AP Calculus AB/BC · Unit 1 · Topic 1.1
Introducing Calculus: Can Change Occur at an Instant?
Use average rates over shrinking intervals to motivate instantaneous change.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Use average rates over shrinking intervals to motivate instantaneous change.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For s(t)=t²+1 at t=2, the average velocity 4+h approaches 4.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
The apparent paradox
An instant has no duration, so the ordinary average-rate formula cannot use the interval \([a,a]\): its denominator would be zero. Calculus resolves this without dividing by zero. Keep a genuine interval with \(h\ne0\), calculate its average rate, and then study the value approached as the interval shrinks.
If these rates approach one finite number from both sides, that number is the instantaneous rate at \(a\):
Two equivalent ways to shrink the interval
The moving endpoint can be called \(x\), or it can be written as \(a+h\). Since \(h=x-a\), the two difference quotients describe the same secant slopes:
- \(f(a+h)-f(a)\) is the output change.
- \(h\) is the nonzero input change.
- The quotient is an average rate and the slope of a secant line.
- The limiting value, when it exists, is an instantaneous rate and the slope of the tangent line.
Numerical evidence from both sides
Let \(p(t)=t^2+2t\). At \(t=1\),
| \(h\) | Interval endpoint \(1+h\) | Average rate \(4+h\) | Approach |
|---|---|---|---|
| \(-0.10\) | \(0.90\) | \(3.90\) | From the left |
| \(-0.01\) | \(0.99\) | \(3.99\) | From the left |
| \(0.01\) | \(1.01\) | \(4.01\) | From the right |
| \(0.10\) | \(1.10\) | \(4.10\) | From the right |
The rates are not required to equal 4 on any nonzero interval. What matters is that the left- and right-side values settle toward the same number, 4.
Connecting representations
- Verbal: output is changing at about 4 output units per input unit at \(t=1\).
- Numerical: average rates over shorter intervals approach 4.
- Graphical: secant lines pivot toward a tangent line with slope 4.
- Analytical: the difference quotient simplifies to \(4+h\), whose limit is 4.
When no instantaneous rate exists
A function value can exist even when nearby secant slopes do not approach one number. For \(q(t)=|t|\) at \(t=0\),
The left-side rates approach \(-1\) and the right-side rates approach \(1\). The graph has a corner, so there is no single two-sided instantaneous rate at 0.
Interpret sign and units
If position is measured in meters and time in seconds, both average and instantaneous velocity have units of meters per second. Positive velocity means position is increasing; negative velocity means position is decreasing. Velocity includes direction, while speed is its nonnegative magnitude. An instantaneous statement describes local behavior at one input and does not claim that the rate stays constant on a surrounding interval.
6. Detailed Worked Example and Error Check
Example. Let \(s(t)=t^3-2t\), where \(s\) is position in meters and \(t\) is time in seconds. Find and interpret the instantaneous velocity at \(t=1\).
Step 1: Build an average velocity over a real interval. Keep the first time fixed at 1 and move the second time to \(1+h\), where \(h\ne0\):
Step 2: Simplify before taking the limit.
The cancellation is valid because every average-rate interval uses \(h\ne0\).
Step 3: Check the numerical trend.
| \(h\) | Average velocity \(1+3h+h^2\) |
|---|---|
| \(-0.10\) | \(0.71\text{ m/s}\) |
| \(-0.01\) | \(0.9701\text{ m/s}\) |
| \(0.01\) | \(1.0301\text{ m/s}\) |
| \(0.10\) | \(1.31\text{ m/s}\) |
Step 4: Take the limit and interpret it.
At exactly 1 second, the object's position is increasing at an instantaneous rate of \(1\) meter per second. Geometrically, the graph of \(s\) has tangent slope 1 at \((1,-1)\), so its tangent line is
Error check: substituting \(h=0\) into the original quotient gives \(0/0\), which is undefined. The answer comes from the nearby quotients' limit after valid algebra for \(h\ne0\), not from evaluating a zero-length interval.
7. AP Reasoning Routine
Read one-sided behavior first, choose a matching limit procedure, and justify conclusions with definitions or theorem conditions.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Let \(A(t)=40+8t-t^2\) be the amount of water in a tank, in liters, \(t\) minutes after monitoring begins.
(a) Write and simplify the average rate from \(t=3\) to \(t=3+h\).
(b) Use \(h=-0.1\) and \(h=0.1\) to give numerical evidence for the instantaneous rate at \(t=3\).
(c) Find the instantaneous rate and interpret its sign and units.
(d) Write the tangent-line model at \(t=3\) and use it to estimate \(A(3.04)\).
(e) Explain why setting \(h=0\) in the original average-rate quotient is not a valid calculation.
Check the solution
\(A(3)=55\) and \(A(3+h)=55+2h-h^2\), so for \(h\ne0\), \(\frac{A(3+h)-A(3)}h=2-h\). The rates are \(2.1\) L/min for \(h=-0.1\) and \(1.9\) L/min for \(h=0.1\), both approaching 2. Thus the water amount is increasing at \(2\) L/min at \(t=3\). The tangent model is \(L(t)=55+2(t-3)\), giving \(A(3.04)\approx55.08\) L. At \(h=0\), the original quotient is \(0/0\), so the instantaneous rate must be obtained as a limit of quotients with \(h\ne0\).