Interpolation and Extrapolation

2 questions

A group of microbiologists measured the enzymatic reaction rate ( $R$ , in $\text{nmol/min}\cdot\text{mg}$ ) of a newly discovered halophilic bacterial strain across various salinity levels ( $S$ , in parts per thousand, ppt) at three fixed temperatures ( $20^\circ\text{C}$ , $30^\circ\text{C}$ , and $40^\circ\text{C}$ ). The results are recorded in the table below.

Salinity ( $S$ , ppt)	Rate at $20^\circ\text{C}$	Rate at $30^\circ\text{C}$	Rate at $40^\circ\text{C}$
10	14.0	22.0	10.0
20	20.0	32.0	16.0
30	26.0	42.0	22.0
50	38.0	62.0	34.0

Based on linear interpolation and extrapolation of the data trends, arrange the four estimated reaction rates described below in order from lowest to highest.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest estimated reaction rate is: the estimated rate at

30^\circ\text{C}

and

5\text{ ppt}

(

17.0\text{ nmol/min}\cdot\text{mg}

), followed by the estimated rate at

40^\circ\text{C}

and

25\text{ ppt}

(

19.0\text{ nmol/min}\cdot\text{mg}

), followed by the estimated rate at

20^\circ\text{C}

and

60\text{ ppt}

(

44.0\text{ nmol/min}\cdot\text{mg}

), and finally the estimated rate at

30^\circ\text{C}

and

40\text{ ppt}

(

52.0\text{ nmol/min}\cdot\text{mg}

To place the estimated rates in correct order from lowest to highest, calculate each value using the constant linear rates of change shown in the table. At

30^\circ\text{C}

and

5\text{ ppt}

, extrapolating below

10\text{ ppt}

yields

17.0\text{ nmol/min}\cdot\text{mg}

. At

40^\circ\text{C}

and

25\text{ ppt}

, interpolating halfway between

20

and

30\text{ ppt}

yields

19.0\text{ nmol/min}\cdot\text{mg}

. At

20^\circ\text{C}

and

60\text{ ppt}

, extrapolating above

50\text{ ppt}

yields

44.0\text{ nmol/min}\cdot\text{mg}

. At

30^\circ\text{C}

and

40\text{ ppt}

, interpolating halfway between

30

and

50\text{ ppt}

yields

52.0\text{ nmol/min}\cdot\text{mg}

. Comparing these numerical values (

17.0 < 19.0 < 44.0 < 52.0

) produces the correct sequence.

Step-by-Step Solution

Determine the linear relationship (slope) for each temperature column.

For

20^\circ\text{C}

, the rate increases by

0.6\text{ nmol/min}\cdot\text{mg}

per ppt salinity. For

30^\circ\text{C}

, the rate increases by

1.0\text{ nmol/min}\cdot\text{mg}

per ppt salinity. For

40^\circ\text{C}

, the rate increases by

0.6\text{ nmol/min}\cdot\text{mg}

per ppt salinity.

Establishing the rate of change allows calculation of values between existing data points (interpolation) and beyond existing data points (extrapolation).

Calculate the numerical value for each item.

Item 1 (

30^\circ\text{C}, 40\text{ ppt}

42.0 + 10(1.0) = 52.0\text{ nmol/min}\cdot\text{mg}

.
Item 2 (

20^\circ\text{C}, 60\text{ ppt}

38.0 + 10(0.6) = 44.0\text{ nmol/min}\cdot\text{mg}

.
Item 3 (

40^\circ\text{C}, 25\text{ ppt}

16.0 + 5(0.6) = 19.0\text{ nmol/min}\cdot\text{mg}

.
Item 4 (

30^\circ\text{C}, 5\text{ ppt}

22.0 - 5(1.0) = 17.0\text{ nmol/min}\cdot\text{mg}

Obtaining exact numerical estimates enables precise relative comparison.

Sort the items by calculated value from lowest to highest.

17.0

(Item 4) <

19.0

(Item 3) <

44.0

(Item 2) <

52.0

(Item 1).

Direct numerical comparison determines the correct sequence.

Key Concept

Interpolation and Extrapolation of Data Trends

Estimated Time:1m 30s

Question 2Question →

Materials engineers measured the electrical conductivity ( $\sigma$ , in $10^6\text{ S/m}$ ) of a copper-nickel alloy at various temperatures ( $T$ , in K). The measured values are recorded in the table below:

Temperature ( $T$ , K)	Electrical Conductivity ( $\sigma$ , $10^6\text{ S/m}$ )
300	14.2
400	11.8
500	9.4
600	7.0
700	4.6

Based on the data, if the linear relationship between temperature and electrical conductivity continues beyond $700\text{ K}$ , what is the predicted electrical conductivity of the alloy at $800\text{ K}$ ?

2.2 \times 10^6\text{ S/m}

3.4 \times 10^6\text{ S/m}

4.6 \times 10^6\text{ S/m}

7.0 \times 10^6\text{ S/m}

Show answer & explanation

Answer: $2.2 \times 10^6\text{ S/m}$

Answer

The predicted electrical conductivity at

800\text{ K}

2.2 \times 10^6\text{ S/m}

The electrical conductivity decreases by a constant rate of

2.4 \times 10^6\text{ S/m}

for every

100\text{ K}

increase in temperature. At

700\text{ K}

, the conductivity is

4.6 \times 10^6\text{ S/m}

. Extrapolating to

800\text{ K}

requires subtracting

2.4 \times 10^6\text{ S/m}

from

4.6 \times 10^6\text{ S/m}

, giving

2.2 \times 10^6\text{ S/m}

Step-by-Step Solution

Determine the change in electrical conductivity per

100\text{ K}

temperature increase.

Between

300\text{ K}

and

400\text{ K}

, conductivity decreases by

14.2 - 11.8 = 2.4 \times 10^6\text{ S/m}

. Checking other intervals confirms a constant rate of

-2.4 \times 10^6\text{ S/m}

per

100\text{ K}

Establishing the linear rate of change is necessary for linear extrapolation.

Apply the linear trend to extrapolate from

700\text{ K}

800\text{ K}

4.6 - 2.4 = 2.2 \times 10^6\text{ S/m}

Extrapolation involves extending the observed trend by one temperature interval (

100\text{ K}

) beyond the measured range.

Key Concept

Linear Extrapolation

Estimated Time:1m 0s

Interpolation and Extrapolation

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes