Possible typing error in the book












2












$begingroup$


One of the side effects of learning on your own is to seriously doubt if the book has an error or your reasoning is incorrect. My book on multivariable calculus (Rogawski 3rd Edition of Calculus Multivariable) has a problem at the end of Chapter 15:




Let $g(u, v) = f(u^3-v^3, v^3 - u^3)$. Prove that,
$$ v^2
frac{partial g}{partial u} - u^2 frac{partial g}{partial v} = 0
$$




I proceed as following:
Consider the functions $x(u, v) = u^3 - v^3$ and $y(u,v) = v^3 - u^3$. I can now say $g(u, v) = f(x, y)$. And therefore,
$$
frac{partial g}{partial u} = frac{partial f}{partial x}frac{partial x}{partial u} + frac{partial f}{partial y}frac{partial y}{partial u} \
frac{partial g}{partial v} = frac{partial f}{partial x}frac{partial x}{partial v} + frac{partial f}{partial y}frac{partial y}{partial v}
$$



Now,
$$
frac{partial x}{partial u} = 3u^2 \
frac{partial x}{partial v} = -3v^2 \
frac{partial y}{partial u} = -3u^2 \
frac{partial y}{partial v} = 3v^2
$$
Therefore,



begin{align}
& frac{partial g}{partial u} = 3u^2frac{partial f}{partial x} - 3u^2 frac{partial f}{partial y} \
implies & frac{1}{3u^2} frac{partial g}{partial u} = frac{partial f}{partial x} - frac{partial f}{partial y} \
& frac{partial g}{partial v} = -3v^2frac{partial f}{partial x} + 3v^2frac{partial f}{partial y} \
implies & frac{1}{3v^2} frac{partial g}{partial v} = -frac{partial f}{partial x} + frac{partial f}{partial y}
end{align}



Adding the two up I get $displaystyle v^2frac{partial g}{partial u} + u^2frac{partial g}{partial v} = 0$ which is different from what I am asked to prove by a minus sign.



Am I wrong or is there a typing error in the book?






multivariable-calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which book are you referring to? Please edit the question to include the details.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:18






  • 1




    $begingroup$
    Let's do an example. Say $f(x,y)=y$. in that case $g(u,v)=v^3-u^3$. Then $frac {partial g}{partial u}=-3u^2$ and $frac {partial g}{partial v}=3v^2$. The text then claims that $-3v^2u^2-(3u^2v^2)=0$ which is false.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:28








  • 2




    $begingroup$
    General note: as you surmise, math texts tend to have lots of typos and sometimes even deeper errors. Keeps you on your toes.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:29










  • $begingroup$
    Thanks @lulu. Your example pretty much answers my question. You should post it as an answer and I will accept it.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47






  • 1




    $begingroup$
    @Shaun, edited as requested.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47
















2












$begingroup$


One of the side effects of learning on your own is to seriously doubt if the book has an error or your reasoning is incorrect. My book on multivariable calculus (Rogawski 3rd Edition of Calculus Multivariable) has a problem at the end of Chapter 15:




Let $g(u, v) = f(u^3-v^3, v^3 - u^3)$. Prove that,
$$ v^2
frac{partial g}{partial u} - u^2 frac{partial g}{partial v} = 0
$$




I proceed as following:
Consider the functions $x(u, v) = u^3 - v^3$ and $y(u,v) = v^3 - u^3$. I can now say $g(u, v) = f(x, y)$. And therefore,
$$
frac{partial g}{partial u} = frac{partial f}{partial x}frac{partial x}{partial u} + frac{partial f}{partial y}frac{partial y}{partial u} \
frac{partial g}{partial v} = frac{partial f}{partial x}frac{partial x}{partial v} + frac{partial f}{partial y}frac{partial y}{partial v}
$$



Now,
$$
frac{partial x}{partial u} = 3u^2 \
frac{partial x}{partial v} = -3v^2 \
frac{partial y}{partial u} = -3u^2 \
frac{partial y}{partial v} = 3v^2
$$
Therefore,



begin{align}
& frac{partial g}{partial u} = 3u^2frac{partial f}{partial x} - 3u^2 frac{partial f}{partial y} \
implies & frac{1}{3u^2} frac{partial g}{partial u} = frac{partial f}{partial x} - frac{partial f}{partial y} \
& frac{partial g}{partial v} = -3v^2frac{partial f}{partial x} + 3v^2frac{partial f}{partial y} \
implies & frac{1}{3v^2} frac{partial g}{partial v} = -frac{partial f}{partial x} + frac{partial f}{partial y}
end{align}



Adding the two up I get $displaystyle v^2frac{partial g}{partial u} + u^2frac{partial g}{partial v} = 0$ which is different from what I am asked to prove by a minus sign.



