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### Section 3-2 : Interpretation of the Derivative

13. Suppose that the volume of water in a tank for$$0 \le t \le 6$$ is given by $$Q\left( t \right) = 10 + 5t - {t^2}$$.

1. Is the volume of water increasing or decreasing at $$t = 0$$?
2. Is the volume of water increasing or decreasing at $$t = 6$$?
3. Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing?

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a Is the volume of water increasing or decreasing at $$t = 0$$? Show Solution

We know that the derivative of a function gives us the rate of change of the function and so we’ll first need the derivative of this function. We computed this derivative in Problem 4 from the previous section and so we won’t show the work here. If you need the practice you should go back and redo that problem before proceeding.

So, from our previous work we know that the derivative is,

$Q'\left( t \right) = 5 - 2t$

Now all that we need to do is to compute : $$Q'\left( 0 \right) = 5$$. This is positive and so we know that the volume of water in the tank must be increasing at $$t = 0$$.

b Is the volume of water increasing or decreasing at $$t = 6$$? Show Solution

Again, all we need to do is compute a derivative and since we’ve got the derivative written down in the first part there’s no reason to redo that here.

The evaluation is : $$Q'\left( 6 \right) = - 7$$. This is negative and so we know that the volume of water in the tank must be decreasing at $$t = 6$$.

c Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop changing? Show Solution

Here all that we’re really asking is if the derivative is ever zero. So we need to solve,

$Q'\left( t \right) = 0\hspace{0.25in} \to \hspace{0.25in}5 - 2t = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}t = \frac{5}{2}$

So, the volume of water will stop changing at $$\frac{5}{2}$$.