Finding curvature at time $t$, and tangential and normal components

Let $$r(t) = e^t \cos(t)\ \hat{i} + e^t\sin(t)\ \hat{j} + e^t\ \hat{k}$$ be a vector valued function.

Interpret $r$ as the position of a moving object at time $t$: find the curvature of $r$ at time $t$, and determine the tangential and normal components of acceleration.

Any help AT ALL?

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You've asked quite a few questions in the past hour or so. Can you show any work you've done? Where exactly are you stuck? – EuYu Aug 28 '12 at 2:54
How can the position of an object have a curvature? – anthus Aug 28 '12 at 3:00
Do you have a formula for how to find the curvature? If so, what is it? Which parts of that formula are you having issues with? – Thomas Aug 28 '12 at 3:01
A problem with this question is that it's phrased as if the poster is passing on to us verbatim an exercise written by someone other than the poster, instead of asking his own question about that exercise. – Michael Hardy Aug 28 '12 at 3:13

Hint: use this formula to find the curvature:

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

For the other two things, try to take a look here.

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What symbol is that? Kappa? So once the curvature is found the other stuff is tedious work? – Nick Aug 28 '12 at 3:17
@Nick: Yes, the symbol on the left is a kappa that is often used for curvature. Try and take a look at the link that I provided so see how you can find the other things. – Thomas Aug 28 '12 at 13:24