Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.1 Exercises: 15

Answer

The solutions are: a) $g(4,0)=-4.$ b) $g(4,1)=-6.$ c) $g(4,3/2) = -25/4.$ d) $g(3/2,0)=9/4.$

Work Step by Step

Firstly we will calculate the integral and state the function $g(x,y)$ explicitly: $$g(x,y) = \int_x^y (2t-3)dt = 2\int_x^ytdt - 3\int_x^ydt=2\left.\frac{t^2}{2}\right|_x^y-3t\left.\right|_x^y=y^2-x^2-3(y-x).$$ Noting that $y^2-x^2=(x+y)(y-x)$ we can rewrite the previous as: $$g(x,y) = (x+y)(y-x)-3(y-x)= (y-x)(x+y-3).$$ Now we will substitute for $x$ the first and for $y$ the second coordinate from the given point into the upper expression for $g(x,y)$. a) $g(4,0)=(0-4)(4+0-3)=-4.$ b) $g(4,1)=(1-4)(4+1-3)=-6.$ c) $g(4,3/2)=(3/2-4)(4+3/2-3) = -25/4.$ d) $g(3/2,0)=(0-3/2)(3/2+0-3)=9/4.$
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