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 960: 46

Answer

$${H_p}\left( {p,q,r} \right) = 2p\sqrt {q + r} ,{\text{ }}{H_q}\left( {p,q,r} \right) = \frac{{{p^2}}}{{2\sqrt {q + r} }}{\text{ and}}\,\,\,\,{H_r}\left( {p,q,r} \right) = \frac{{{p^2}}}{{2\sqrt {q + r} }}$$

Work Step by Step

$$\eqalign{ & H\left( {p,q,r} \right) = {p^2}\sqrt {q + r} \cr & {\text{Find the partial derivatives }}{H_p}\left( {p,q,r} \right){\text{,}}\,\,{H_q}\left( {p,q,r} \right){\text{ and }}{H_r}\left( {p,q,r} \right) \cr & {H_p}\left( {p,q,r} \right) = \frac{\partial }{{\partial p}}\left[ {{p^2}\sqrt {q + r} } \right] \cr & {\text{treat }}q{\text{ and }}r{\text{ as a constant}} \cr & {H_p}\left( {p,q,r} \right) = \sqrt {q + r} \frac{\partial }{{\partial p}}\left[ {{p^2}} \right] \cr & {\text{then}} \cr & {H_p}\left( {p,q,r} \right) = \sqrt {q + r} \left( {2p} \right) \cr & {H_p}\left( {p,q,r} \right) = 2p\sqrt {q + r} \cr & \cr & {H_q}\left( {p,q,r} \right) = \frac{\partial }{{\partial q}}\left[ {{p^2}\sqrt {q + r} } \right] \cr & {\text{treat }}q{\text{ and }}r{\text{ as a constant}} \cr & {H_q}\left( {p,q,r} \right) = {p^2}\frac{\partial }{{\partial q}}\left[ {{{\left( {q + r} \right)}^{1/2}}} \right] \cr & {\text{then by the chain rule}} \cr & {H_q}\left( {p,q,r} \right) = {p^2}\left( {\frac{1}{2}} \right){\left( {q + r} \right)^{ - 1/2}}\frac{\partial }{{\partial q}}\left[ {q + r} \right] \cr & {H_q}\left( {p,q,r} \right) = {p^2}\left( {\frac{1}{2}} \right){\left( {q + r} \right)^{ - 1/2}} \cr & {H_q}\left( {p,q,r} \right) = \frac{{{p^2}}}{{2\sqrt {q + r} }} \cr & \cr & and \cr & \cr & {H_r}\left( {p,q,r} \right) = \frac{\partial }{{\partial r}}\left[ {{p^2}\sqrt {q + r} } \right] \cr & {\text{treat }}q{\text{ and }}r{\text{ as a constant}} \cr & {H_r}\left( {p,q,r} \right) = {p^2}\frac{\partial }{{\partial r}}\left[ {{{\left( {q + r} \right)}^{1/2}}} \right] \cr & {\text{then by the chain rule}} \cr & {H_r}\left( {p,q,r} \right) = {p^2}\left( {\frac{1}{2}} \right){\left( {q + r} \right)^{ - 1/2}}\frac{\partial }{{\partial r}}\left[ {q + r} \right] \cr & {H_r}\left( {p,q,r} \right) = {p^2}\left( {\frac{1}{2}} \right){\left( {q + r} \right)^{ - 1/2}} \cr & {H_r}\left( {p,q,r} \right) = \frac{{{p^2}}}{{2\sqrt {q + r} }} \cr} $$
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