Find the curvature of the curve

[undrcvrbro]

Homework Statement

Find the curvature $$\kappa(t)$$ of the curve $$\\r(t)=(2sint)i +(2sint)j +(3cost)k$$

Homework Equations

$$\\\k(t)= (\left|T'(t)\right|) / (\left|r'(t)\right|)$$

The Attempt at a Solution

I found $$\\\\r'(t)= (2cost)i + (2cost)j + (-3sint)k$$

$$\\\\\<br /> \left|r'(t)\right|= sqrt((2cost)^2 + (2cost)^2 + (-3sint)^2<br /> <br /> \left|r'(t)\right|=sqrt((4cost)^2+(-3sint)^2)<br /> \left|r'(t)\right|=sqrt(4+-3)<br /> \left|r'(t)\right|=sqrt(1)$$

I think this is where I'm getting caught up. I won't go any further becuas I'm positive I messed up the sin cos relationship when finding the magnitude of r'(t). For all I know, I could have made another mistake along the way.

I understand the equations we are using in this course(Calc III), but I almost always find myself getting caught up on the basic mathematics.

Can any help lead me in the right direction for this problem?

 

[jbunniii]

I don't understand your equation

$$(t) = (|T'(t)|) / (|r'(t)|)$$

The curvature I am familiar with is

$$\kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3}$$

Can you write down the definition of $T(t)$, so we can check whether our definitions are the same? Also, what does the $(t)$ on the left-hand side refer to? I guess it's a typo for $\kappa(t)$?

 

[undrcvrbro]

I don't understand your equation

$$(t) = (|T'(t)|) / (|r'(t)|)$$

The curvature I am familiar with is

$$\kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3}$$

Can you write down the definition of $T(t)$, so we can check whether our definitions are the same? Also, what does the $(t)$ on the left-hand side refer to? I guess it's a typo for $\kappa(t)$?

They are the same.

T(t)= r'(t)/ abs(r(t))

Just a different way of writing it I guess. And yeah, that's k(t).I'm sort of learning latex as I go.

 

[jbunniii]

OK, so let's proceed from there. You made several errors in your calculation of $r'(t)$. It should be

$$\begin{align*}|r'(t)| &= \sqrt{(2 \cos t)^2 + (2 \cos t)^2 + (-3 \sin t)^2} \\<br /> &= \sqrt{4 \cos^2 t + 4 \cos^2 t + 9 \sin^2 t} \\<br /> &= \sqrt{8 \cos^2 t + 9 \sin^2 t}\end{align*}$$

which can't easily be simplified further. (It certainly doesn't equal $\sqrt{8 + 9}$!)

Now can you calculate $T(t)$?

P.S. In case you didn't already know, you can click on any typeset equation in these forums to see the Latex code that produced it. Very useful while learning.

 

[jbunniii]

OK, you can make a bit more simplification than what I wrote:

$$\begin{align*}|r'(t)| &= \sqrt{8 \cos^2 t + 9 \sin^2 t} \\<br /> &= \sqrt{8 \cos^2 t + 8 \sin^2 t + \sin^2 t} \\<br /> &= \sqrt{8 (\cos^2 t + \sin^2 t) + \sin^2 t} \\<br /> &= \sqrt{8 + \sin^2 t}\end{align*}$$

 

[undrcvrbro]

Now can you calculate $T(t)$?

So then would $T(t)$
1/(sqrt(8(sint)^2)) * (2cost, 2cost, -3sint)


So then to find T'(t) you would have to use the quotient rule, right?

P.S. In case you didn't already know, you can click on any typeset equation in these forums to see the Latex code that produced it. Very useful while learning.

Thanks for the tip. Sorry if the normal text is to hard to read. I'm getting frustrated trying to figure out how to put this into latex. It seems to only complicate things more. I'll eventually get the hang of it.