Wemazi

Topic 2.1

Calculate an instantaneous rate

Builds on calculating an average rate from a formula or table. 7 minutes.

A falling ball at 2 seconds

A ball falls from a tower, and after \(t\) seconds it has fallen \(5t^2\) meters. The orange point \(P\) shows the ball at 2 seconds, after a fall of 20 meters. The point \(Q\) shows the ball at \(2 + h\) seconds.

The slope of the orange line through \(P\) and \(Q\) is the average rate over the interval from 2 to \(2 + h\). For \(h = 1\), the green leg shows that the ball falls 25 meters in 1 second, so the average rate is 25 meters per second.

Shrink the interval. The average rate is 22.5 for h = 0.5, 20.5 for h = 0.1 and 20.05 for h = 0.01, so the averages approach 20.

This number is the instantaneous rate, the rate at one instant instead of over an interval. At 2 seconds, the ball falls at 20 meters per second.

Any curve, any input

The curve can be any function \(f\), and \(P\) can be any point, at the input \(a\). The point \(Q\) is at the input \(a + h\).

The line through \(P\) and \(Q\) is a secant line. Its slope is the average rate over the interval, the rise divided by the run \(h\). The rise is green when \(Q\) is higher than \(P\) and red when \(Q\) is lower.

Shrink \(h\) toward 0, from either side. The secant turns about \(P\) and approaches one line, the tangent line at \(P\). This line touches the curve at \(P\) and points in the direction of the curve there.

The slope of the tangent is the instantaneous rate. It is the number that the average rates approach, called their limit as \(h\) approaches 0.

Notation

  1. The rise is \(f(a+h)-f(a)\), so the average rate from \(a\) to \(a + h\) is this fraction.

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

  2. The instantaneous rate at \(a\) is the limit of these averages as \(h\) approaches 0. It is written \(f'(a)\) and read "f prime of a".

    \[\cy{f'(a)} = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]

    It is also the slope of the tangent line at P. It exists only if the averages approach one number.

  3. To find the limit, simplify the fraction until \(h\) cancels. Then let \(h\) approach 0.

    In the average, \(h\) is not 0, so you can cancel it.

When it fails

If \(h\) is 0, then \(P\) and \(Q\) are the same point, and the secant has no direction. The fraction is \(0 \div 0\), so substituting \(h = 0\) does not find the rate.

A table of averages only suggests the limit. The values 20.5 and 20.05 point to 20, but only the algebra shows that the limit is 20.

Example 1: a quadratic

Find the instantaneous rate of change of \(f(x) = x^2 - 3x\) at \(x = 4\).

  1. Because \(f(4) = 16 - 12 = 4\), the average rate from 4 to \(4 + h\) is this fraction.

    \[\frac{(4+h)^2-3(4+h)-4}{h}\]

  2. Expand the numerator and combine like terms. Every term that remains has \(h\) as a factor.

    \[\begin{aligned} &\frac{16+8h+h^2-12-3h-4}{h} \\ &\quad=\frac{5h+h^2}{h} \end{aligned}\]

  3. Cancel \(h\), which is allowed because \(h\) is not 0.

    \[5+h\]

  4. Let \(h\) approach 0. The averages approach 5.

    \[\cy{f'(4)=5}\]

    As a check, for \(h = 0.1\) the average is 5.1.

Example 2: a cube

Find the instantaneous rate of change of \(g(x) = x^3\) at \(x = 2\).

  1. Because \(g(2) = 8\), the average rate from 2 to \(2 + h\) is this fraction.

    \[\frac{(2+h)^3-8}{h}\]

  2. A tempting wrong move: Substitute \(h = 0\) into this fraction.

    \[\cr{\frac{(2+0)^3-8}{0}=0}\]

    The red line through \(P\) has slope 0, but the secants for \(h = 0.1\) and \(h = 0.01\) have slopes 12.61 and 12.0601, so the averages approach 12. Also, \(0 \div 0\) has no value.

  3. Expand the cube first. The 8s cancel, so every term in the numerator has \(h\) as a factor, and \(h\) cancels.

    \[\begin{aligned} &\frac{8+12h+6h^2+h^3-8}{h} \\ &\quad=12+6h+h^2 \end{aligned}\]

  4. Let \(h\) approach 0. The averages approach 12.

    \[\cy{g'(2)=12}\]

Practice

A circle has radius \(r\) meters and area \(A(r) = \pi r^2\) square meters. Find the instantaneous rate of change of \(A\) with respect to \(r\) at \(r = 3\).

The slope of the orange secant is \(\frac{A(3+h) - A(3)}{h}\). Expand the numerator and cancel \(h\) before you let \(h\) approach 0.

Solution

  1. Expand the change in area, \(A(3+h) - A(3)\).

    \[\begin{aligned} &\pi(3+h)^2-9\pi \\ =\ &\pi(9+6h+h^2)-9\pi \\ =\ &\pi(6h+h^2) \end{aligned}\]

  2. Divide by \(h\) and cancel.

    \[\frac{\pi(6h+h^2)}{h}=\pi(6+h)\]

  3. Let \(h\) approach 0. The area grows at \(6\pi\) square meters for each meter of radius.

    \[\cy{A'(3)=6\pi}\]

Questions

Questions 1 and 2 use the graph of \(f(x) = \frac{x^2}{4}\). Question 3 needs a calculation.

  1. The orange tangent touches the curve at \(x = 4\) and passes through \((2, 0)\) and \((6, 8)\). What is the instantaneous rate of change of \(f\) at \(x = 4\)?

    • 2
    • 4
    • 2.5
    Answer

    2

  2. The two chalk secants start at \(P\) and end at the points where \(x = 5\) and \(x = 6\). Which average rate is closer to the instantaneous rate at \(x = 4\)?

    • The average over \([4, 5]\)
    • The average over \([4, 6]\)
    • They are equally close
    Answer

    The average over \([4, 5]\)

  3. Find \(f'(2)\) for \(f(x) = 3x^2 - x\).

    • 11
    • 14
    • 10
    • 0
    Answer

    11

Summary

You write the average rate from \(a\) to \(a + h\) and simplify until \(h\) cancels.

You let \(h\) approach 0, because the instantaneous rate is the number that the averages approach.

You use it to find a rate at one moment or one input, such as the speed of a ball at 2 seconds.

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