# Thread: find te rate of volume

1. ## find te rate of volume

Hello
i have two questions about calculus. I hope you can help me to solve it. Thanks

A container formed by a cylinder at the bottom surrounded by a heisphere. The radius decreases at the rate of 3 cm/sec and trhe height increases at the rate of 2 cm/sec. Find the rate at which the volume of the contaner changes when the radius is 5 and the height is 10 cm.
I try to solve by using : 2r+ h= 8?

Let U(x,y) be a differentiable and let (r,θ) be the polar coordinates ( that is, x=rcos θ, y=rsinθ)
Compute Ur and Uθ and show that lldedivative of Ull^2= (Ur)^2 + (Uθ)^2 1/r^2

2. These are both exercises in using the chain rule for function of two variables.
Originally Posted by rai2003
Hello
i have two questions about calculus. I hope you can help me to solve it. Thanks

A container formed by a cylinder at the bottom surrounded by a heisphere. The radius decreases at the rate of 3 cm/sec and trhe height increases at the rate of 2 cm/sec. Find the rate at which the volume of the contaner changes when the radius is 5 and the height is 10 cm.
I try to solve by using : 2r+ h= 8?
I take it that "a cylinder at the bottom surrounded by a hemisphere" means that the cylinder has a hemisphere on top of it. The volume of a cylinder of radius r and height h is $\pi r^2h$ and the volume of a hemisphere of radius r is $\frac{2}{3}\pi r^3$. The volume of the figure is the sum of those two: $V= \pi r^2h+ \frac{2}{3}\pi r^3$. Find dV/dt in terms of dr/dt and dh/dt. $\frac{dV}{dt}= \frac{\partial V}{\partial r}\frac{dr}{dt}+ \frac{\partial V}{\partial h}\frac{dh}{dt}$.

Let U(x,y) be a differentiable and let (r,θ) be the polar coordinates ( that is, x=rcos θ, y=rsinθ)
Compute Ur and Uθ and show that lldedivative of Ull^2= (Ur)^2 + (Uθ)^2 1/r^2
Use the chain rule:
$\frac{\partial U}{\partial r}= \frac{\partial U}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial U}{\partial y}\frac{\partial y}{\partial r}$ and
$\frac{\partial U}{\partial \theta}= \frac{\partial U}{\partial x}\frac{\partial x}{\partial \theta}+ \frac{\partial U}{\partial y}\frac{\partial y}{\partial \theta}$

In the first question how can we dr/dt? so i try to get dr, dh and i replcae it by r=5 and h=10. Since i got dr and dh i will multiply it with like dr*(-3) and dh(2) . I got negative answer. I do not know i got the right answer or not.

4. Originally Posted by rai2003
A container formed by a cylinder at the bottom surrounded by a heisphere.
No that description does not explain the shape clearly enough (for me to understand what it is).

CB

5. Originally Posted by rai2003
A container formed by a cylinder at the bottom surrounded by a heisphere.
"Surrounded"? I think you mean that the cylinder has a hemisphere on top.
The volume of a cylinder of radius r and height h is $\pi r^2h$ and the volume of a hemisphere of radius r is $\frac{2}{3}\pi r^3$. The volume together is $\pi r^2h+ \frac{2}{3}\pi r^3$/

The radius decreases at the rate of 3 cm/sec and trhe height increases at the rate of 2 cm/sec. Find the rate at which the volume of the contaner changes when the radius is 5 and the height is 10 cm.
i try to get dr, dh and i replcae it by r=5 and h=10. Since i got dr and dh i will multiply it with like dr*(-3) and dh(2) . I got negative answer. I do not know i got the right answer or not.
What do you mean you tried to get "dr" and "dh"? You are told that $dr/dh= -3$ and that $dh/dt= 2$. You are asked to find dV/dt. Use the chain rule.