Lexie Brown
Related Rates
We have been discussing derivatives and how to go about calculating them. Now, we are doing related rates, which use rates of one thing as it relates to the rates of one or more things. The rates are found using the calculus we already know how to do, such as find derivatives and some simple algebra. When you are finding related rates, there is a basic process you should follow.
The process includes reading the problem, finding what you’re trying to find, drawing a picture, labeling all variables, determining how you are going to solve the problem, solving the problem, and asking if your solution makes sense. Although, this is a basic problem solving process, it will be especially useful when computing related rates. Using these steps we will solve the problem about the faucet that is filling the hemispherical basin.
The problem states that it’s a hemispherical basin with a diameter of 60 cm. The faucet’s filling this basin at a rate of 12 L/min and we are finding the rate at which the water is rising in the basin when it’s half full. This is enough information to draw a picture and label variables.
V = volume
d = diameter = 60 cm
r = radius = = 30cm
h = height of water when basin is half full
= rate of change in volume with respect to time = 12 L/min = 12000 /min (1L = 1000 )
In this case because it is a hemispherical basin, the height of the basin is equal to the radius. Therefore when the basin is half full, the height of the water in the basin is equal to the radius divided by two.
The equation for the volume of the portion of the sphere with radius r to a height h is:
Since the radius is a constant and will not change with time, that can be entered into the equation in place of r.
Now you will find the derivatives, which will give you the rates of change with respect to time.
A big thing to remember, is that V and h are functions of time and the equation could also be written...