**Can anyone help with a calculus problem that includes the integral?**

*The problem begins with a cylinder with a radius of inches. h is the depth* and measurement is at time t. the volume is changing at a rate h-5π ^ (1 / 2). I have to show that dh / dt = – √ h / 5. Help? If the radius is 5 inches sorry!

You left what radio is. I will write r and put in what is supposed to be. For a cylinder, V = π r ² h r Assuming that does not change with time dv / dt = (dV / dh) (dh / dt) = π r ² dh / dt Solving dh / dt, dh / dt = (dV / dt) / (π r ²) You're bearing in mind that dV / dt = -5π √ h, the inclusion of this in the above equation dh / dt = (h √ 5π) / (π r ²) = – (5 / r ²) √ h, so is r = 5?

**Calculus Help: Integrals 1**

Thanks for good stuff