Average Rate of Change

Find the average rate of change of the function $f\left(x\right)=2x^2-1$ as x varies from 1 to 3.

Solution:

Step 1: Calculate the value of the function at the endpoints, 1 and 3:

$f\left(1\right)=2{\left(1\right)}^2-3=-1$

$f\left(3\right)=2{\left(3\right)}^2-3=15$

Step 2: Find the change in x: $3-1=2$.

Step 3: Take the ratio of the change in function to the change in x:

$\frac{\Delta y}{\Delta x}=\frac{f\left(3\right)-f\left(1\right)}{3-1}$

$\frac{\Delta y}{\Delta x}=\frac{15+1}{2}=8$

Hence the average rate of change of the function w.r.t. x between the points 3 and 1 is 8.

Find the average rate of change of the function$y=e^{2t}-\frac{t^3}{2}$ between the values of t:$0\mathrm{and}\frac{3}{2}$.

Solution:

Step 1: Calculate the value of the function at the points given:

$f\left(0\right)=e^{2\left(0\right)}-\frac{{\left(0\right)}^3}{2}=1-0=1$

$f\left(\frac{3}{2}\right)=e^{2.\frac{3}{2}}-\frac{1}{2}{\left(\frac{3}{2}\right)}^3=e^3-\frac{27}{16}$

Step 2: Calculate the difference in the t coordinates: $\frac{3}{2}-0=\frac{3}{2}$.

Step3: Take the ratio the the change in the function and the x coordiantes:

$\frac{\Delta y}{\Delta t}=\frac{f\left(\frac{3}{2}\right)-f\left(0\right)}{\frac{3}{2}-0}$

$\frac{\Delta y}{\Delta x}=\frac{e^3-\frac{27}{16}-1}{\frac{3}{2}}$$=\frac{2e^3}{3}-\frac{27}{24}-\frac{2}{3}$$=\frac{16e^3-27-16}{24}$

$\frac{\Delta y}{\Delta t}=\frac{16e^3-43}{24}$

Hence the average rate of change of the function is $\frac{16e^3-43}{24}$ between the points $0\mathrm{and}\frac{3}{2}$.

A bus covers a distance of 60 km from its first stop to its last stop. It takes the bus about 2 hours for the whole journey. The bus driver says that he had several stops in between and drove at different speeds at different times. The bus driver has to report the average speed he was driving at so that the authorities can be assured that he was driving under the speed limit on average. The speed limit is 35 km/h. What is his average speed under it?

Solution:

Step 1: Convert the given data into mathematical information. Let d be the distance covered by the bus and t be the time it takes to cover the distance.

Step 2: Find the change in net distance covered: $\Delta d=60-0=60$ km. And the time taken to cover that distance: $\Delta t=2hrs$.

Step 3: The average speed of an object is given by the ratio of the net distance covered by the bus to the time taken to cover the distance, hence:

$V_{av}=\frac{\Delta d}{\Delta t}$$=\frac{60}{2}\frac{km}{hr}$

$V_{av}=30km/hr$

Hence the average speed at which the driver was driving was 30 km/hr, which is below the average speed limit of 35 km/hr.