# Rates of Change - Hemispherical bowl filled with water

#### Aminta_1900

##### New member
You are given:

Vs = 1/3 * pi * h^2 * (3a - h)

assume the radius of the liquid circular cross-section = r

from the equation of circle

(a-h)^2 + r^2 = a^2

(a-h)^2 = a^2 - r^2

h = a - (a^2 - r^2)^(1/2) .......................................(1)

Now continue.....

• Aminta_1900
I am working with dV/dt = 3*pi*a^3 and dV/da = 2pi*a^2.

dh/dt = (3/2)a

I can see that the height at which it is filled is also the radius a, but I'm not sure how to use the volume given for the shaded part of the hemisphere. Maybe that I should replace h with a/2.....?
How did you get dV/da = 2pi*a^2? And why do you want that? Since a is a constant, dV/da is irrelevant.

Also, this is not true, unless you mean merely that a is the depth of the water when the bowl is full; that isn't relevant either.

You want to relate the rate of change of volume (V) to the rate of change of depth (h).

To do that, they've given you the volume as a function of depth, so just differentiate that with respect to time. (Radius a is a constant, but you'll use the chain rule to incorporate dh/dt, which is what you'll ultimately solve for.)

(By the way, the radius of the surface of the water isn't relevant, unless you don't want to use the volume formula they gave you.)

• Aminta_1900
I find these related rates problems easier if you symbolise everything.

You are given
$$\displaystyle \frac{dV}{dt} = 3 \pi a^3$$ and $$\displaystyle V=\frac{1}{3}\pi h^2(3a-h)$$

You want to find
$$\displaystyle \frac{dh}{dt}$$ when $$\displaystyle h=\frac{1}{2} a$$

You can work out an expression for $$\displaystyle \frac{dV}{dh}$$ since V is a function of h (remembering that a is a constant).

Now, can you write a chain rule involving $$\displaystyle \frac{dV}{dt}$$, $$\displaystyle \frac{dV}{dh}$$ and $$\displaystyle \frac{dh}{dt}$$ ?

Thanks for the replies. I'll read them carefully and attempt this again.

• blamocur