# Curvature, More of a cross product question really.

I've been using various methods of finding curvature, and using the forumla

$$K=\frac{||\vec{r}'X \vec{r}''||}{||\vec{r}'||^3}$$

I took the cross product of my two vectors and came up with (just assume this is correct, if my question isn't answered i'll post all related work)

$$128 \pi^3 sin(2 \pi t)^2 - -128 \pi^3 cos(2 \pi t)^2$$

My question is: Can I factor out the 128$\pi^3$ and then use a trig identity to make sin^2 + cos^2 =1?

I assumed that since these were components of vectors I couldn't do this. However, if I do, I get the correct answer. I left them as is, and then took the magnitude of the resulting cross product.

Which produced

$$128 \pi^3 \sqrt{sin(2 \pi t)^4+cos(2 \pi t)^4}$$

This would also produce the correct answer if sin(x)^n +cos(x)^n =1, but I'm fairly certain it doesn't.

Any help would be greatly appreciated. Thanks

## Answers and Replies

Dick
Science Advisor
Homework Helper
Well, sin(x)^4+cos(x)^4 is NOT 1. Yes, I think you'll need to post more work.

HallsofIvy
Science Advisor
Homework Helper
I've been using various methods of finding curvature, and using the forumla

$$K=\frac{||\vec{r}'X \vec{r}''||}{||\vec{r}'||^3}$$

I took the cross product of my two vectors and came up with (just assume this is correct, if my question isn't answered i'll post all related work)

$$128 \pi^3 sin(2 \pi t)^2 - -128 \pi^3 cos(2 \pi t)^2$$
Do you mean you got this as one of the components of the cross product or do you mean, rather, something like
$$128\pi^3 sin(2\pi t)^2\vec{i} - 128 \pi^2 cos(2\pi t)^2\vec{j}$$?
If the latter, you can certainly factor out the $128\pi^3$ but not combine the different components.

[/itex]My question is: Can I factor out the 128$\pi^3$ and then use a trig identity to make sin^2 + cos^2 =1?

I assumed that since these were components of vectors I couldn't do this. However, if I do, I get the correct answer. I left them as is, and then took the magnitude of the resulting cross product.

Which produced

$$128 \pi^3 \sqrt{sin(2 \pi t)^4+cos(2 \pi t)^4}$$

This would also produce the correct answer if sin(x)^n +cos(x)^n =1, but I'm fairly certain it doesn't.

Any help would be greatly appreciated. Thanks

The original question is: Find the curvature of

$$\vec{r}(t) = 4cos(2 \pi t)\vec{i} + 4sin(2 \pi t)\vec{j}$$

I have a solutions guide and found the curvature using

$$K = \frac{||\vec{T}'(t)||}{||\vec{r}'(t)||}$$

but I wanted to try the cross product method to see if it is easier, and just for some review.

$$\vec{r}'(t) = -8 \pi sin(2 \pi t)\vec{i} + 8 \pi cos(2 \pi t)\vec{j}$$

$$\vec{r}''(t) = -16 \pi^2 cos(2 \pi t)\vec{i} - 16 \pi^2 sin(2 \pi t)\vec{j}$$

$$\vec{r}'(t) X \vec{r}''(t) = 128 \pi^3 sin(2 \pi t )^2\vec{i} - - 128 \pi^3 cos(2 \pi t)^2\vec{j}$$

$$|| \vec{r}'(t) \crcoss \vec{r}''(t) || = \sqrt{(128 \pi^3 sin(2 \pi t )^2)^2+(28 \pi^3 cos(2 \pi t)^2)^2}$$

$$||\vec{r}'(t)|| = \sqrt{(-8 \pi sin(2 \pi t))^2+(8 \pi cos(2 \pi t))^2} = 8 \pi$$

$$K=\frac{||\vec{r}'(t)X \vec{r}'(t)||}{||\vec{r}'(t)||^3}=\frac{128 \pi^3 \sqrt{sin(2 \pi t)^4+cos(2 \pi t)^4}}{(8 \pi)^3}$$

As I typed all this out, I realized that I took the cross product of a 2 X 2 matrix, and probably should have taken a 3 X 3 cofactor expansion with vectors $\vec{i},\vec{j}$ and $\vec{k}$

so i'll try that and see if it works, until then this is the problem in its entirety.

Actually, looking back over the problem, I think I assumed my 2 X 2 cross product was a cofactor expansion, where it should have been a constant. This would have allowed me to simplify that result and get the $128 \pi^3$ I needed in the numerator.

Cofactor expansion results in $0 \vec{i} + 0\vec{j} + 128 \pi^3 \vec{k}$ which if I take the magnitude of that vector I get the result I need.

Do I have this somewhat straightened out not?

gabbagabbahey
Homework Helper
Gold Member
Actually, looking back over the problem, I think I assumed my 2 X 2 cross product was a cofactor expansion, where it should have been a constant. This would have allowed me to simplify that result and get the $128 \pi^3$ I needed in the numerator.

Cofactor expansion results in $0 \vec{i} + 0\vec{j} + 128 \pi^3 \vec{k}$ which if I take the magnitude of that vector I get the result I need.

Do I have this somewhat straightened out not?

That looks much better, yes.