Question

U=x^3+y^3 x= acost y=bsint then du/dt

Answer 1

Answer:

dt

Step-by-step explanation:

and follow please

Answer 2

Answer:

The value of dU/dt is equal to 3cost.sint (b²sint-a²cost).

Step-by-step explanation:

We have given the expression:

$U=x^{3}+y^{3}$                                                            ................(1)

Differentiate the equation (1) on both sides w.r.t. 't':

We get,  $\frac{dU}{dt}=\frac{d}{dt}(x^{3}+y^{3})$

⇒     $\frac{dU}{dt}=\frac{d}{dt}(x^{3})+\frac{d}{dt}( y^{3})$

⇒     $\frac{dU}{dt}=3x^{2}\frac{dx}{dt}+3y^{2}\frac{dy}{dt}$                                      .....................(2)

Now, given x = acost and y= bsint

Therefore, $\frac{dx}{dt}=\frac{d}{dt}(acost) =a(-sint)=-asint$

and, $\frac{dy}{dt}=\frac{d}{dt}(asint) =a(cost)=acost$

Put the values of x, y, dx/dt and dy/dt in equation (1):

$\frac{dU}{dt}=3(acost)^{2}(-asint)+3(bsint)^{2}(bcost)$

$\frac{dU}{dt}=-3a^{2}cos^{2}t. sint+ 3b^{2}sin^{2}t.cost$

$\frac{dU}{dt}=3cost. sint(b^{2}sint-a^{2}cost)$

Therefore, the derivative of U with respect to 't' is equal to 3cost.sint (b²sint-a²cost).