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: 12

Answer

$P(m)=(\frac{1}{m^2},\frac{1}{m})$

Work Step by Step

$\mathrm{See\: the\:figure\:below}$. Let's say $\ P(a,b)\ $ is any point on the graph of $\ f(x)=\sqrt{x}\ $. If we input $\ x\ $ into the function $\ f(x)=\sqrt{x}\ $, we get the corresponding value of $\ y\ $ to be $\ \sqrt{x}\ $. And, since $\ P\ $ is on the graph of the function $\ \sqrt{x}\ $, we have a relation for its coordinates. $P=(x,\sqrt{x})$ The slope of the line joining the point $\ P\ $ and the origin is $\ m=\frac{\sqrt{x}}{x}.$ After rearranging the equation, we have: $\ m=\frac{\sqrt{x}}{x}$ $\Rightarrow\ m=\frac{1}{\sqrt{x}}$ $\Rightarrow\ \sqrt{x}=\frac{1}{m}$ $\Rightarrow\ x=\frac{1}{m^2}$ So, the coordinates of $\ P\ $ as a function of the slope $\ m\ $ can be written as: $P(m)=(\frac{1}{m^2},\frac{1}{m})$ where $\ m\ $ represents the slope of the line.
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