At t = 6 the tangent drops 2 units for each hour to the right. Which statement is correct?

The amount decreases at 2 milligrams per hour.

The amount decreases at 2 milligrams per minute.

A balloon's altitude is A(t) = 500 - \tfrac12 t^2 meters after t seconds. Find A'(10) and explain what it means.

A'(10) = -10: the altitude decreases at 10 meters per second.

A'(10) = 10: the altitude increases at 10 meters per second.

A'(10) = -10: the altitude is −10 meters.

Topic 2.1

Interpret the sign and units

Builds on calculating an instantaneous rate. 6 minutes.

A tank that fills and drains

A tank holds \(V(t) = 12t - t^2\) liters of water after \(t\) minutes. The curve is green where the volume increases and red where it decreases.

Go to 2 minutes. The orange line is the tangent, which touches the curve and points in its direction. Its slope is \(V'(2) = 8\), so at that moment the volume increases at 8 liters per minute. The unit is liters per minute because \(V\) is in liters and \(t\) is in minutes.

Go to 6 minutes. The tangent is flat and \(V'(6) = 0\), so at this instant the volume neither increases nor decreases.

Go to 9 minutes. The tangent slopes down and \(V'(9) = -6\), so the volume decreases at 6 liters per minute.

Any curve

The curve can be any function \(f\), and the slope of the tangent at the input \(a\) is the rate \(f'(a)\).

Where the tangent slopes up, \(f'(a)\) is positive and \(f\) increases. Where it slopes down, \(f'(a)\) is negative and \(f\) decreases. Where it is flat, \(f'(a)\) is 0.

The size of \(f'(a)\), its value without the sign, tells how fast \(f\) changes. A steep tangent, up or down, gives a large size, and a flat tangent gives a size near 0.

Notation

The sign of the rate gives the direction of change at the input \(a\).
\[\begin{aligned} \cg{f'(a)>0}&:\ f \text{ increases} \\ \cr{f'(a)<0}&:\ f \text{ decreases} \\ f'(a)=0&:\ \text{flat} \end{aligned}\]
The unit of the rate is the unit of the output divided by the unit of the input.
\[\cy{\frac{\text{unit of } f}{\text{unit of } x}}\]
For the tank, liters and minutes give liters per minute.
To explain \(f'(a) = k\), write one sentence with the input \(a\), the direction from the sign of \(k\), and the size \(|k|\) with its units. For the tank, at 9 minutes the volume decreases at 6 liters per minute.
If \(f\) is a position, the sign of \(f'\) gives the direction of motion. The size \(|f'(a)|\) is the speed.

When it fails

Go to the top of the curve. There the rate is 0, which does not mean that the output is 0. The orange bar shows that \(f\) is still well above the axis, and the curve is flat only at this instant.

A rate gives the direction at one input only. A moment later, the slope is negative and the output decreases.

Example 1: profit and loaves

A bakery's profit is \(P(n)\) dollars when it bakes \(n\) loaves. It is known that \(P'(100) = 0.4\). Explain what this means.

The input is loaves and the output is dollars, so the unit of \(P'\) is dollars per loaf.
\[\cy{\text{dollars per loaf}}\]
The value 0.4 is positive, so the tangent slopes up and the profit increases at \(n = 100\).
\[\cg{P'(100)=0.4>0}\]
At 100 loaves, each extra loaf adds about 0.40 dollars to the profit.
As a check, \(P(101) - P(100) = 0.399\) dollars.

Example 2: a cooling cup

A cup of coffee has temperature \(T(t) = 20 + \dfrac{80}{t+1}\) degrees Celsius after \(t\) minutes. It is known that \(T'(3) = -5\). Explain what this means.

The input is minutes and the output is degrees Celsius, so the unit of \(T'\) is degrees per minute.
\[\cy{\text{degrees per minute}}\]
A tempting wrong move: Since \(T'(3) = -5\), the temperature at 3 minutes is −5 degrees.
\[\cr{T(3)=-5\ ?}\]
The point on the graph at 3 minutes is 40 degrees high, and the curve never goes below 20. The number −5 is the slope of the tangent, not a height.
The slope is negative, so the temperature decreases. At 3 minutes, the coffee cools at a rate of 5 degrees Celsius per minute.
\[\cr{T'(3)=-5<0}\]
The size 5 tells how fast, and the minus sign tells the direction.

Practice

\(F(d)\) is the number of liters of fuel in a car's tank after \(d\) kilometers. It is known that \(F'(60) = -0.11\). Explain what this means.

Find the unit of the rate first. Then drag the point to 60 kilometers and read the sign from the tangent.

Solution

Fuel is in liters and distance is in kilometers, so the unit of \(F'\) is liters per kilometer.
\[\cy{\text{liters per kilometer}}\]
The value is negative and the tangent slopes down, so the amount of fuel decreases.
\[\cr{F'(60)=-0.11<0}\]
At 60 kilometers, the car uses fuel at a rate of 0.11 liters per kilometer.

Questions

Questions 1 and 2 use the graph of the amount \(M(t)\) of a medicine in the blood, in milligrams, \(t\) hours after a dose. Question 3 needs a calculation.

At \(t = 6\) the tangent drops 2 units for each hour to the right. Which statement is correct?
- The amount decreases at 2 milligrams per hour.
- The amount is −2 milligrams.
- The amount decreases at 2 milligrams per minute.
Answer
The amount decreases at 2 milligrams per hour.
At which marked time is the amount neither increasing nor decreasing?
- \(t = 4\)
- \(t = 8\)
- \(t = 1\)
Answer
\(t = 4\)
A balloon's altitude is \(A(t) = 500 - \tfrac12 t^2\) meters after \(t\) seconds. Find \(A'(10)\) and explain what it means.
- \(A'(10) = -10\): the altitude decreases at 10 meters per second.
- \(A'(10) = 10\): the altitude increases at 10 meters per second.
- \(A'(10) = -10\): the altitude is −10 meters.
Answer
\(A'(10) = -10\): the altitude decreases at 10 meters per second.

Summary

You find the unit of a rate by dividing the output unit by the input unit.

You read the sign from the tangent, because a positive slope means that the output increases and a negative slope means that it decreases.

You use both to explain what a rate means in a situation, in one sentence with a direction and units.

This is one of four skills in topic 2.1 of Which & Why: AP Calculus.