Meaning of a Derivative in Context (College Board AP® Calculus BC): Revision Note

Water is leaking from a cylindrical storage tank. The amount of water in the tank at time $t$ minutes is modeled by a differentiable function $W$, where $W\left(t\right)$ is measured in gallons. Selected values of $W\left(t\right)$ are given in the table above.

Use the data in the table to approximate $W\prime \left(7\right)$. Show the computations that lead to your answer. Using correct units, interpret the meaning of your answer in the context of this problem.

Answer:

Find the average rate of change using the two values of $t$ closest to $t=7$

$\begin{array}{rcl}W\text{'}\left(7\right)& \approx & \frac{W\left(10\right)-W\left(4\right)}{10-4}\\ & =& \frac{24-42}{10-4}\\ & =& -3\end{array}$

The units for $W$ are gallons

The units for $t$ are minutes

Therefore, the units for $W\text{'}\left(t\right)$ are gallons per minute

The rate of change is negative which means $W$ is decreasing

$W\text{'}\left(7\right)\approx -3$ gallons per minute

This means that the water in the tank is decreasing at a rate of 3 gallons per minute when $t=7$ minutes

These questions are usually worth two points.

One point for the calculation of the rate of change. You must show the difference quotient in your calculation.

One point for the interpretation. You must give all points in context:

• $t=7$ means 7 minutes

• $W\text{'}\left(t\right)$ means the rate at which the water in the tank is changing

• A negative rate of change means the quantity is decreasing

  • Do not say it is decreasing at a rate of -6

  • The negative is not needed when you say it is decreasing

Do not use the word "velocity" to describe rates of change in non-motion contexts. You will lose points for this in FRQs.

The depth of the water in a harbor, measured in feet, is modeled by the function $f\left(t\right)$. The variable $t$ represents the number of hours after midnight.

$f\left(t\right)=6\cos \left(\frac{\pi }{6}\left(t-10\right)\right)+16,0\le t<24$

(a) State the maximum depth of the water in the harbor according to the model.

(b) Find the rate at which the depth of the water in the harbor is changing at 6 am. State appropriate units for your answer.

(c) It is given that $f^{\text{'}}\left(12\right)=-2.721$ and $f^{\text{'}\text{'}}\left(12\right)=-0.822$. Explain the meaning of these two values in the context of the model.

Answer:

(a)

$f\left(t\right)$ models the depth of the water, so we need to find the maximum value of $f\left(t\right)$

The maximum of $\cos \left(\frac{\pi }{6}\left(t-10\right)\right)$ will be 1

Use this to find the maximum of the function

$6\left(1\right)+16=22$

Maximum depth = 22 feet

(b)

The rate of change of the depth will be given by $f^{\text{'}}\left(t\right)$

Differentiate $f\left(t\right)$, using the chain rule for $6\cos \left(\frac{\pi }{6}\left(t-10\right)\right)$

$f^{\text{'}}\left(t\right)=-6\sin \left(\frac{\pi }{6}\left(t-10\right)\right)·\frac{\pi }{6}=-\pi \sin \left(\frac{\pi }{6}\left(t-10\right)\right)$

6 am is 6 hours after midnight, so substitute $t=6$

Make sure your calculator is set to use radians as the angle measure

$f^{\text{'}}\left(6\right)=-\pi \sin \left(\frac{\pi }{6}\left(6-10\right)\right)=\frac{\pi \sqrt{3}}{2}=2.7206...$

Depth is in feet, and time is in hours, so the units will be feet per hour

2.721 feet per hour (to 3 decimal places)

(c)

$t=12$ is 12 hours after midnight, so noon

$f^{\text{'}}\left(t\right)$ is the rate of change of the depth

$f^{\text{'}}\left(12\right)=-2.721$ means that at noon, the depth of water in the harbor is decreasing at a rate of 2.721 feet per hour

$f^{\text{'}\text{'}}\left(t\right)$ is the rate of change of the rate of change of the depth

$f^{\text{'}\text{'}}\left(12\right)=-0.822$ means that at noon, the rate at which the depth of water in the harbor is changing, is decreasing at rate of 0.822 feet per hour per hour

The rate of change of the volume of water in a container is modeled by the function $r\left(t\right)$.

$r\left(t\right)$ is measured in gallons per minute and $t$ is measured in minutes.

(a) Explain the meaning of $r\left(0.1\right)=2$ in the context of the model.

(b) At a particular time, $r\left(t\right)$ is positive and $r^{\text{'}}\left(t\right)$ is negative. Explain what this means in the context of the model.

(c) State the units for the quantity found by calculating$\int r\left(t\right)dt$.

Answer:

(a)

Note that in this problem, $r\left(t\right)$ is modelling a rate, rather than an amount

The volume of water in the container at t=0.1 minutes (6 seconds) is increasing at a rate of 2 gallons per minute

(b)

$r\left(t\right)$ models the rate of change of volume, while $r^{\text{'}}\left(t\right)$ models the rate of change of the rate of change (how fast it is increasing or decreasing)

The volume of water in the container is increasing, but at a decreasing rate

(c)

Integrating the rate of change of a quantity will produce an expression for the change in the quantity

I.e. $\int \frac{dy}{dx}dx=y+C$

So in this context we are integrating gallons per minute, with respect to minutes

The units of $\int r\left(t\right)dt$ will be gallons