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I have no idea how to calculate $z_x+z_y$ at a point $\left( \frac{\pi +3}{3}, \frac{\pi+1}{2}\right)$, if $z=uv^2$ and $x=u+sinv$, $y=v+cosu$. $z$ is not expressed in terms of $x$ and $y$. Maybe it is meant to be solved as $x_u=1$ and $x_v=cosv$, $y_u=-sinu$ and $y_v=1,$ then $x_u=cos^2v$ and $y_u=-sinu$.

$$z_u=v^2 \Rightarrow v^2= \frac{\partial z}{\partial x} \cdot 1+\frac{\partial z}{\partial y} \cdot (-sinu)$$

$$z_v=2uv \Rightarrow 2uv=\frac{\partial z}{\partial x} cosv+\frac{\partial z}{\partial y}\cdot 1$$

$$\Rightarrow \frac{\partial z}{\partial x}=v^2+sin(u) \frac{\partial z}{\partial y}$$

$$\Rightarrow 2uv=cos(v)\left( v^2+sin(u)\frac{\partial z}{\partial y} \right)+\frac{\partial z}{\partial y}$$

$$\Rightarrow 2uv-v^2cos(v)=\frac{\partial z}{\partial y}\left(sin(u)+1 \right)$$

$$\Rightarrow \frac{\partial z}{\partial y}=\frac{2uv-v^2cos(v)}{sin(u)+1}$$

I know that what I have done looks just confusing.

$$v^2=\frac{\partial z}{\partial x}+\frac{\sqrt{3}}{2}\Rightarrow \frac{\partial z}{\partial x}=\frac{\pi^2}{9}-\frac{\sqrt{3}}{2}$$

$$2uv=\frac{\partial z}{\partial y} \Rightarrow \frac{\partial z}{\partial y}=\frac{\pi^2}{18}$$

$$z_x+z_y=\frac{3\pi^2-9\sqrt{3}}{18}$$

Quanto
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user
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  • I take it $z_x$ means "partial derivative of $z$ with respect to $x$." There's a chain rule which says, among other things, $z_x=z_uu_x+z_vv_x$, but this may be of limited use since you aren't given $x$ and $y$ as functions of $u$ and $v$. – Gerry Myerson Jan 12 '20 at 15:03
  • I think you might be overcomplicating things. Notice that when $(u,v)=\left(\pi/3, \pi/2 \right)$ then $(x,y)=\left( \pi/3+1, \pi/2 +1/2 \right)$. Instead of extracting the desired partial derivatives and then substituting, substitute the appropriate $u$ and $v$ and then try to extract from the new equations the term $z_x+z_y$. – Keen-ameteur Jan 12 '20 at 17:17
  • @Keen-ameteur The result, that I got still doesn't look nice. I wonder how did you arrived at the right $u,v$ values? – user Jan 12 '20 at 21:54

2 Answers2

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Differentiate $u+\sin v=x$ and $v+\cos u=y$ with respect to $x$ and $y$, respectively,

$$u_x+\cos v\> v_x=1,\>\>\>\>\> v_x-\sin u\> u_x=0,\>\>\>\>\> u_y+\cos v\> v_y = 0,\>\>\>\>\> v_y-\sin u\> u_y = 1$$

Solve the linear equations,

$$u_x= \frac1{1+\sin u\cos v},\>\>\>\>\>v_x= \frac{\sin u}{1+\sin u\cos v}$$ $$u_y= -\frac{\cos v}{1+\sin u\cos v},\>\>\>\>\>v_y= \frac{1}{1+\sin u\cos v}$$

Then, from $z=uv^2$, evaluate

$$z_x+z_y =( z_u u_x+z_v v_x)+( z_u u_y+z_v v_y)=\frac{v^2(1-\cos v)+2uv(1+\sin u)}{1+\sin u\cos v}\tag 1$$

Given the point $(x,y)=(\frac{\pi+3}3,\frac{\pi+1}2)$, set

$$\frac{\pi+3}3=u+\sin v,\>\>\>\>\>\frac{\pi+1}2 = v+\cos u$$

to get $(u,v) = (\frac\pi3,\frac\pi2)$. Plug the point into the expression (1) to obtain the result

$$z_x+z_y =\frac{7+2\sqrt3}{12}\pi^2$$

Quanto
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Hint:

Define $Z(u,v)=z\big( x(u,v),y(u,v) \big)$. By the chain rule (assuming the conditions hold) you can write:

$$ \frac{\partial Z}{\partial u}(u,v)= \frac{\partial z}{\partial x}\big( x(u,v),y(u,v) \big) \cdot \frac{\partial x}{\partial u}(u,v)+ \frac{\partial z}{\partial y}\big( x(u,v),y(u,v) \big) \cdot \frac{\partial y}{\partial u}(u,v)$$

and:

$$ \frac{\partial Z}{\partial v}(u,v)= \frac{\partial z}{\partial x}\big( x(u,v),y(u,v) \big) \cdot \frac{\partial x}{\partial v}(u,v)+ \frac{\partial z}{\partial y}\big( x(u,v),y(u,v) \big) \cdot \frac{\partial y}{\partial v}(u,v) $$

And using this set of linear equations you want to find what is:

$$ \frac{\partial z}{\partial x}\left( \frac{\pi +3}{3}, \frac{\pi+1}{2}\right)+ \frac{\partial z}{\partial y}\left( \frac{\pi +3}{3}, \frac{\pi+1}{2}\right) $$

Keen-ameteur
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  • @Keen_ameteur I still don't get how do I partially derive $z$ by $x$, if $z$ is not given as a function of $x$? – user Jan 12 '20 at 15:18
  • @user: do as Keen-amateur says, plug in $x_u, y_u, x_v, y_v $ and then solve his two equations for $z_x, z_y$ in terms of $Z_u, Z_v$ (which you know, $Z=uv^2$). Then you've got $z_x + z_y$ and can plug in the specific point. – ancient mathematician Jan 12 '20 at 16:02
  • @ancientmathematician Look up what I have done, it seems totally wrong though. – user Jan 12 '20 at 16:28