Integrating an Expression with d(h³)/dt

[Marcos Domingues]

Homework Statement

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I have this expression: dV/dt = F0 - K*h^(1/2); it describes a variation in time of a fluid volume V in a cone-shaped tank of total volume H*pi*R²/3;

By a trigonometric relation we get V = (pi*R²/3*H²)*h³; since tan a = H/R = h/r
where: R = radius of the tank; H = height of the tank; h = variable height of the fluid; r = radius related to a specific height h

So, when we replace V, we got the general expression (pi*R²/3*H²) d(h³)/dt = F0 - K*h^(1/2);

This d(h³) term is the main problem on the integration. How do I integrate this expression?

Homework Equations

(pi*R²/3*H²) d(h³)/dt = F0 - K*h^(1/2)

The Attempt at a Solution

Maybe consider d(h³) = dh*dh*dh and solve it as multiple integrals?

 

[eys_physics]

So, when we replace V, we got the general expression (pi*R²/3*H²) d(h³)/dt = F0 - K*h^(1/2);

No, this wrong. You need to use the chain rule, i.e.
$$\frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt}$$,
since $V=V(h(t))$.

 

[Marcos Domingues]

No, this wrong. You need to use the chain rule, i.e.
$$\frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt}$$,
since $V=V(h(t))$.

That's right. Thanks.