Learning Outcomes
- Model and analyze scenarios using related rates.
Section 3.3 Related rates (AD3)
Subsection 3.3.1 Activities
Remark 3.3.1.
Activity 3.3.3.
Activity 3.3.5.
A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the radius of the balloon changing at the instant the balloon’s diameter is 12 inches? Is the radius changing more rapidly when \(d = 12\) or when \(d = 16\text{?}\) Why?
(a)
Draw several spheres with different radii, and observe that as volume changes, the radius, diameter, and surface area of the balloon also change.
(b)
Recall that the volume of a sphere of radius \(r\) is \(V = \frac{4}{3} \pi r^3\text{.}\) Note as well that in the setting of this problem, both \(V\) and \(r\) are changing with time \(t\text{.}\) Hence both \(V\) and \(r\) may be viewed as implicit functions of \(t\text{,}\) with respective derivatives \(\frac{dV}{dt}\) and \(\frac{dr}{dt}\text{.}\) Differentiate both sides of the equation \(V = \frac{4}{3} \pi r^3\) with respect to \(t\) (using the chain rule on the right) to find a formula for \(\frac{dV}{dt}\) that depends on both \(r\) and \(\frac{dr}{dt}\text{.}\)
(c)
At this point in the problem, by differentiating we have “related the rates” of change of \(V\) and \(r\text{.}\) Recall that we are given in the problem that the balloon is being inflated at a constant rate of 20 cubic inches per second. Is this rate the value of \(\frac{dr}{dt}\) or \(\frac{dV}{dt}\text{?}\) Why?
(d)
From part (c), we know the value of \(\frac{dV}{dt}\) at every value of \(t\text{.}\) Next, observe that when the diameter of the balloon is 12, we know the value of the radius. In the equation \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\text{,}\) substitute these values for the relevant quantities and solve for the remaining unknown quantity, which is \(\frac{dr}{dt}\text{.}\) How fast is the radius changing at the instant \(d = 12\text{?}\)
(e)
How is the situation different when \(d = 16\text{?}\) When is the radius changing more rapidly, when \(d = 12\) or when \(d = 16\text{?}\)
Remark 3.3.6.
Remark 3.3.7. A guide to solving related rated problems.
Remark 3.3.8. Volume formulas.
- A sphere of radius \(r\) has volume \(V= \frac{4}{3} \pi r^3 \)
- A vertical cylinder of radius \(r\) and height \(h\) has volume \(V= \pi r^2 h\)
- A cone of radius \(r\) and height \(h\) has volume \(V= \frac{\pi}{3} r^2 h \)
Activity 3.3.9.
A vertical cylindrical water tank has a radius of 1 meter. If water is pumped out at a rate of 3 cubic meters per minute, at what rate will the water level drop?
(a)
Draw a figure to represent the situation. Introduce variables that measure the radius of the water’s surface, the water’s depth in the tank, and the volume of the water. Label your diagram.
(b)
What information about rates of changes does the problem give you?
(c)
Recall that the volume of a cylinder of radius \(r\) and height \(h\) is \(V = \pi r^2 h\text{.}\) What is the related rates equation in the context of the vertical cylindrical tank? What derivative rules did you use to find this equation?
- \(\displaystyle \frac{dV}{dt}= \pi 2 r \frac{dh}{dt}\)
- \(\displaystyle \frac{dV}{dt}= \pi r^2 \frac{dh}{dt}\)
- \(\displaystyle \frac{dV}{dt}= \pi \frac{dr}{dt} h\)
- \(\displaystyle \frac{dV}{dt}= \pi 2r \frac{dr}{dt} h + \pi r^2 \frac{dh}{dt}\)
- \(\displaystyle \frac{dV}{dt}= \pi 2r h + \pi r^2 \)
(d)
Which variable(s) have a constant value in this situation? Why?
- The variable measuring the radius of the water’s surface
- The variable measuring the depth of the water
- The variable measuring the volume of the water
(e)
Which variable(s) have a constant rate of change in this situation? Why?
- The variable measuring the radius of the water’s surface
- The variable measuring the depth of the water
- The variable measuring the volume of the water
(f)
Using your finding above, find at what rate the water level is dropping.
(g)
If the full tank contains 12 cubic meters of water, how long does it take to empty the tank?
(h)
Confirm your finding in the previous part by finding the initial water level for 12 cubic centimeters of water and determine how long it takes for the water level to reach 0.
Activity 3.3.10.
A water tank has the shape of an inverted circular cone (the cone points downwards) with a base of radius 6 feet and a depth of 8 feet. Suppose that water is being pumped into the tank at a constant instantaneous rate of 4 cubic feet per minute.
(a)
Draw a picture of the conical tank, including a sketch of the water level at a point in time when the tank is not yet full. Introduce variables that measure the radius of the water’s surface and the water’s depth in the tank, and label them on your figure.
(b)
Say that \(r\) is the radius and \(h\) the depth of the water at a given time, \(t\text{.}\) Notice that at any point of time there is a fixed proportion between the depth and the radius of the volume of water, forced by the shape of the tank. What proportional equation relates the radius and height of the water, and why?
(c)
Determine an equation that gives the volume of water in the tank as a function of only the depth \(h\) of the water (so eliminate the radius from the volume equation using the previous part).
(d)
Through differentiation, find an equation that relates the instantaneous rate of change of water volume with respect to time to the instantaneous rate of change of water depth at time \(t\text{.}\)
(e)
Find the instantaneous rate at which the water level is rising when the water in the tank is 3 feet deep.
(f)
When is the water rising most rapidly?
- \(\displaystyle h = 3\)
- \(\displaystyle h = 4\)
- \(\displaystyle h = 5\)
- The water level rises at a constant rate
Remark 3.3.11.
Recall that in a right triangle with sides \(a,b\text{,}\) and hypotenuse \(c\) we have the relationship
\begin{equation*}
a^2 + b^2 = c^2.
\end{equation*}
You might know this as the Pythagorean theorem, even though this relationship was known by the Babylonians already a thousand years before his time!
