University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 12: 13

Answer

$L(x)=\frac{\sqrt{20x^2-20x+25}}{4}$

Work Step by Step

$\mathrm{See\:the\:figure\:below.}$ When you have any point in a plane, you can use the Pythagorean theorem to determine the distance to the origin. $L=\sqrt{x^2+y^2}$ From the given function $\ 2y+4y=5,\ $ we can express $\ y\ $ as a function of $\ x.$ $2x+4y=5$ $\Rightarrow\ 4y=5-2x$ $\Rightarrow\ y=\frac{5-2x}{4}$ Plug in the expression for $\ y\ $ to get $\ L\ $ as a function of $\ x.$ $L=\sqrt{x^2+(\frac{5-2x}{4})^2}$ $\Rightarrow\ L=\frac{\sqrt{20x^2-20x+25}}{4}$
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