Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 13 - Vector Functions - 13.3 Exercises - Page 884: 28

Answer

$\dfrac{2 \sec^2 x \tan x}{(1+\sec^4 x)^{3/2}}$

Work Step by Step

Given: $y=\tan x$ Consider $f(x)=y=\tan x$ In order to find the curvature we will have to use formula 11, such that $\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ $y'=\sec^2 x$ and $y''=2 \sec^2 x \tan x$ $|y''|=2 \sec^2 x \tan x$ $\kappa(x)=\dfrac{|2 \sec^2 x \tan x|}{[1+(\sec^2 x)^2]^{3/2}}$ Hence, $\kappa(x)=\dfrac{2 \sec^2 x \tan x}{(1+\sec^4 x)^{3/2}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.