The Student Room Group

Partial Differentiation

Question is:



Define

u = x^n * y^n
x = cos(at)
y = sin(at)


find du/dt by substituting for x and y

I've started by putting x and y (the cos and sine functions) raised to the power of n into the equation u, do I just proceed to partially differentiate as normal once I've done this?
Any help appreciated

Cheers
Reply 1
Original post by SwimGood
Question is:



Define

u = x^n * y^n
x = cos(at)
y = sin(at)


find du/dt by substituting for x and y

I've started by putting x and y (the cos and sine functions) raised to the power of n into the equation u, do I just proceed to partially differentiate as normal once I've done this?
Any help appreciated

Cheers


Are you sure you're expected to use partial differentiation for this?

Surely once you've substituted for x and y into the definition of u you can find du/dy using the normal chain rule.
Reply 2
The question is part of a 'Partial Differentiation' problem sheet, so I assume so.

The next part of the question is solving via implicit differentiation.
Reply 3
Original post by SwimGood
The question is part of a 'Partial Differentiation' problem sheet, so I assume so.

The next part of the question is solving via implicit differentiation.


OK, so you should have an equivalent rule in your book for working out du/dt in terms of partial derivatives of u w.r.t.x and w.r.t.y together with the derivatives of x and y w.r.t,t
Reply 4
Original post by SwimGood
Question is:



Define

u = x^n * y^n
x = cos(at)
y = sin(at)


find du/dt by substituting for x and y

I've started by putting x and y (the cos and sine functions) raised to the power of n into the equation u, do I just proceed to partially differentiate as normal once I've done this?
Any help appreciated

Cheers


Just write down the partial dervatives of u ux\frac{\partial u}{\partial x} and uy\frac{\partial u}{\partial y}

get the derivatives of x and y w.r.t t dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}

For du/dt
dudt=uxdxdt\frac{du}{dt}=\frac{\partial u}{\partial x}\cdot \frac{dx}{dt}
or
dudt=uydydt\frac{du}{dt}=\frac{\partial u}{\partial y}\cdot \frac{dy}{dt}

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