Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 12 - Functions of Several Veriables - Review Exercises - Page 961: 53

Answer

$$\frac{{\partial w}}{{dr}} = \frac{{3r + s}}{{r\left( {r + s} \right)}}$$ $$\frac{{\partial w}}{{ds}} = \frac{{r + 3s}}{{r\left( {r + s} \right)}}$$ $$\frac{{\partial w}}{{dt}} = \frac{1}{t}$$

Work Step by Step

$$\eqalign{ & w = \ln \left( {x{y^2}} \right),\,\,\,\,\,\,\,x = rst,\,\,\,\,\,\,\,\,y = r + s \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial w}}{{\partial x}},\,\,\,\frac{{\partial w}}{{\partial y}} \cr & \frac{{\partial w}}{{\partial x}} = \frac{{{y^2}}}{{x{y^2}}} = \frac{1}{x} \cr & \frac{{\partial w}}{{\partial y}} = \frac{{2xy}}{{x{y^2}}} = \frac{2}{y} \cr & \cr & {\text{Calculate the derivatives }}\frac{{dx}}{{dr}},\,\,\,\frac{{dy}}{{dr}}{\text{, }}\frac{{dx}}{{ds}},\,\,\,\frac{{dy}}{{ds}}{\text{and }}\frac{{dx}}{{dt}},\,\,\,\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dr}} = st,\,\,\,\,\,\,\frac{{dy}}{{dr}} = 1 \cr & \frac{{dx}}{{ds}} = rt,\,\,\,\,\,\,\frac{{dy}}{{ds}} = 1 \cr & \frac{{dx}}{{dt}} = rs,\,\,\,\,\,\,\frac{{dy}}{{dt}} = 0 \cr & \cr & {\text{Use the chain rule }}\frac{{\partial w}}{{dr}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dr}} + \,\,\frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dr}} \cr & \frac{{\partial w}}{{dr}} = \left( {\frac{1}{x}} \right)\left( {st} \right) + \left( {\frac{2}{y}} \right)\left( 1 \right) \cr & \frac{{\partial w}}{{dr}} = \frac{{st}}{x} + \frac{2}{y} \cr & {\text{Write in terms of }}rst \cr & \frac{{\partial w}}{{dr}} = \frac{{st}}{{rst}} + \frac{2}{{r + s}} \cr & \frac{{\partial w}}{{dr}} = \frac{1}{r} + \frac{2}{{r + s}} \cr & \frac{{\partial w}}{{dr}} = \frac{{3r + s}}{{r\left( {r + s} \right)}} \cr & \cr & {\text{Use the chain rule }}\frac{{\partial w}}{{ds}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{ds}} + \,\,\frac{{\partial w}}{{\partial y}}\frac{{dy}}{{ds}} \cr & \frac{{\partial w}}{{ds}} = \left( {\frac{1}{x}} \right)\left( {rt} \right) + \left( {\frac{2}{y}} \right)\left( 1 \right) \cr & \frac{{\partial w}}{{ds}} = \frac{{rt}}{x} + \frac{2}{y} \cr & {\text{Write in terms of }}rst \cr & \frac{{\partial w}}{{dr}} = \frac{{rt}}{{rst}} + \frac{2}{{r + s}} \cr & \frac{{\partial w}}{{dr}} = \frac{1}{s} + \frac{2}{{r + s}} \cr & \frac{{\partial w}}{{dr}} = \frac{{r + 3s}}{{r\left( {r + s} \right)}} \cr & \cr & {\text{Use the chain rule }}\frac{{\partial w}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \,\,\frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} \cr & \frac{{\partial w}}{{dt}} = \left( {\frac{1}{x}} \right)\left( {rs} \right) + \left( {\frac{2}{y}} \right)\left( 0 \right) \cr & \frac{{\partial w}}{{dt}} = \frac{{rs}}{x} \cr & {\text{Write in terms of }}rst \cr & \frac{{\partial w}}{{dt}} = \frac{{rs}}{{rst}} \cr & \frac{{\partial w}}{{dt}} = \frac{1}{t} \cr} $$
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