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Wize High School Grade 12 Pre-Calculus Textbook > Rates of Change

Average Rate of Change & the Secant Line

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Average Rate of Change & the Secant Line


The average rate of change for some function f(x)f(x) over the closed interval [x1,x2][x_1,x_2] is defined as the slope of the secant line that passes through the points (x1,f(x1))(x_1,f(x_1)) and (x2,f(x2))(x_2,f(x_2)) and it is expressed as:

Average Rate of Change=msec=f(x2)−f(x1)x2−x1\text{Average Rate of Change}=m_{sec}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}


A secant line is a line that passes through at least two points on a curve.
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Example

Let h(t)=−4.9t2+16t+28h(t)=-4.9t^2+16t+28 represent the height, in meters, of a cannon when t≥0t\geq{0} (tt is in seconds).

The average rate of change over each of the following intervals is:


0≤t≤2‾:\underline{0\leq{t}\leq{2}}:

msec=f(x2)−f(x1)x2−x1=−4.9(2)2+16(2)+28−282−0=20.2 m/s\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{-4.9(2)^2+16(2)+28-28}{2-0}\\\\ &=&20.2~\text{m/s} \end{array}



2≤t≤4‾:\underline{2\leq{t}\leq{4}}:
msec=f(x2)−f(x1)x2−x1=(−4.9(4)2+16(4)+28)−(−4.9(2)2+16(2)+28)4−2=−13.4 m/s\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{(-4.9(4)^2+16(4)+28)-(-4.9(2)^2+16(2)+28)}{4-2}\\\\ &=&-13.4~\text{m/s} \end{array}


Example: Average Rate of Change & the Secant Line


The volume of an air balloon is given by V(t)=109t+7V(t)=\dfrac{10}{9t+7}. Compute the average rate of change of the air balloon between t=0t=0 and the following values of tt.
  • t=1t=1
  • t=0.5t=0.5
  • t=0.1t=0.1
  • t=0.01t=0.01
  • t=0.001t=0.001
t=1‾:\underline{t=1}:
msec=f(x2)−f(x1)x2−x1=109(1)+7−109(0)+71−0=58−107=−4556≈−0.8036\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{\frac{10}{9(1)+7}-\frac{10}{9(0)+7}}{1-0}\\\\ &=&\dfrac{5}{8}-\dfrac{10}{7}\\\\ &=&-\dfrac{45}{56}\\\\ &\approx&-0.8036 \end{array}

t=0.5‾:\underline{t=0.5}:
msec=f(x2)−f(x1)x2−x1=109(0.5)+7−109(0)+70.5−0=(2023−107)0.5=−180161≈−1.118\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{\frac{10}{9(0.5)+7}-\frac{10}{9(0)+7}}{0.5-0}\\\\ &=&\dfrac{\Bigg(\dfrac{20}{23}-\dfrac{10}{7}\Bigg)}{0.5}\\\\ &=&-\dfrac{180}{161}\\\\ &\approx&-1.118 \end{array}

t=0.1‾:\underline{t=0.1}:
msec=f(x2)−f(x1)x2−x1=109(0.1)+7−109(0)+70.1−0=10079−1070.1=−900553≈−1.6275\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{\frac{10}{9(0.1)+7}-\frac{10}{9(0)+7}}{0.1-0}\\\\ &=&\dfrac{\dfrac{100}{79}-\dfrac{10}{7}}{0.1}\\\\ &=&-\dfrac{900}{553}\\\\ &\approx&-1.6275 \end{array}

t=0.01‾:\underline{t=0.01}:
msec=f(x2)−f(x1)x2−x1=109(0.01)+7−109(0)+70.01−0=1000709−1070.01=−90004963≈−1.8134\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{\frac{10}{9(0.01)+7}-\frac{10}{9(0)+7}}{0.01-0}\\\\ &=&\dfrac{\dfrac{1000}{709}-\dfrac{10}{7}}{0.01}\\\\ &=&-\dfrac{9000}{4963}\\\\ &\approx&-1.8134 \end{array}

t=0.001‾:\underline{t=0.001}:
msec=f(x2)−f(x1)x2−x1=109(0.001)+7−109(0)+70.001−0=100007009−1070.001≈−1.8344\begin{array}{rcl} m_{sec}&=&\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\\\\ &=&\dfrac{\frac{10}{9(0.001)+7}-\frac{10}{9(0)+7}}{0.001-0}\\\\ &=&\dfrac{\dfrac{10000}{7009}-\dfrac{10}{7}}{0.001}\\\\ &\approx&-1.8344 \end{array}

Practice: Average Rate of Change & the Secant Line

Determine the average rate of the change of the function g(t)=5tt−1g(t)=\dfrac{5t}{t-1} over the interval 1.5≤t≤31.5\leq{}t\leq{}3.

Practice: Average Rate of Change & the Secant Line

The population, PP, of a certain species over time tt (in years) is given by the table of values below.

t01234P10204080160\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{P}&10&20&40&80&160\\\hline \end{array}

Determine the following average rates of change over the following intervals:

Practice: Average Rate of Change & the Secant Line

Germaine and his family purchased a home in 2000 for $400,000\$400,000. 1515 years later, they sold the house for $550,000\$550,000. What is the average annual rate of change of the value of the home?
Extra Practice
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
The population of Vancouver in 2000 was 450,000. In 2020, the population was 740,000. What is the average annual rate of change of the population of Vancouver?
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Sherry bought a car in 2010 for $60,000\$60,000. 5050 years later, she sold the car as a collector car for $98,000\$98,000. What is the average annual rate of change of the value of the car? (Ignore units)
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
An investment, PP, in $, grows with interest over time tt (in years) is given by the table of values below.
t01234P10001305152817741936\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{P}&1000&1305&1528&1774&1936\\\hline \end{array}
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
The amount of bacteria, AA, in mg, of a certain species over time tt (in minutes) is given by the table of values below.
t01234A150183204247283\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{A}&150&183&204&247&283\\\hline \end{array}
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Determine the average rate of the change of the function g(t)=−3t2t+5g(t)=\dfrac{-3t}{2t+5} over the interval 0≤t≤50\leq{}t\leq{}5.
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Determine the average rate of the change of the function g(t)=t2t2−6g(t)=\dfrac{t^2}{t^2-6} over the interval 0.5≤t≤20.5\leq{}t\leq{}2.