Question

I have two calculus 3 questions help please
1)Find the curvature k(t) of the curve r(t)=(4sint)I+(4sint)j+(4cost)k?
2)find the arclength of the curve r(t)=<8sqrt(2t),e^(8t),e^(-8t)>,0 10 years ago

Answer 1

I think k(t)=|r^1(t)xr^11(t)|/|r^1(t)|^3 is for the curvature
but I keep getting it wrong

Answer 2

You think \[
\kappa (t) = \frac{\|\mathbf r'(t)\times \mathbf r''(t)\|}{\|\mathbf r'(t)\|^3}
\]?

Answer 3

What did you get for \(\mathbf r'(t)\) and \(\mathbf r''(t)\)?

Answer 4

for r^1(t) I got =4cos(t)I+4cos(t)j-4sin(t)k
for r^11(t)=-4sin(t)I-4sin(t)j+4cos(t)k

Answer 5

that looks right...

Answer 6

do you need help?

Answer 7

yes i need help on the second one

Answer 8

I think the vector in the second problem might supposed to be\[r(t)=<8\sqrt{2}t,e ^{8t},e ^{-8t}>\]

Answer 9

\[Arclength = \sqrt{(\frac{ dr _{x} }{ dt })^{2}+(\frac{ dr _{y} }{ dt })^{2}+(\frac{ dr _{z} }{ dt })^{2}}\]

Answer 10

r'(t) = (-4 cos t, -4 cos t, 4 sin t)
r''(t) = (4 sin t, 4 sin t, 4 cos t)

r' x r'' = (-16, 16, 0).

==> k = ||r' x r''|| / ||r'||^3
k = (16 sqrt(2)) / (4^3 sqrt(1 + cos^2(t))) = sqrt(2)/(4sqrt(1 + cos^2(t))).

Answer 11

Integrate my previous post over your bounds of t to find the whole arclength