Daniel_Feldman
Full Member
- Joined
- Sep 30, 2005
- Messages
- 252
Solved. Thanks.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Partial deriv's Q (Solved)
- Thread starter Daniel_Feldman
- Start date
Sometimes it may help to construct one of those tree diagrams you've seen.
\(\displaystyle \L\\s=x(t,u), \;\ t=y(s,u), \;\ u=z(s,t)\)
\(\displaystyle \L\\G(x(t,u), y(s,u), z(s,t))\)
\(\displaystyle \L\\\frac{\partial{du}}{\partial{ds}}=\frac{\partial{du}}{\partial{dz}}\cdot\frac{\partial{dz}}{\partial{ds}}\)
\(\displaystyle \L\\\frac{\partial{dt}}{\partial{du}}=\frac{\partial{dt}}{\partial{dy}}\cdot\frac{\partial{dy}}{\partial{du}}\)
\(\displaystyle \L\\\frac{\partial{ds}}{\partial{dt}}=\frac{\partial{ds}}{\partial{dx}}\cdot\frac{\partial{dx}}{\partial{dt}}\)
\(\displaystyle \L\\s=x(t,u), \;\ t=y(s,u), \;\ u=z(s,t)\)
\(\displaystyle \L\\G(x(t,u), y(s,u), z(s,t))\)
\(\displaystyle \L\\\frac{\partial{du}}{\partial{ds}}=\frac{\partial{du}}{\partial{dz}}\cdot\frac{\partial{dz}}{\partial{ds}}\)
\(\displaystyle \L\\\frac{\partial{dt}}{\partial{du}}=\frac{\partial{dt}}{\partial{dy}}\cdot\frac{\partial{dy}}{\partial{du}}\)
\(\displaystyle \L\\\frac{\partial{ds}}{\partial{dt}}=\frac{\partial{ds}}{\partial{dx}}\cdot\frac{\partial{dx}}{\partial{dt}}\)
Daniel_Feldman
Full Member
- Joined
- Sep 30, 2005
- Messages
- 252
Wouldn't that yield -1, not 1? Or am I not seeing something?