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: 14

Answer

$L(y)=\sqrt{y^4-y^2+1}$

Work Step by Step

Let's say, $\ (x,y)\ $ is some random point on the graph of the given function $\ y=\sqrt{x-3}.\ $ Since this random point is on the graph of the function $\ y=\sqrt{x-3}\ $, we can write the coordinates of it as, $\ (x,\sqrt{x-3}).$ Now use the distance formula to find the distance between point $\ (x,\sqrt{x-3})\ \ \mathrm{and}\ \ (4,0).$ $L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $\Rightarrow\ L=\sqrt{(4-x)^2+(0-\sqrt{x-3})^2}$ $\Rightarrow\ L=\sqrt{16-8x+x^2+x-3}$ $\Rightarrow\ L=\sqrt{x^2-7x+13}$ Now write the given function in terms of $\ y\ $. $\ y=\sqrt{x-3}$ $\Rightarrow\ y^2=x-3$ $\Rightarrow\ x=y^2+3$ Put this obtained function into $\ L\ $ to get: $L=\sqrt{(y^2+3)^2-7(y^2+3)+13}$ $\Rightarrow\ L=\sqrt{y^4+6y^2+9-7y^2-21+13}$ $\Rightarrow\ L=\sqrt{y^4-y^2+1}$ $L(y)=\sqrt{y^4-y^2+1}$
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