1.2 Functions: Modeling Relationships
Activity 1.1.2
Consider a spherical tank of radius [latex]4 \hspace{0.5mm} m[/latex] that is filling water. Let [latex]V[/latex] be the volume of water in the tank (in cubic meters) at a given time, and [latex]h[/latex] the depth of the water (in meters) at the same time. It can be show using calculus that [latex]V[/latex] is a function of [latex]h[/latex] according to the rule
[latex]V = f(h) = \frac{\pi}{3}h^2(12-h)[/latex]
a. What values of [latex]h[/latex] make sense to consider in the context of this function? What values of [latex]V[/latex] make sense in the same context?
b. What is the domain of this function [latex]f[/latex] in the context of the spherical tank? Why? What is the corresponding codomain? Why?
c. Determine and interpret (with appropriate units) the values [latex]f(2)[/latex], [latex]f(4)[/latex], and [latex]f(8)[/latex]. What is the important about the value of [latex]f(8)[/latex]?
d. Consider the claim: “since [latex]f(9) = \frac{\pi}{3}(12-9) = 81\pi \approx 254.47[/latex], when the water is [latex]9[/latex] meters deep, there is about [latex]254.47[/latex] cubic meters of water in the tank”. Is this claim valid? Why or why not? Further, does it make sense to observe that “[latex]f(13) = -\frac{169\pi}{3}[/latex]“. Why or why not?
e. Can you determine a value of [latex]h[/latex] for which [latex]f(h) = 300[/latex] cubic meters?
Show Solution
a. The values of h that make sense in context is [latex]0[/latex] to [latex]8[/latex] meters because we cannot have negative height and the maximum height of the water is when the height of the water matches the height of the tank. We know the tank is full at [latex]8[/latex] meters. So, the maximum volume is [latex]f(8) = \frac{\pi}{3}(8^2)(12-8) = \frac{256\pi}{3} \approx 268.08 \hspace{0.5mm} m^3[/latex]. Thus, the values of [latex]V[/latex] that make sense in context is [latex]0[/latex] to [latex]268.08[/latex].
b. The domain of [latex]f[/latex] is the possible heights which is [latex][0,8][/latex]. The codomain is all positive amount of volume since in theory we we can plug in any height and get out any volume amount.
c.
[latex]f(2) = \frac{\pi}{3}(2)^2(12-2) = \frac{40\pi}{3} \approx 41.89[/latex]. When the depth of the the water is [latex]2[/latex] meters, the volume of the tank is [latex]41.89[/latex] meters cubed.
[latex]f(4) = \frac{\pi}{3}(4)^2(12-4) \frac{128\pi}{3} \approx 134.04[/latex]. When the depth of the water is [latex]4[/latex] meters, the volume of the tank is [latex]134.04[/latex] meters cubed.
[latex]f(8) = \frac{\pi}{3}(8)^2(12-8) \frac{265\pi}{3} \approx 2668.09[/latex]. When the depth of the water is [latex]8[/latex] meters, the volume of the tank is [latex]269.09[/latex] meters cubed. [latex]f(8)[/latex] is the maximum amount of volume in the tank.
d. The claim [latex]f(9)[/latex] is not valid because when [latex]h = 9[/latex] is greater than the overall height of the tank and so the the water would be overflowing. [latex]f(13) = \frac{-169\pi}{3}[/latex] does not make sense because we cannot have negative volume.
Activity 1.2.3
Consider a spherical tank of radius [latex]4[/latex] that is completely full of water. Suppose that the tank is being drained by regulating an exit valve in such a way that the height of the water in the tank is always decreasing at a rate of [latex]0.5[/latex] meters per minute. Let [latex]V[/latex] be the volume of water in the tank (in cubic meters) at a given time [latex]t[/latex] (in minutes), and [latex]h[/latex] the depth of the water (in meters) at the same time. It can be shown using calculus that [latex]V[/latex] is a function of t according to the model.
[latex]V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t)[/latex]
In addition, let [latex]h = q(t)[/latex] be the function whose output is the depth of the water in the tank at time [latex]t[/latex].
a. What is the height of the water when [latex]t = 0[/latex]? When [latex]t = 1[/latex]? When [latex]t = 2[/latex]? How long will it take to completely drain? Why?
b. What is the domain of the model [latex]h = q(t)[/latex]? What is the domain of the model [latex]V = p(t)[/latex]?
c. How much water is in the tank when the tank is full? What is the range of the model [latex]h = q(t)[/latex]? What is the range of the model [latex]V = p(t)[/latex]?
d. We will frequently use a graphing utility to help us understand function behavior, and strongly recommend Desmos because it is intuitive, online, and free.
In this prepared Desmos worksheet, you can see how we enter the (abstract) function [latex]V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t)[/latex], as well as the corresponding graph the program generates. Make as many observations as you can about the model [latex]V = p(t)[/latex]. You should discuss its shape and overall behavior, its domain, its range, and more.
e. How does the model [latex]V = p(t) = \frac{256\pi}{3} - \frac{\pi}{24}t^2(24-t)[/latex] differ from the abstract function [latex]y = r(x) = \frac{256\pi}{3} - \frac{\pi}{24}x^2(24-x)[/latex]? In particular, how do the domain and range of the model differ from those of the abstract funtion, if at all?
f. How should the graph of the height function [latex]h = q(t)[/latex] appear? Can you determine a formula for [latex]q[/latex]? Explain your thinking.
Show Solution
a. When [latex]t = 0[/latex] the height of the water is [latex]8[/latex] meters because that is the length of the overall tank. Since the tank is decreasing at a rate of [latex]0.5[/latex] meters per minute then when [latex]t = 1[/latex] the height of the water is [latex]7.5[/latex] meters. When [latex]t = 2[/latex], the height of the water is at [latex]7[/latex] meters.
b. Domain of [latex]p(t)[/latex]: [latex][0,16][/latex], Domain of [latex]q(t)[/latex]: [latex][0,16][/latex]
c. The tank is full at [latex]t=0[/latex]. So the amount of water when the tank is full is [latex]V = p(0) = \frac{256\pi}{3} - \frac{\pi}{4}(0^2)(24-0) = \frac{256\pi}{3} \approx 268.08[/latex] cubic meters.
Range of [latex]h = q(t)[/latex]: [latex][0,8][/latex]
Range of [latex]V = p(t)[/latex]: [latex][0,256.08][/latex]
d. The domain and range of the abstract function [latex]V[/latex] is all real numbers. The shape looks similar to a cubic function. It increases to start and then after [latex]t = 0[/latex] the model decreases until it gets to [latex]t = 16[/latex] and then it increases again. There is a [latex]y[/latex]-intercept at [latex]256.08[/latex] and there are two x-intercepts at [latex]-8[/latex] and [latex]16[/latex].
e. The domain and range of [latex]r(x)[/latex] the abstract function are both all real numbers compared to the domain and range of [latex]p(t)[/latex]. In [latex]p(t)[/latex] the domain is [latex][0,16][/latex] since we cannot consider negative height and the range is [latex][0,256.08][/latex] since it does not make sense to have negative volume or to consider the volume before it is full.
f. The graph of the height function would be a linear function since it is decreasing at a constant rate. So the graph would start at [latex]t = 8[/latex] and then slowly decrease in a linear fashion until it gets to [latex]t = 16[/latex]. The formula is [latex]h = q(t) = 8-0.5t[/latex].