involved derivative

• Jan 12th 2009, 10:32 PM
tombrownington
involved derivative
Let x, y and z be given parametrically by the arbitrary variables n & m:

x = ncos(m) + f'(m)sin(m);
y = nsin(m) - f'(m)cos(m);
z = n + f(m);

where f is a function of m, assumed to be derivable twice.

Then what are the partial derivates of z, with respect to the two variables x and y?
• Jan 12th 2009, 10:50 PM
Mathstud28
Quote:

Originally Posted by tombrownington
Let x, y and z be given parametrically by the arbitrary variables n & m:

x = ncos(m) + f'(m)sin(m);
y = nsin(m) - f'(m)cos(m);
z = n + f(m);

where f is a function of m, assumed to be derivable twice.

Then what are the partial derivates of z, with respect to the two variables x and y?

Maybe someone more knowedgable will come and answer this, but I have a question of clarification. You do not have $\displaystyle z$ explicitly as a function of $\displaystyle x,y$. Is that intentional? Are we supposed to solve for $\displaystyle z$?
• Jan 12th 2009, 11:10 PM
tombrownington
yup. that's the way it is. z is given in function of u and v, not x and y.
solving for u and v is impracticable since the function f(m) is introduced in z while its derivative is given in x and y.
• Jan 13th 2009, 01:57 AM
tombrownington
Can one at least write:

$\displaystyle \frac {\partial z} {\partial x} = \frac {\partial z} {\partial u} . \frac {\partial u} {\partial x} + \frac {\partial z} {\partial v} . \frac {\partial v} {\partial x}$