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Wize AP Calculus (BC) Textbook > Differentiation 1: Basics & Fundamental Properties

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?