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# RobertTalbert/Activity 1.2.2 input.md Last active Sep 4, 2019

This week you'll begin working on Assigned Activities in groups, and posting those to CampusWire. Here's an example of a fully worked-out Activity from the textbook, Activity 1.2.2 in Section 1.2. Read through it, ask questions in the follow-up comments, and keep it handy as a guide for writing your own solutions. This is a complete and correct solution, but your group does NOT need a complete or correct solution to start with -- just post a good faith initial attempt, and then you'll be responsible for making it complete and correct by the end of the week.

Some things to note as you read this: Each part of the solution begins by restating the part; all solutions are given in complete English sentences with an appropriate mix of English and math; all math is formatted properly (see #2 ); the solution is written with a view toward teaching the reader.

Activity: Consider a spherical tank of radius 4m that is filling with water. Let $$V$$ be the volume of water in the tank (in cubic meters) at a given time, and $$h$$ the depth of the water (in meters) at the same time. It can be shown using calculus that $$V$$ is a function of $$h$$ according to the rule

$$V = f(h) = \frac{\pi}{3}h^2 (12-h)$$

(a) What values of $$h$$ make sense to consider in the context of this function? What values of $$V$$ make sense in the same context?

Since the radius of the tank is 4 meters the height (distance from bottom to top) is 8 meters. So the values of $$h$$ that make sense are in the interval $$[0,8]$$. The values of $$V$$ that make sense are just the volume when the tank is full and when it's empty: $$f(0) = \frac{\pi}{3}0^2 (12-0) = 0$$, and $$f(8) = \frac{\pi}{3}8^2 (12-8) = \frac{256\pi}{3} \approx 268.08$$.

(b) What is the domain of the function $$f$$ in the context of the spherical tank? Why? What is the corresponding codomain? Why?

The domain of the function is the set of all inputs to $$f$$ that can be computed. Although as an "abstract function", there is no value of $$h$$ that cannot be plugged in, in the context of the tank we can only use values between 0 and 8 inclusive based on what we said in part (a). So the domain is $$[0,8]$$. The codomain is the set of potential outputs of this function, which we determined above were between 0 and $$\frac{256\pi}{3}$$ --- as an interval, that's $$\left[ 0, \frac{256\pi}{3}\right]$$.

(c) Determine and interpret (with appropriate units) the values $$f(2)$$, $$f(4)$$, and $$f(8)$$. What is important about the value of $$f(8)$$?

Using our formula we get:

$$f(2) = \frac{\pi}{3}2^2(12-2) = \frac{40\pi}{3} \approx 41.89$$ cubic meters

$$f(4) = \frac{\pi}{3}4^2(12-4) = \frac{128\pi}{3} \approx 134.04$$ cubic meters

$$f(8) = \frac{\pi}{3}8^2 (12-8) = \frac{256\pi}{3} \approx 268.08$$. cubic meters

The value of $$f(8)$$ is important because this is the volume of the entire tank.

(d) Consider the claim: "Since $$f(9) = \frac{\pi}{3}9^2(12-9) = 81\pi \approx 254.47$$, when the water is 9 meters deep there is about $$254.47$$ cubic meters of water in the tank." Is this claim valid? Why or why not? Further, does it makes sense to observe that "$$f(13) = -\frac{169\pi}{3}$$"? Why or why not?

Although we can compute $$f(9)$$ mathematically, this doesn't make sense in context because the tank is only 8 meters deep, so we can't consider a depth of 9 meters. Therefore although the math here is correct, the claim is not valid. Similarly, $$f(13)$$ doesn't make sense in context because you can't have a depth of 13 meters (or a negative volume!). The statement that $$f(13) = -\frac{169\pi}{3}$$ is mathematically correct but not valid given the physical situation.

(e) Can you determine a value of $$h$$ for which $$f(h) = 300$$ cubic meters?

There are at least three ways to answer this question. [Note: Ordinarily you do not need to present all possible solution approaches, just one that works. But for teaching purposes I am putting in different approaches here that would all be considered "Correct". -rt]

Approach 1: Set up the following equation and solve for $$h$$:

$$\frac{\pi}{3}h^2 (12-h) = 300$$

But this seems like difficult algebra, so let's look at a graphical approach first. Using Desmos to graph $$f$$, our default view looks like this:

(Here's the direct link to the Desmos graph.). If we adjust the viewing window to match the domain and codomain that we determined in part (a) --- putting the input between 0 and 8 and the outputs between 0 and roughly 300 --- we get a better view:

(Direct link) We can see from the graph that the volume never quite makes it to 300. So the answer to the original question is NO, we cannot determine a value of $$h$$ where the volume is 300 cubic meters.

Approach 2: Set up the following equation and solve for $$h$$:

$$\frac{\pi}{3}h^2 (12-h) = 300$$

We'll do this by going into Wolfram|Alpha and asking it to solve. Click here to see the input and the results.

It tells us we have three results, but they all involve imaginary numbers! So NO, we cannot determine a value of $$h$$ where the volume is 300 cubic meters.

Approach 3: (The simplest one!) NO we cannot determine a value of $$h$$ where the volume is 300 cubic meters, because we already determined that the volume of the tank when it is completely full is only 268.08 cubic meters. It doesn't get any fuller than that, so there's no solution.