University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.1 - Rates of Change and Tangents to Curves - Exercises - Page 56: 5



Work Step by Step

$R(\theta)=\sqrt{4(\theta)+1} , [0,2]$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{R(\theta_{2})-R(\theta_{1})}{\theta_{2}-\theta_{1}}$$ $R(\theta_{2})=R(2)=\sqrt{4(2)+1}$ $R(2)=\sqrt{9}$ $R(2)=3$ $R(\theta_{1})=R(0)=\sqrt{4(0)+1}$ $R(0)=\sqrt{1}$ $R(0)=1$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{R(2)-R(0)}{\theta_{2}-\theta_{1}}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{3-1}{2-0}$$ $$\frac{\Delta{y}}{\Delta{x}}=\frac{2}{2}=1$$
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.