0 \Big>$
$r'(t) \: X \: r''(t) = \big< \frac{ -125 \pi sin( \frac{2 \pi}{16}t )}{32} , \frac{ -25 \pi ^2 cos( \frac{2 \pi}{16}t )}{32} , \Big[ \frac{25 \pi ^3}{8} \Big]\Big[cos(\frac{2 \pi}{16}t) \Big]^2 + \frac{125 \pi ^2}{8} \Big>$

So I know the next step will be to get the magnitude of this. Does this look right so far? I have to admit, this is probably the hardest problem (algebra-wise) I have ever done in college so far.

 

[LCKurtz]

I didn't check your work, but I can tell you that I certainly would have started with$$
\vec r(t) = \langle a\cos(bt),a\sin(bt),ct\rangle$$and put the constants in at the end.

 

[dlacombe13]

Well sure, I'm positive I have done the derivatives right. I was told by someone that the 't' will cancel somewhere, and the answer will simply be a number. I just can't see how 't' will cancel. I was also told it was related to the identity sin^2(t) + cos^2(t) = 1, but when I did take the magnitude of this it was horrendous.

 

[LCKurtz]

I will just repeat my advice: do it again with $a,~b,~c$. It will cut down mistakes and make it apparent where to use the sine-cosine identity. It isn't "horrendous".

 

[dlacombe13]

Okay I took your advice, which helped a ton. I'm just hoping I didn't make any stupid arithmetic errors. When I crossed r'(t) X r''(t) I got:
$\big< ab^2c\:sin(bt), -ab^2c\:cos(bt),a^2b^3 \big>$
I took the magnitude and got:
$ab^2\sqrt{a^b+b^2+1}$
I then took the magnitude of r'(t) cubed and got:
$(a^2 + b^2 + c^2)^{3/2}$
I plugged everything back in for a,b and c and got 0.01. Seems kind of low, I'm not sure if i have made some kind of arithmetic error

 

[LCKurtz]

Okay I took your advice, which helped a ton. I'm just hoping I didn't make any stupid arithmetic errors. When I crossed r'(t) X r''(t) I got:
$\big< ab^2c\:sin(bt), -ab^2c\:cos(bt),a^2b^3 \big>$
I took the magnitude and got:
$ab^2\sqrt{a^b+b^2+1}$

I agree with the cross product, but check that last line. I get a different value.