Question

Calculate the arc length parameter function
s(t) for the space curve r(t). Look at attachment.

Answer 1

I can't get the answer in red. I keep getting (t^2)-(t/(2sqrt(2)))

Answer 2

This is the formula I have to use.

Answer 3

I'm not sure this makes it easier, but let A= t^2 so you have
\[ \frac{1}{2\sqrt{2}} < \cos A, \sin A, A> \]
\[ v= \frac{1}{2\sqrt{2}} <-\sin A , \cos A, 1>\]
\[ |v|^2 = \frac{1}{8}( \sin^2 A + \cos^2 A + 1) = \frac{1}{4} \]
and
\[ |v|= \frac{1}{2} \]

Answer 4

\[\frac{1}{2} \int_0^{t^2} \ dA = \frac{1}{2} t^2 \]

Answer 5

but aren't the boundries 0 to t and not t^2

Answer 6

I did a change of variables
A= t^2

if t goes from 0 to T
then A goes from 0^2 to T^2

if you like we can keep the t^2
\[ \frac{1}{2\sqrt{2}} <\cos t^2 , \sin t^2 , t^2>\\
v= \frac{1}{2\sqrt{2}} <-2t \sin t^2 , 2t \cos t^2 , 2t>\\
= \frac{t}{\sqrt{2}}<-\sin t^2 , \cos t^2 , 1>
\]
\[ |v|^2 = \frac{t^2}{2}(\sin^2 t^2 + \cos^2 t^2 + 1)= t^2 \\|v|= t\]

now integrate
\[ \int_0^t t \ dt = \frac{t^2}{2} \]

Answer 7

Thank you!! I see what you did, I appreciate the help.

Answer 8

yw