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

Rates of Change of Exponential Functions

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Rates of Change of Exponential Functions

The rate of change of an exponential function can be solved used both algebraic and graphical methods.

Example

The population growth of a Vancouver, BC is roughly 1.16% annually. If the current population in Vancouver is approximately 2.5 million people, how fast is the population increasing 10 years from now?

Let P(t)=2.5(1.0116)tP(t)=2.5(1.0116)^t model the above situation and P(t)P(t) is in millions. Then,

is a rough sketch of P(t)P(t).

We are looking for the instantaneous rate of change when t=10.t=10.

Then, in a graphing calculator, we can find the IRC using the tangent line:
2nd + PRGM + 5: Tangent + ENTER\boxed{2^{\text{nd}}}~+~\boxed{\text{PRGM}}~+~\boxed{\text{5: Tangent}}~+~\boxed{\text{ENTER}}

The tangent line is approximately y=0.02589x+1.99y=0.02589x+1.99.


This means that the population of Vancouver, BC will increase at a rate of 0.02589 million people/year 10 years from now.
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Example: Rates of Change of Exponential Functions

A radioactive substance has a half-life of 2.25 seconds and a mass of 150mg.
  1. What is the instantaneous rate of change of the substance when t=20t=20 seconds.
  2. If the half-life changed to 4.354.35 seconds, what would the instantaneous rate of change be when t=20t=20 seconds.

Algebraic Solution


1. Let's find a function to describe this situation.

A(t)=150(12)t2.25A(t)=150\Big(\dfrac{1}{2}\Big)^{\frac{t}{2.25}}

Then, when t=20t=20 seconds, then instantaneous rate of change is:

minst=150(12)t+h2.25−150(12)t2.25(t+h)−t=150((12)20+0.0012.25−(12)202.25)0.001=−0.097 mg/second\begin{array}{rcl} m_{inst}&=&\dfrac{150\Big(\dfrac{1}{2}\Big)^{\frac{t+h}{2.25}}-150\Big(\dfrac{1}{2}\Big)^{\frac{t}{2.25}}}{(t+h)-t}\\\\ &=&\dfrac{150\Bigg(\Big(\dfrac{1}{2}\Big)^{\frac{20+0.001}{2.25}}-\Big(\dfrac{1}{2}\Big)^{\frac{20}{2.25}}\Bigg)}{0.001}\\\\ &=&-0.097~\text{mg}/\text{second} \end{array}



2.
minst=150(12)t+h4.35−150(12)t4.35(t+h)−t=150((12)20+0.0014.35−(12)204.35)0.001=−0.9871 mg/second\begin{array}{rcl} m_{inst}&=&\dfrac{150\Big(\dfrac{1}{2}\Big)^{\frac{t+h}{4.35}}-150\Big(\dfrac{1}{2}\Big)^{\frac{t}{4.35}}}{(t+h)-t}\\\\ &=&\dfrac{150\Bigg(\Big(\dfrac{1}{2}\Big)^{\frac{20+0.001}{4.35}}-\Big(\dfrac{1}{2}\Big)^{\frac{20}{4.35}}\Bigg)}{0.001}\\\\ &=&-0.9871~\text{mg}/\text{second} \end{array}

Graphical Solution


1. We are looking for the instantaneous rate of change (IRC) when t=20t=20.

Then, in a graphing calculator, we can find the IRC using the tangent line:

Y=150(12)t4.35 + GRAPH + 2nd + PRGM + 5: Tangent + x=20+ +ENTER\boxed{Y=150\Big(\dfrac{1}{2}\Big)^{\frac{t}{4.35}}}~+~\boxed{\text{GRAPH}}~+~\boxed{2^{\text{nd}}}~+~\boxed{\text{PRGM}}~+~\boxed{\text{5: Tangent}}~+~\boxed{x=20}+~+\boxed{\text{ENTER}}

The tangent line is approximately y=−0.0975x+2.266y=-0.0975x+2.266.

Therefore, the IRC is approximately −0.0975 mg/second-0.0975~\text{mg}/\text{second}



2. We are looking for the instantaneous rate of change (IRC) when t=20t=20.

Then, in a graphing calculator, we can find the IRC using the tangent line:

Y=150(12)t4.35 + GRAPH + 2nd + PRGM + 5: Tangent + x=20+ +ENTER\boxed{Y=150\Big(\dfrac{1}{2}\Big)^{\frac{t}{4.35}}}~+~\boxed{\text{GRAPH}}~+~\boxed{2^{\text{nd}}}~+~\boxed{\text{PRGM}}~+~\boxed{\text{5: Tangent}}~+~\boxed{x=20}+~+\boxed{\text{ENTER}}

The tangent line is approximately y=−0.987x+25.937y=-0.987x+25.937.

Therefore, the IRC is approximately −0.987 mg/second-0.987~\text{mg}/\text{second}

Practice: Rates of Change of Exponential Functions

Jake decides to invest $4000 compounded monthly at an interest rate of 3.25%3.25\%.

Practice: Rates of Change of Exponential Functions

The population of a town in BC is decreasing 5%5\% annually. The town had a size of 150,000150,000 people in the year 2010.

Practice: Rates of Change of Exponential Functions

If the instantaneous rate of change for the function f(t)=3(1.25)tf(t)=3(1.25)^t is 44, what is tt? Let h=0.001h=0.001.



Extra Practice
Rates of Change
The instantaneous rate of change for f(x)=2xf(x)=2^{x} is 32\dfrac{3}{2}.
If h=0.001h=0.001, what is xx?
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If the instantaneous rate of change for the function f(t)=−6(0.45)t2f(t)=-6(0.45)^{\frac{t}{2}} is 55, what is tt? Let h=0.001h=0.001.
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