Interpret the result

A tank that drains

Water drains from a cylindrical tank of radius 2 meters at 3 cubic meters each minute. The teal region is the water, and the orange bar is its depth h.

The volume is \(V = 4\pi h\), so \(dV/dt = 4\pi\,dh/dt\) and \(dh/dt\) is \(-3/(4\pi)\), about \(-0.24\). This number needs a sign, a unit and a name.

Move the time slider. The surface moves down, so the depth decreases and its rate is negative, which the red arrows mark.

The depth is in meters and the time is in minutes, so the rate is in meters per minute.

The 3 cubic meters each minute is the rate of the volume, and 0.24 meters each minute is the rate of the depth. The two rates are related, but different.

Sign, unit and name

The graph shows a quantity q against the time t. The slope of the orange tangent is the rate dq/dt.

Drag the point. The curve is green where q increases, so the rate is positive. It is red where q decreases, so the rate is negative.

The unit of the rate is the unit of q divided by the unit of time. Each rate belongs to one named quantity.

Notation

1. A positive rate means that the quantity increases, and a negative rate means that it decreases.

   \[\frac{dh}{dt} = \cr{-0.24}\ \text{m/min}\]

   The direction belongs to the quantity as you defined it.

2. The unit is the unit of the quantity over the unit of time. If the question asks for another unit, convert it.

   \[0.24\ \text{m/min} \approx 24\ \text{cm/min}\]

3. Write one sentence that names the quantity, says whether it increases or decreases, and gives the size with its unit.

   \[\begin{gathered} \text{The depth decreases} \\ \text{at about } 0.24\ \text{m/min.} \end{gathered}\]

4. Keep related rates apart, because each one belongs to its own quantity.

   \[\begin{gathered} \frac{dV}{dt} = -3\ \text{m}^3/\text{min} \\ \frac{dh}{dt} \approx -0.24\ \text{m/min} \end{gathered}\]

When it fails

The sign depends on how you define the quantity. Let d be the distance from the rim down to the water surface.

When the water drains, \(d\) increases, so \(dd/dt\) is \(+0.24\) m/min. The water moves in the same way, so only the definition changes the sign.

Example 1: A melting cube

A cube of ice melts and stays a cube. When its side \(s\) is 4 cm, a solution gives \(ds/dt = -0.25\) cm/min and \(dV/dt = -12\) cm³/min. Explain what each number says.

1. Both rates are negative, so both quantities decrease. The red layers show the ice that melts in one minute.

   \[\begin{gathered} \frac{ds}{dt} = \cr{-0.25}\ \text{cm/min} \\ \frac{dV}{dt} = \cr{-12}\ \text{cm}^3/\text{min} \end{gathered}\]

2. The side loses 0.25 cm each minute, and the volume loses 12 cm³ each minute.

3. The two rates are related. Three faces of area \(4^2 = 16\) cm² each lose a layer 0.25 cm thick, so the volume loses 12 cm³ each minute.

   \[3 \cdot 16 \cdot 0.25 = 12\]

Example 2: A plane and a radar station

A plane flies at a height of 3 km, directly toward a radar station, at a horizontal speed of 600 km/h. Let \(x\) be the horizontal distance to the point above the station, and \(s\) the distance to the station. Find \(ds/dt\) when \(x = 4\) km and when \(x = 1\) km.

1. The relation \(s^2 = x^2 + 9\) gives \(s\,ds/dt = x\,dx/dt\). The plane approaches the station, so \(dx/dt = -600\). The red arrow shows \(ds/dt\), the part of the motion along the line of sight.

   \[\frac{ds}{dt} = \frac{x}{s}\,\frac{dx}{dt} = -600\,\frac{x}{s}\]

2. A tempting wrong move: The plane flies at 600 km/h, so the distance to the station decreases at 600 km/h.

   \[\frac{ds}{dt} = -600\]

   At x = 1 the line of sight is steep, so the plane moves almost across it. The dashed red arrow shows this claim, but the solid red arrow, the true rate, is much shorter.

3. At \(x = 4\), \(s = 5\), so the distance decreases at 480 km/h.

   \[\frac{ds}{dt} = -600 \cdot \frac{4}{5} = \cr{-480}\ \text{km/h}\]

4. At \(x = 1\), \(s = \sqrt{10}\), so the distance decreases at about 190 km/h.

   \[\frac{ds}{dt} = -\frac{600}{\sqrt{10}} \approx \cr{-190}\ \text{km/h}\]

5. Both answers are negative, so the distance decreases, but more slowly than the plane moves.

Practice

A rope runs from a boat through a pulley on a dock 7 m above the boat. A winch pulls in the rope at 2 m/s. The boat is 24 m from the dock, and the rope is 25 m long. Find \(dx/dt\) for the distance \(x\) from the boat to the dock, and say what it means.

In the picture, x, the rope s and the 7 m height form a right triangle.

Questions

The graph shows the depth \(D\) of a pond, in centimeters, against the time \(t\) in minutes. The orange tangent at \(t = 4\) has a slope of about \(-1.5\).

1. At \(t = 4\), is the depth increasing or decreasing?

   • Decreasing
   • Increasing
   • Neither, because the depth is not zero
   Answer

   Decreasing

2. What does the slope \(-1.5\) say?

   • At \(t = 4\), the depth decreases at about 1.5 cm each minute.
   • At \(t = 4\), the depth is \(-1.5\) cm.
   • From \(t = 0\) to \(t = 4\), the depth decreases by 1.5 cm.
   Answer

   At \(t = 4\), the depth decreases at about 1.5 cm each minute.

3. The area of a circle changes at \(dA/dt = 8\pi\) cm²/s when \(r = 4\) cm. Find \(dr/dt\) and say what it means.

   • \(dr/dt = 1\) cm/s: the radius increases by 1 cm each second.
   • \(dr/dt = 2\) cm/s: the radius increases by 2 cm each second.
   • \(dr/dt = 8\pi\) cm/s: the radius increases by \(8\pi\) cm each second.
   • \(dr/dt = 1\) cm/s: the area increases by 1 cm² each second.
   Answer

   \(dr/dt = 1\) cm/s: the radius increases by 1 cm each second.

Summary

The sign of a rate shows whether its quantity increases or decreases, and its unit is the unit of the quantity over time.

This works because a rate is the slope of one quantity against time, so its sign and unit depend on how you define that quantity.

You use it to state each answer as a sentence that names its quantity, so you keep related rates apart.

This is one of four skills in topic 4.5 of Which & Why: AP Calculus.