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$\displaystyle \bigg(\frac{\partial u}{\partial y}\bigg)_{x}$ at point $(u,x,y,z)=(5,1-3,1)$ . If it is given $u=x^2y^2+yz-z^3$ and $x^2+y^2+z^2=11$

What i try ::

$\displaystyle \frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\bigg(x^2y^2+yz-z^3\bigg)=2xy+z$

$\displaystyle \bigg(\frac{\partial u}{\partial y}\bigg)_{x}=\frac{\partial }{\partial x}\bigg(2x^2y+z\bigg)=4xy=4(1)(-3)=-12$

I did not understand where i have use the relation $x^2+y^2+z^2=11$ also please tell me is my solution is right

Thanks

jacky
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  • As you already know, $$\bigg( \frac{∂u}{∂y} \bigg)x$$ means the partial derivative of $u$ with respect to $y$, treating $x$ as a constant. Your last computation is $$\bigg( \frac{∂u}{∂y} \bigg){x,z}$$ this being where we'd need the second equation. – Darius Chitu Aug 01 '20 at 18:54
  • Can u please explain me second expression $\displaystyle \bigg(\frac{\partial u}{\partial y}\bigg)_{x,z}$ and How can i use second equation here.Thanks – jacky Aug 01 '20 at 19:03
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    $\bigg( \frac{∂u}{∂y} \bigg)_{x,z}$ is the partial derivative of $u$ as a function of $(x,y,z)$ with respect to $y$ and treating both $x$ and $z$ as constants. You can simply write $z$ as $$z= \sqrt{11-x^2-y^2}$$ and then plug that into $u$. You'll see now that $u$ is just a function of $x$ and $y$ so when you compute the partial w.r.t. $y$ the only variable that'll be treated as a constant will be $x$, thus obtaining the required answer. – Darius Chitu Aug 01 '20 at 19:17
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    In a more general case, say you have a function $$f:U \subseteq \mathbb{R}^n \to V \subseteq \mathbb{R},$$ then $$ \bigg( \frac{∂f}{∂x_i}(x_1,x_2, ..., x_n) \bigg){x_1,x_2, ..., x{i-1}, x_{i+1}, ..., x_n}$$ is the partial derivative of $f$ w.r.t $x_i$ treating ${x_1,x_2, ..., x_{i-1}, x_{i+1}, ..., x_n}$ as constants. (for some $i \in {1,2,...,n }$). – Darius Chitu Aug 01 '20 at 19:21
  • Thanks Darius Chitu got it. – jacky Aug 01 '20 at 19:44

1 Answers1

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Asking for $$\left({\partial u\over \partial y}\right)_x$$ means that you are considering the value of $u$ as a function of $x$ and $y$. This is allowed since in the neighborhood of $(x,y,z):=(1,-3,1)$ the constraint $x^2+y^2+z^2=11$ defines $z$ as $z=\sqrt{11-x^2-y^2}$. It follows that you are actually looking at the function $$u=\psi(x,y):=x^2y^2+y(11-x^2-y^2)^{1/2}-(11-x^2-y^2)^{3/2}\ .$$ Compute now the partial derivative $${\partial\psi\over\partial y}\biggr|_{(1,-3)}\ .$$ During this calculation $x$ is held constant $=1$.

Christian Blatter
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