Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.1 - Functions of Several Variables - 14.1 Exercise: 53


The graph will look like this:

Work Step by Step

Sketch $f(x,y)=x^2 + 9y^2$. We know $f(x,y)=z$ so $z = x^2 + 9y^2$ When looking at this function we know that it will be parabolic. To find the coordinate where the graph crosses the $z$-axis, fill in x = 0 and y = 0. This gives: $z = 0^2 + 9*0^2=0$ So it crosses at $z=0$ Make $x$ and $y$ clear: $x= \sqrt{z-9y^2}$ $y=\sqrt{ \frac{z-x^2}{9}}$ Plugging in 0 for z and y gives $x= \sqrt{0-9\times 0^2}=\sqrt0=0$ Plugging in 0 for z and x gives $y=\sqrt{ \frac{0-0^2}{9}} = \sqrt 0 = 0$ The graph crosses the origin! By filling in more numbers for $x$, $y$, and $z$, you could plot the whole graph. The final graph looks like this:
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