Calculus
Calculus · Lesson Review · August 17, 2026

Estimating Instantaneous Rate from a Table

When no function formula is available, use average rates immediately before and after the target to estimate the tangent slope.

8-minute edited lesson7 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captionsSRT captions
0:0001Estimating Instantaneous Rate from a Table1:2902Instantaneous Rate as a Tangent Slope2:3603Why the Formula-Based Method Is Unavailable3:4204Use Secant Slopes on Both Sides4:5805Compute the Neighboring Average Rates6:1806Average the Two Rate Estimates7:3807When to Use This Method

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02 · Understand

Learning targets

03 · Vocabulary

Key definitions

Connect the calculation to its graph and units.

01Instantaneous rate of change
The output's rate of change at one exact input value. On a graph, it is the slope of the tangent line.
02Average rate of change
Change in output divided by change in input over an interval. On a graph, it is the slope of a secant line.
03Tangent line
A line that matches a curve's direction at one point. Its slope represents the instantaneous rate.
04Secant line
A line through two points on a curve. Its slope represents an average rate over an interval.
05Two-sided estimate
An estimate that uses a rate just before and a rate just after the target, then combines them.
06Compound units
Units written as output units per input unit, such as gallons per minute.

04 · Apply

Worked example

Estimate the pool's instantaneous volume rate at 20 minutes.

Given data

Water volume over time

Recorded volume of water in the pool
Time (min)Volume (gal)
102,000
152,500
203,500
255,000
3010,000

There is no formula for V(t)V(t), so we cannot evaluate inputs such as 19.9919.99 and 20.0120.01.

Step 1

Find the average rate just before 20

  1. AROC⁡[15,20]=3500−250020−15\operatorname{AROC}_{[15,20]}=\frac{3500-2500}{20-15}
  2. =10005=200 gal/min=\frac{1000}{5}=200\ \mathrm{gal/min}

This is the green secant slope through the data at 15 and 20 minutes.

Step 2

Find the average rate just after 20

  1. AROC⁡[20,25]=5000−350025−20\operatorname{AROC}_{[20,25]}=\frac{5000-3500}{25-20}
  2. =15005=300 gal/min=\frac{1500}{5}=300\ \mathrm{gal/min}

This is the orange secant slope through the data at 20 and 25 minutes.

Step 3

Average the rates on both sides

  1. 200+3002=250 gal/min\frac{200+300}{2}=250\ \mathrm{gal/min}

Estimated instantaneous rate at 20 minutes: 250 gallons per minute.

05 · Avoid

Common mistakes

Averaging the volumes.

Average the two calculated rates, 200 and 300—not the table's volume values.

Using only one side.

A left or right secant alone is an average rate. Using both sides gives a better estimate at the target.

Dropping the units.

Volume divided by time produces gallons per minute, not gallons or minutes alone.

Pretending a formula exists.

Do not invent values between table entries. Use the information actually provided.

06 · Check yourself

Quick check

Try each question before opening the answer.

1 What does the tangent-line slope represent?

The instantaneous rate of change at the point of tangency.

2 Why can we not use values such as 19.99 and 20.01?

No function formula is given. The table only supplies five recorded inputs.

3 What is the average rate on [15,20][15,20]?

200 gal/min. Compute 3500−250020−15\frac{3500-2500}{20-15}.

4 What is the average rate on [20,25][20,25]?

300 gal/min. Compute 5000−350025−20\frac{5000-3500}{25-20}.

5 What is the final instantaneous-rate estimate at 20 minutes?

250 gal/min. Average the two neighboring rates: 200+3002\frac{200+300}{2}.

07 · Revisit

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