Use shorter intervals to estimate a rate

A growing cube

The curve shows the volume \(V\) of a cube, in cubic centimeters, against the length \(x\) of its edge, in centimeters. The first orange dot is at \(x = 2\), where \(V = 8\).

Drag the second dot, at \(x = 2 + h\), to change the gap \(h\) between the dots.

The line through the dots is a secant line, and its slope is the average rate over the gap. At \(h = 1\) the average is 19, and at \(h = 0.1\) it is 12.61.

Try a negative \(h\), which puts the second dot on the left. At \(h = -0.1\) the average is 11.41.

As the gap shrinks, the averages from both sides approach 12, and the secant comes closer to the faint tangent line at the first dot. The rate at \(x = 2\) is the slope of that tangent, 12 cm³ per cm.

Any curve

The curve can be any curve, and the first dot can be anywhere. Move \(h\) toward 0 from the right and from the left, and the secant line turns.

As the gap shrinks, the secant slope approaches one number from both sides. This number is the slope of the tangent line, and it is the instantaneous rate, which is the rate at that one input.

The gap comes close to 0 but is never 0, because a secant line needs two different dots.

Notation

1. Choose a number \(h\) that is not 0. The second dot is at \(a + h\), and the average rate between the dots is this quotient.

   \[\dfrac{f(a + h) - f(a)}{h}\]

2. If \(h\) is positive, the second dot is on the right of the first. If \(h\) is negative, it is on the left.

3. Make the size of \(h\) smaller. If the averages from both sides approach one number, that number \(\cy{L}\) is the instantaneous rate at \(a\).

   \[\dfrac{f(a + h) - f(a)}{h} \to \cy{L} \quad \text{as } h \to 0\]

   The arrow means approaches. The number \(h\) approaches 0, but it is not 0.

4. A list of averages gives an estimate of \(\cy{L}\), but it does not prove the exact value.

When it fails

When h = 0, the second dot is on top of the first dot, so the top and the bottom of the quotient are both 0.

The red 0 ÷ 0 has no value, and one point does not fix a secant line, so the rate comes from values of h near 0.

Example 1: the cube at 3 cm

A cube has volume \(V(x) = x^3\) cubic centimeters for an edge of \(x\) centimeters. Estimate the rate of change of \(V\) at \(x = 3\).

1. Take h = 0.1 on the right. The volume at x = 3.1 is 29.791.

   \[\dfrac{29.791 - 27}{0.1} = 27.91\]

2. Take h = −0.1 on the left. The volume at x = 2.9 is 24.389.

   \[\dfrac{24.389 - 27}{-0.1} = 26.11\]

3. Shrink the gap. With h = 0.01 the average is 27.0901, and with h = −0.01 it is 26.9101.

4. The averages from both sides approach 27, so the rate is about 27 cm³ per cm.

   \[\cy{\approx 27 \text{ cm}^3 \text{ per cm}}\]

Example 2: a melting block of ice

A block of ice has mass \(M(t) = 200 - 2t^2\) grams after \(t\) minutes. Estimate the rate of change of the mass at \(t = 5\).

1. At t = 5 the mass is M(5) = 150 g. The rate will be negative, because the mass falls.

2. A tempting wrong move: Use one long stretch, from t = 5 to t = 10, and report its average.

   \[\dfrac{0 - 150}{5} = -30\]

   This secant is much steeper than the curve at t = 5, because the ice melts faster late in the stretch. Its slope is the average over five minutes.

3. Shrink the gap to h = 0.1 on the right and h = −0.1 on the left.

   \[\begin{aligned} \dfrac{147.98 - 150}{0.1} &= -20.2 \\ \dfrac{151.98 - 150}{-0.1} &= -19.8 \end{aligned}\]

4. With h = 0.01 and h = −0.01 the averages are −20.02 and −19.98. They approach −20, so the ice loses mass at about 20 g per minute.

Practice

A shop earns \(R(q) = 40q - q^2\) dollars from selling \(q\) items. Estimate the rate of change of \(R\) at \(q = 10\).

Find the slope of the secant for \(h = 0.1\) and for \(h = -0.1\). Then shrink the gap to 0.01 on both sides.

Questions

The graph shows \(f(x) = x^2/2\) with the first dot at \(x = 2\) and a second dot that moves along the curve.

1. The second dot moves from \(x = 4\) toward \(x = 2\) from the right. What happens to the slope of the secant?

   • It decreases and approaches 2
   • It stays at 3
   • It increases without limit
   Answer

   It decreases and approaches 2

2. Which line has a slope closer to the rate at \(x = 2\)?

   • The secant through \(x = 2\) and \(x = 2.5\)
   • The secant through \(x = 2\) and \(x = 4\)
   • They are equally close
   Answer

   The secant through \(x = 2\) and \(x = 2.5\)

3. Near \(t = 3\), the average rate is 7.98 for \(h = -0.01\) and 8.02 for \(h = 0.01\). Estimate the rate at \(t = 3\).

   • 8
   • 7.98
   • 0.04
   Answer

   8

Summary

You find average rates over shorter and shorter gaps on both sides of the input, and you read the number they approach.

This works because the secant line comes closer to the tangent line as the gap shrinks.

Use it to estimate a rate from a formula, a table or a graph.

This is one of four skills in topic 1.1 of Which & Why: AP Calculus.