How Does It Rate?

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Water is being poured into a large tub in such a way that the amount of water in the tub $t$ minutes after the pouring started is: $f(t) = 7t+19$ gallons. We want to find out how fast water is being poured in at time $t = 3$ minutes. That is, we want $f'(3)$ .

How much water is in the tub when water started being poured in?

How much water is in the tub at 3 minutes?

We want to find the rate of pouring at $3$ minutes. Since we can’t tell right away what our rate is at one instance of time, let’s go back to middle school algebra to find some average rates (and try to answer the rate at 3 minutes in part (b)):

(a): Find the average rate of pouring between $t=2$ and $t=5$ minutes. Between $t = 7$ and $t=23$ minutes. Between $t=4.26$ and $t=2.938$ minutes. Between $t=3$ and $t = 3.01$ minutes. How did you calculate these? What is the unit for the rates? Why are you getting the results you are? What is another “Algebra I” name for the rates you are finding?
(b): What do you think the rate at $3$ minutes is? What do you think the rate at $t$ minutes is? How confident are you with your guess (and why)?

Let’s now take a huge step and deal with an “amount function” that is not linear, which means the rate is constantly changing (like you change velocity on the highway- we’re not on cruise control!). Let’s say the amount of water in the tub $t$ minutes after the pouring started is $f(t) = t^2+5t+17$ .

(a): Find the average rate of pouring between $t=2$ and $t=5$ minutes. Between $t = 7$ and $t=23$ minutes. Between $t=4.26$ and $t=2.938$ minutes. Between $t=3$ and $t=5$ minutes. Between $t=3$ and $t=4$ minutes. Between $t=3$ and $t=3.5$ minutes. Between $t=3$ to $t=3.1$ minutes. Between $t=3$ and $t=3.01$ minutes. Between $t=3$ and $t=3.001$ minutes. (Can you make this process easier by taking advantage of the function and table feature of your graphing calculator?)
(b): What is the issue we have to wrestle with that we didn’t in Question 3? What are you observing? How does this help you make a conjecture for the precise rate of pouring ( $f'(3)$ )? Is there a rate at time $t=3$ minutes?

From a graphical perspective, what were you doing in the previous question?

Write a general algebraic formula for the process you did in Question 4 (Hint: Could you describe the process to a friend?). What would we change about it to find, say, $f'(7)$ ? How would the process change if the function was really complicated, such as and we wanted to estimate \[ f(t) = \sqrt{\frac{e^{t^{37}}+\sin (t+4)-6}{\log t + \frac{67}{\sqrt [54]{t^9-89}}}} \] (Don’t do the process for that monster, but would the essential technique change (recall you have a graphing calculator).

Free Space

Back to the boring quadratic we we’ve been dealing with. How could we use the algebraic formula from Question 6 to “speed up” the process to find, say, $f'(7)$ (i.e., so we won’t have to plug in numbers after numbers like in Question 4)? Do so. Do it again to find $f'(5.26)$ . (Note: What mathematical weirdness are you forced to deal with here? (Hint: What does $\frac {0}{0}$ mean?))

What could we do to speed up the process done in Question 8 (i.e., so we won’t have to go through the process of Question 8 for each and every time for which we want the rate)? This is what we’ll do in the next activity.

Note: A graphing calculator with a “derivative” capability can calculate a numerical derivative. For TI’s, this is the nDeriv feature under the MATH menu (in the MATH portion of that menu). To find the derivative at a point, enter nDeriv(function, $x$ , time for which you want the rate). For example, for $f'(3)$ above, you would type in nDeriv( $x^2+5x+17$ , $x$ , $3$ ). If you already have the function in the “ $Y=$ ” menu (say, as $Y_4$ ), you can type in nDeriv( $Y_4$ , $x$ , $3$ ), where you can find “ $Y_4$ ” by going to VARS, then Y-VARS, then FUNCTION, then select $Y_4$ from the list.

2024-10-10 13:51:01