Question

Find the curvature K of the curve: r(t) = 4cos(2pit)i + 4sin(2pit)j

I'm confused on how they got -sin(2pit)i+cos(2t)j for T(t)

Answer 1

\[k=\frac{|r'xr' '|}{|r'|}\]sounds familiar to me

Answer 2

\[r(t)=4\cos2\pi t +4\sin2\pi t\]

Answer 3

The tangent vector T(t) is the normal derivative of r(t); i.e.,
\[ T(t) = \frac{r'(t)}{||r'(t)||} \] Calculate that out and you'll see how they arrived at their function T(t).

Answer 4

(and notice you have a small typo in your T(t), having dropped a pi in the argument of the coefficient of the unit vector j)

Answer 5

*The tangent vector T(t) is the \( \it normalized \) derivative of r(t)

Answer 6

Recognize the formula for the tangent vector T(t)?

Answer 7

yes

Answer 8

I got ||r'(t)|| = 8 and
r'(t) =\[ -8\sin2\pi ti + 8 \cos2\pi t j\]

Answer 9

Hence when you normalize r'(t) you get the T(t) they are using.