Question

If u=f(x,y) where x=(e^s)(cos(t)) and y=(e^s)(sin(t)), show that (du/dx)^2 + (du/dy)^2 = (e^(-2s))((du/ds)^2+(du/dt)^2))

Answer 1

start with the right hand side..expand with

\[\frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\]
and
\[\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\]

Answer 2

then what?

Answer 3

then calculate \[\frac{\partial x}{\partial s},\frac{\partial x}{\partial t},\frac{\partial y}{\partial s},\text{ and }\frac{\partial y}{\partial t}\] Substitute into your equation and simplify.

Answer 4

okay I've tried getting it but at some point i get lost and no where near proving what I'm supposed to prove

Answer 5

did you get

\[\frac{\partial x}{\partial s}=e^s\cos(t),\frac{\partial x}{\partial t}=-e^s\sin(t)\]
\[\frac{\partial y}{\partial s}=e^s\sin(t),\frac{\partial y}{\partial t}=e^s\cos(t)\]

Answer 6

yeah i got to that step

Answer 7

\[(u_xx_t+u_yy_t)^2=(u_xx_t)^2+2u_xu_yx_ty_t+(u_yy_t)^2\]
\[=(u_x(-e^s\sin(t)))^2+2u_xu_y(-e^s\sin(t))e^s\cos(t)+(u_ye^s\cos(t))^2\]
\[(u_xx_s+u_yy_s)^2=(u_xe^s\cos(t))^2+2u_xu_ye^s\cos(t)e^s\sin(t)+(u_ye^s\sin(t))^2\]

add


\[(u_xx_t+u_yy_t)^2+(u_xx_s+u_yy_s)^2\]
\[=(u_x(-e^s\sin(t)))^2+(u_ye^s\cos(t))^2+(u_xe^s\cos(t))^2+(u_ye^s\sin(t))^2\]

\[=u_x^2e^{2s}[\sin^2(t)+\cos^2(t)]+u_y^2e^{2s}[\sin^2(t)+\cos^2(t)]\]

Answer 8

okay i see where i made the mistake now...thanks!

Answer 9

np