Unit 2 · Topic 2.3 Beta

Estimating a Derivative at a Point

With data instead of a formula, estimate f′(a) with a difference quotient from two table inputs on either side of a , as close as possible, and give the units.

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

A nurse records a patient's temperature every few hours, and a lab logs a reaction's concentration every few seconds. Nobody hands you a formula. Yet the question is still "how fast is it changing right now?" With a table, a graph or a calculator, you can estimate a derivative from the data you actually have.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The derivative f′(a)f'(a) is the limit of

  1. f(a+h)−f(a)h\frac{f(a + h) - f(a)}{h} as h→0h \to 0
  2. f(a+h)f(a + h) as h→0h \to 0
  3. f(a)a\frac{f(a)}{a} as a→0a \to 0
  4. f(a+h)−f(a)f(a + h) - f(a) as h→0h \to 0
Show the answer

A derivative is a limit of difference quotients: average rates on shrinking intervals.

  • Correct: f(a+h)−f(a)h\frac{f(a + h) - f(a)}{h} as h→0h \to 0:
  • f(a+h)f(a + h) as h→0h \to 0:
  • f(a)a\frac{f(a)}{a} as a→0a \to 0:
  • f(a+h)−f(a)f(a + h) - f(a) as h→0h \to 0:

2. If T(t)T(t) is a temperature in degrees Celsius and tt is in minutes, the units of T(9)−T(5)9−5\frac{T(9) - T(5)}{9 - 5} are

  1. degrees Celsius per minute
  2. degrees Celsius
  3. minutes per degree Celsius
  4. degrees Celsius per minute per minute
Show the answer

A change in temperature over a change in time.

  • Correct: degrees Celsius per minute:
  • degrees Celsius:
  • minutes per degree Celsius:
  • degrees Celsius per minute per minute:

Part 4 · See it

See it first

Five measured temperatures B(t) of a cooling loaf of bread, in degrees Fahrenheit, at t = 0, 4, 10, 14 and 20 minutes: 180, 152, 121, 105 and 88. A dashed secant line joins the points at t = 10 and t = 14, on either side of t = 12. Its slope, (105 minus 121) over (14 minus 10), is minus 4 degrees Fahrenheit per minute, the estimate of B prime of 12.
From a table: the secant line through the data at t=10t = 10 and t=14t = 14 estimates B′(12)B'(12) as 105−12114−10\frac{105 - 121}{14 - 10}=−4= -4 degrees Fahrenheit per minute. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Only a table of values is knownthe limit in the definition cannot be taken, so use a difference quotient with the data
  2. The two table inputs sit on either side of aa and close to ittheir difference quotient is a good estimate of f′(a)f'(a)
  3. The quotient divides output units by input unitsthe estimate is a rate with units, such as degrees per minute
  4. A formula is known but hard to work with by handa calculator's numerical derivative estimates f′(a)f'(a) with a tiny symmetric difference quotient

Part 6 · Key ideas

Key ideas

  • From a table, estimate f′(a)f'(a) by f(b)−f(c)b−c\frac{f(b) - f(c)}{b - c} with c<a<bc < a < b: the two closest data points on either side of aa.
  • At the end of a table, only one side exists: use the closest one-sided quotient.
  • The symmetric difference quotient f(a+h)−f(a−h)2h\frac{f(a + h) - f(a - h)}{2h} averages the left and right quotients and is usually the most accurate.
  • A numerical derivative on a graphing calculator gives a decimal estimate of f′(a)f'(a); report it to three decimal places.
  • From a graph, read the slope of the tangent line from two points on it.

Part 7 · Misconception

A common mistake

The wrong idea: To estimate P′(4)P'(4) from P(3)=20P(3) = 20 and P(5)=34P(5) = 34, average the values: P′(4)≈27P'(4) \approx 27.

What actually happens: Averaging values estimates P(4)P(4), a value. A derivative is a rate: P′(4)P'(4)≈34−205−3\approx \frac{34 - 20}{5 - 3}=7= 7, in output units per input unit.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A cup of coffee cooling on a desk

Temperature H(t)H(t) of the coffee, in degrees Celsius, at selected times tt, in minutes
tt (minutes)025912
H(t)H(t) (degrees Celsius)9285776964

1. Use the data in the table to estimate H′(7)H'(7). Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

t=7t = 7 lies between the data at t=5t = 5 and t=9t = 9: H′(7)H'(7)≈H(9)−H(5)9−5\approx \frac{H(9) - H(5)}{9 - 5}=69−774= \frac{69 - 77}{4}=−2= -2 degrees Celsius per minute.

  • Answer: -2 degrees Celsius per minute

2. Which sentence interprets H′(7)≈−2H'(7) \approx -2 in context?

  1. At t=7t = 7 minutes, the coffee's temperature is decreasing at about 2 degrees Celsius per minute.
  2. At t=7t = 7 minutes, the coffee's temperature is about −2-2 degrees Celsius.
  3. From t=5t = 5 to t=9t = 9 minutes, the coffee's temperature drops a total of 2 degrees Celsius.
  4. At t=7t = 7 minutes, the temperature is decreasing at about 2 degrees Celsius per minute per minute.
Show the answer

A derivative is a rate at one instant; the negative sign means the temperature is going down.

