Question

Find the curvature of r(t)=ti+(1/2)t^2*j+t^2*k.

Answer 1

@hartnn @myininaya @thomaster @iambatman @Compassionate

Answer 2

@Hero @Luigi0210

Answer 3

\[k = \dfrac{|\vec r'(t) \times \vec r ''(t)|}{|r'(t)|^3}\]In your case, \(\vec r' (t) = \hat i + t\hat j + 2t \hat k\) and \(\vec r''(t) = \hat j + 2\hat k\).

Answer 4

So I got r'(t)=<1, t, 2t> and r"(t)=<0, 1, 2>, now what do I do?

Answer 5

Plug that into the formula.

Answer 6

So r'(t)xr"(t)=<0, t, 4t>?

Answer 7

Well, I haven't tried it yet, but if you say so.

Answer 8

So what's \[\left| r'(t)*r"(t) \right|\]

Answer 9

Just follow what the formula tells you, that's really what's to it

Answer 10

That's what I did and got r'(t)xr"(t)=<0, t, 4t>.

Answer 11

It's the magnitude of the vector.

Answer 12

So how do I find the magnitude of the vector?

Answer 13

How do you find the magnitude of vectors?

Answer 14

Determinant, lol.

Answer 15

Holy hell, I just noticed that Idealist didn't calculate the cross product either.

Answer 16

Yup

Answer 17

Do you know how to do the cross product?

Answer 18

No.

Answer 19

You will have a 3 by 3 determinant