- #1
jamesbob
- 63
- 0
This is annoying me as i have the answer on the tip of my pen, just can't write it down. I'm not 100% sure i understand what the question is asking me to do.
Consider the quantity [tex] u = e^{-xy} [/tex] where (x,y) moves in time t along a path:
Use a method based on partial derivatives to calculate [tex] \frac{du}{dt} [/tex] as a function of x, y and t.
My answer:
I partially differentiated u, getting:
[tex] \frac{\delta{u}}{\delta{x}} = -ye^{-xy} [/tex]
[tex] \frac{\delta{u}}{\delta{y}} = -xe^{-xy} [/tex]
So does this mean [tex] du = -ye^{-xy} + -xe^{-xy} ? [/tex]
I though that i would get du from the part iv just explained, then get dt from differentiating x and y. But this ofcourse leaves me with expressions for dx/dt and dy/dt. Where do i go from here?
Consider the quantity [tex] u = e^{-xy} [/tex] where (x,y) moves in time t along a path:
[tex]x = \cosh{t}, \mbox{ } y = \sinh{t}[/tex]
Use a method based on partial derivatives to calculate [tex] \frac{du}{dt} [/tex] as a function of x, y and t.
My answer:
I partially differentiated u, getting:
[tex] \frac{\delta{u}}{\delta{x}} = -ye^{-xy} [/tex]
[tex] \frac{\delta{u}}{\delta{y}} = -xe^{-xy} [/tex]
So does this mean [tex] du = -ye^{-xy} + -xe^{-xy} ? [/tex]
I though that i would get du from the part iv just explained, then get dt from differentiating x and y. But this ofcourse leaves me with expressions for dx/dt and dy/dt. Where do i go from here?