Biophysicists conducted an experiment measuring the action potential conduction velocity (… — ACT

Biophysicists conducted an experiment measuring the action potential conduction velocity ( $v$ , in $\text{m/s}$ ) in unmyelinated giant nerve fibers as a function of fiber diameter ( $d$ , in $\mu\text{m}$ ) at a constant temperature of $20^\circ\text{C}$ . The recorded data is presented in Table 1.

Fiber Diameter ( $d$ , $\mu\text{m}$ )	Conduction Velocity ( $v$ , $\text{m/s}$ )
100	5.0
200	8.0
400	14.0
600	18.0
900	21.0

Based on Table 1, what is the average rate of change in conduction velocity, in $\text{m/s}$ per $100\,\mu\text{m}$ increase in fiber diameter, over the interval from $d = 200\,\mu\text{m}$ to $d = 600\,\mu\text{m}$ ?

Answer: 2.5 m/s per 100 µm

Answer

The average rate of change in conduction velocity over the specified interval is

2.5\,\text{m/s per }100\,\mu\text{m}

increase in fiber diameter.

To find the average rate of change in conduction velocity per

100\,\mu\text{m}

increase in fiber diameter between

d = 200\,\mu\text{m}

and

d = 600\,\mu\text{m}

, subtract the initial velocity (

8.0\,\text{m/s}

) from the final velocity (

18.0\,\text{m/s}

) to get

\Delta v = 10.0\,\text{m/s}

. Divide by the change in diameter

\Delta d = 600 - 200 = 400\,\mu\text{m}

to obtain

0.025\,\text{m/s per }\mu\text{m}

. Multiplying by

100

yields

2.5\,\text{m/s per }100\,\mu\text{m}

Step-by-Step Solution

Locate data points for

d = 200\,\mu\text{m}

and

d = 600\,\mu\text{m}

in Table 1.

d = 200\,\mu\text{m}

v = 8.0\,\text{m/s}

. At

d = 600\,\mu\text{m}

v = 18.0\,\text{m/s}

These data points define the boundaries of the interval specified in the question.

Calculate the overall changes in velocity (

\Delta v

) and diameter (

\Delta d

\Delta v = 18.0 - 8.0 = 10.0\,\text{m/s}

and

\Delta d = 600 - 200 = 400\,\mu\text{m}

Calculating the rate of change requires dividing the change in the dependent variable by the change in the independent variable.

Scale the rate of change to a

100\,\mu\text{m}

diameter interval.

\frac{10.0\,\text{m/s}}{400\,\mu\text{m}} \times 100\,\mu\text{m} = 2.5\,\text{m/s per }100\,\mu\text{m}

The question asks specifically for the rate per

100\,\mu\text{m}

increase in diameter.

Key Concept

Calculating average rate of change and trend slopes from quantitative scientific data tables.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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