Am I wrong or is there a typing error in the book?






multivariable-calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which book are you referring to? Please edit the question to include the details.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:18






  • 1




    $begingroup$
    Let's do an example. Say $f(x,y)=y$. in that case $g(u,v)=v^3-u^3$. Then $frac {partial g}{partial u}=-3u^2$ and $frac {partial g}{partial v}=3v^2$. The text then claims that $-3v^2u^2-(3u^2v^2)=0$ which is false.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:28








  • 2




    $begingroup$
    General note: as you surmise, math texts tend to have lots of typos and sometimes even deeper errors. Keeps you on your toes.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:29










  • $begingroup$
    Thanks @lulu. Your example pretty much answers my question. You should post it as an answer and I will accept it.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47






  • 1




    $begingroup$
    @Shaun, edited as requested.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47














2












2








2





$begingroup$


One of the side effects of learning on your own is to seriously doubt if the book has an error or your reasoning is incorrect. My book on multivariable calculus (Rogawski 3rd Edition of Calculus Multivariable) has a problem at the end of Chapter 15:




Let $g(u, v) = f(u^3-v^3, v^3 - u^3)$. Prove that,
$$ v^2
frac{partial g}{partial u} - u^2 frac{partial g}{partial v} = 0
$$




I proceed as following:
Consider the functions $x(u, v) = u^3 - v^3$ and $y(u,v) = v^3 - u^3$. I can now say $g(u, v) = f(x, y)$. And therefore,
$$
frac{partial g}{partial u} = frac{partial f}{partial x}frac{partial x}{partial u} + frac{partial f}{partial y}frac{partial y}{partial u} \
frac{partial g}{partial v} = frac{partial f}{partial x}frac{partial x}{partial v} + frac{partial f}{partial y}frac{partial y}{partial v}
$$



Now,
$$
frac{partial x}{partial u} = 3u^2 \
frac{partial x}{partial v} = -3v^2 \
frac{partial y}{partial u} = -3u^2 \
frac{partial y}{partial v} = 3v^2
$$
Therefore,



begin{align}
& frac{partial g}{partial u} = 3u^2frac{partial f}{partial x} - 3u^2 frac{partial f}{partial y} \
implies & frac{1}{3u^2} frac{partial g}{partial u} = frac{partial f}{partial x} - frac{partial f}{partial y} \
& frac{partial g}{partial v} = -3v^2frac{partial f}{partial x} + 3v^2frac{partial f}{partial y} \
implies & frac{1}{3v^2} frac{partial g}{partial v} = -frac{partial f}{partial x} + frac{partial f}{partial y}
end{align}



Adding the two up I get $displaystyle v^2frac{partial g}{partial u} + u^2frac{partial g}{partial v} = 0$ which is different from what I am asked to prove by a minus sign.



Am I wrong or is there a typing error in the book?






multivariable-calculus derivatives







share|cite|improve this question











$endgroup$




One of the side effects of learning on your own is to seriously doubt if the book has an error or your reasoning is incorrect. My book on multivariable calculus (Rogawski 3rd Edition of Calculus Multivariable) has a problem at the end of Chapter 15:




Let $g(u, v) = f(u^3-v^3, v^3 - u^3)$. Prove that,
$$ v^2
frac{partial g}{partial u} - u^2 frac{partial g}{partial v} = 0
$$




I proceed as following:
Consider the functions $x(u, v) = u^3 - v^3$ and $y(u,v) = v^3 - u^3$. I can now say $g(u, v) = f(x, y)$. And therefore,
$$
frac{partial g}{partial u} = frac{partial f}{partial x}frac{partial x}{partial u} + frac{partial f}{partial y}frac{partial y}{partial u} \
frac{partial g}{partial v} = frac{partial f}{partial x}frac{partial x}{partial v} + frac{partial f}{partial y}frac{partial y}{partial v}
$$



Now,
$$
frac{partial x}{partial u} = 3u^2 \
frac{partial x}{partial v} = -3v^2 \
frac{partial y}{partial u} = -3u^2 \
frac{partial y}{partial v} = 3v^2
$$
Therefore,



begin{align}
& frac{partial g}{partial u} = 3u^2frac{partial f}{partial x} - 3u^2 frac{partial f}{partial y} \
implies & frac{1}{3u^2} frac{partial g}{partial u} = frac{partial f}{partial x} - frac{partial f}{partial y} \
& frac{partial g}{partial v} = -3v^2frac{partial f}{partial x} + 3v^2frac{partial f}{partial y} \
implies & frac{1}{3v^2} frac{partial g}{partial v} = -frac{partial f}{partial x} + frac{partial f}{partial y}
end{align}



Adding the two up I get $displaystyle v^2frac{partial g}{partial u} + u^2frac{partial g}{partial v} = 0$ which is different from what I am asked to prove by a minus sign.



Am I wrong or is there a typing error in the book?






multivariable-calculus derivatives




multivariable-calculus derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 11:46







user1player1

















asked Dec 8 '18 at 11:14









user1player1user1player1

284




284












  • $begingroup$
    Which book are you referring to? Please edit the question to include the details.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:18






  • 1




    $begingroup$
    Let's do an example. Say $f(x,y)=y$. in that case $g(u,v)=v^3-u^3$. Then $frac {partial g}{partial u}=-3u^2$ and $frac {partial g}{partial v}=3v^2$. The text then claims that $-3v^2u^2-(3u^2v^2)=0$ which is false.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:28








  • 2




    $begingroup$
    General note: as you surmise, math texts tend to have lots of typos and sometimes even deeper errors. Keeps you on your toes.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:29










  • $begingroup$
    Thanks @lulu. Your example pretty much answers my question. You should post it as an answer and I will accept it.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47






  • 1




    $begingroup$
    @Shaun, edited as requested.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47


















  • $begingroup$
    Which book are you referring to? Please edit the question to include the details.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:18






  • 1




    $begingroup$
    Let's do an example. Say $f(x,y)=y$. in that case $g(u,v)=v^3-u^3$. Then $frac {partial g}{partial u}=-3u^2$ and $frac {partial g}{partial v}=3v^2$. The text then claims that $-3v^2u^2-(3u^2v^2)=0$ which is false.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:28








  • 2




    $begingroup$
    General note: as you surmise, math texts tend to have lots of typos and sometimes even deeper errors. Keeps you on your toes.
    $endgroup$
    – lulu
    Dec 8 '18 at 11:29










  • $begingroup$
    Thanks @lulu. Your example pretty much answers my question. You should post it as an answer and I will accept it.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47






  • 1




    $begingroup$
    @Shaun, edited as requested.
    $endgroup$
    – user1player1
    Dec 8 '18 at 11:47
















$begingroup$
Which book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 8 '18 at 11:18




$begingroup$
Which book are you referring to? Please edit the question to include the details.
$endgroup$
– Shaun
Dec 8 '18 at 11:18




1




1




$begingroup$
Let's do an example. Say $f(x,y)=y$. in that case $g(u,v)=v^3-u^3$. Then $frac {partial g}{partial u}=-3u^2$ and $frac {partial g}{partial v}=3v^2$. The text then claims that $-3v^2u^2-(3u^2v^2)=0$ which is false.
$endgroup$
– lulu
Dec 8 '18 at 11:28






$begingroup$
Let's do an example. Say $f(x,y)=y$. in that case $g(u,v)=v^3-u^3$. Then $frac {partial g}{partial u}=-3u^2$ and $frac {partial g}{partial v}=3v^2$. The text then claims that $-3v^2u^2-(3u^2v^2)=0$ which is false.
$endgroup$
– lulu
Dec 8 '18 at 11:28






2




2




$begingroup$
General note: as you surmise, math texts tend to have lots of typos and sometimes even deeper errors. Keeps you on your toes.
$endgroup$
– lulu
Dec 8 '18 at 11:29




$begingroup$
General note: as you surmise, math texts tend to have lots of typos and sometimes even deeper errors. Keeps you on your toes.
$endgroup$
– lulu
Dec 8 '18 at 11:29












$begingroup$
Thanks @lulu. Your example pretty much answers my question. You should post it as an answer and I will accept it.
$endgroup$
– user1player1
Dec 8 '18 at 11:47




$begingroup$
Thanks @lulu. Your example pretty much answers my question. You should post it as an answer and I will accept it.
$endgroup$
– user1player1
Dec 8 '18 at 11:47




1




1




$begingroup$
@Shaun, edited as requested.
$endgroup$
– user1player1
Dec 8 '18 at 11:47




$begingroup$
@Shaun, edited as requested.
$endgroup$
– user1player1
Dec 8 '18 at 11:47










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