  • Correct: At t=7t = 7 minutes, the coffee's temperature is decreasing at about 2 degrees Celsius per minute.: It names the time, the quantity, the direction and the units.
  • At t=7t = 7 minutes, the coffee's temperature is about −2-2 degrees Celsius.: That reads the derivative as a temperature; it is a rate of change of the temperature.
  • From t=5t = 5 to t=9t = 9 minutes, the coffee's temperature drops a total of 2 degrees Celsius.: Over those 4 minutes the drop is 8 degrees; −2-2 is a rate per minute.
  • At t=7t = 7 minutes, the temperature is decreasing at about 2 degrees Celsius per minute per minute.: That unit is a rate of a rate; H′H' is in degrees Celsius per minute.

3. Using the table, which is the best estimate of H′(10.5)H'(10.5)?

  1. −53-\frac{5}{3} degrees Celsius per minute
  2. −5-5 degrees Celsius per minute
  3. −35-\frac{3}{5} degrees Celsius per minute
  4. −83-\frac{8}{3} degrees Celsius per minute
Show the answer

The closest data on either side of 10.5 are at 9 and 12: 64−6912−9\frac{64 - 69}{12 - 9}=−53= -\frac{5}{3}.

  • Correct: −53-\frac{5}{3} degrees Celsius per minute: Change −5-5 over 3 minutes.
  • −5-5 degrees Celsius per minute: That is the change in temperature; divide by 3 minutes.
  • −35-\frac{3}{5} degrees Celsius per minute: The quotient is upside down.
  • −83-\frac{8}{3} degrees Celsius per minute: That uses t=2t = 2 and t=5t = 5, data that does not contain t=10.5t = 10.5.

4. Which difference quotient should NOT be used to estimate H′(5)H'(5)?

  1. H(12)−H(9)12−9\frac{H(12) - H(9)}{12 - 9}
  2. H(5)−H(2)5−2\frac{H(5) - H(2)}{5 - 2}
  3. H(9)−H(5)9−5\frac{H(9) - H(5)}{9 - 5}
  4. H(9)−H(2)9−2\frac{H(9) - H(2)}{9 - 2}
Show the answer

An estimate of H′(5)H'(5) must use an interval that contains t=5t = 5. The interval [9,12][9, 12] describes a later stretch of time.

  • Correct: H(12)−H(9)12−9\frac{H(12) - H(9)}{12 - 9}: Its interval, [9,12][9, 12], does not contain t=5t = 5.
  • H(5)−H(2)5−2\frac{H(5) - H(2)}{5 - 2}: A one-sided quotient ending at t=5t = 5 is a reasonable estimate.
  • H(9)−H(5)9−5\frac{H(9) - H(5)}{9 - 5}: A one-sided quotient starting at t=5t = 5 is a reasonable estimate.
  • H(9)−H(2)9−2\frac{H(9) - H(2)}{9 - 2}: This interval has t=5t = 5 inside it, so it is reasonable, if wider.

5. A runner has covered D(t)D(t) meters after tt seconds, with D(10)=62D(10) = 62 and D(12)=71D(12) = 71. Which is the best estimate of D′(11)D'(11)?

  1. 4.54.5 meters per second
  2. 99 meters per second
  3. 4.54.5 meters
  4. 29\frac{2}{9} seconds per meter
Show the answer

D(12)−D(10)12−10\frac{D(12) - D(10)}{12 - 10}=92= \frac{9}{2}=4.5= 4.5 meters per second; t=11t = 11 is midway between the data.

  • Correct: 4.54.5 meters per second: Change in distance over change in time, with rate units.
  • 99 meters per second: That is the 9 meters covered; divide by the 2 seconds.
  • 4.54.5 meters: The number is right but a derivative of distance is a rate, in meters per second.
  • 29\frac{2}{9} seconds per meter: That is time over distance, upside down.

6. A bacteria culture has P(t)P(t) thousand bacteria after tt hours, with P(3)=20P(3) = 20 and P(5)=34P(5) = 34. Which is the best estimate of P′(4)P'(4)?

  1. 77 thousand bacteria per hour
  2. 2727 thousand bacteria
  3. 1414 thousand bacteria per hour
  4. 77 thousand bacteria
Show the answer

P(5)−P(3)5−3\frac{P(5) - P(3)}{5 - 3}=142= \frac{14}{2}=7= 7 thousand bacteria per hour.

  • Correct: 77 thousand bacteria per hour: A rate: change in population over change in time.
  • 2727 thousand bacteria: That averages the two values, which estimates P(4)P(4), not P′(4)P'(4).
  • 1414 thousand bacteria per hour: That is the change over 2 hours; divide by 2.
  • 77 thousand bacteria: The number is right but the units are those of PP; a rate is per hour.

Part 9 · Summary

Summary

With data instead of a formula, estimate f′(a)f'(a) with a difference quotient from two table inputs on either side of aa, as close as possible, and give the units. A symmetric quotient is the best simple estimate; a calculator's numerical derivative does the same with a tiny hh, reported to three decimal places.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections