Question

Find the unit tangent vector T and the curvature \kappa for the following parameterized curve.\\ $r(t) = \left( \int_0^t \cos \frac{\pi u^2}{22} du, \int_0^t \sin \frac{\pi u^2}{22} du \right), t > 0$\\ T = $\langle \Box, \Box \rangle$\\ $\kappa = \Box$

          Find the unit tangent vector T and the curvature \kappa for the following parameterized curve.\\
$r(t) = \left( \int_0^t \cos \frac{\pi u^2}{22} du, \int_0^t \sin \frac{\pi u^2}{22} du \right), t > 0$\\
T = $\langle \Box, \Box \rangle$\\
$\kappa = \Box$

Find the unit tangent vector T and the curvature κfor the following parameterized curve.

r(t) = ( ∫0^t cos(π u^2)/(22) du, ∫0^t sin(π u^2)/(22) du ), t > 0

T = ⟨, ⟩

κ =

Added by Joseph F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert James Kiss

12/02/2022

Step 1

The velocity vector v(t) is given by the derivative of r(t) with respect to t. v(t) = (d/dt) (t^2 u cos(tu), sin(tu)) = (2tu cos(tu) - t^2 u sin(tu), u cos(tu)) Show more…

Show all steps

Thanks for your feedback!

Please help as soon as possible!!!! Thank you! Find the unit tangent vector T and the curvature k for the following parameterized curve t tu cos du, sin du),t>0 22 22 0 r(t) =