Lesson 7: Instantaneous Rates of Change
Instantaneous Rates of Change
Graph Description: A coordinate plane showing a curved function \(f(x)\) that increases from left to right with decreasing steepness. Two points are marked on the curve: \((x, f(x))\) at the lower left and \((x+h, f(x+h))\) at the upper right. A secant line connects these two points, showing the average rate of change over the interval from \(x\) to \(x+h\). The graph illustrates the geometric interpretation of the average rate of change formula.
Recall that the slope of the secant line to \(f(x)\) at the points \(x\) and \(x + h\) is
\[\frac{f(x + h) - f(x)}{h}\]Definition: The average rate of change of the function \(f(x)\) over the interval \([x, x + h]\) is
\[\frac{f(x + h) - f(x)}{h}\]Definition: The (instantaneous) rate of change of the function \(f(x)\) at the point \(x\) is
\[\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = f'(x)\]Example 1: Bacterial Population Growth
The population of a culture of bacteria is given by \(p(t) = 5t^2 + 9t + 1\), where \(t\) is time in hours.
(a) Find the equation for the rate of change of the population after \(t\) hours.
Solution:
The rate of change of the population is the derivative of \(p(t)\):
\[p'(t) = \frac{d}{dt}(5t^2 + 9t + 1) = 10t + 9\](b) What is the rate of change after 3 hours?
Solution:
Evaluate \(p'(t)\) at \(t = 3\):
\[p'(3) = 10\cdot 3 + 9 = 30 + 9 = 39\]The population is increasing at a rate of 39 bacteria per hour after 3 hours.
Velocity
Definition: The velocity is the rate of change of the position.
Notation: If \(s(t)\) is a function giving the position of an object at time \(t\), then the velocity of that object at time \(t\) is \(v(t) = s'(t)\).
Example 2: Velocity of a Thrown Ball
The height of a ball \(t\) seconds after being thrown into the air is given by \(s(t) = -10t^2 + 60t\).
(a) Find the velocity function \(v(t)\).
Solution:
The velocity is the derivative of the position function:
\[v(t) = s'(t) = \frac{d}{dt}(-10t^2 + 60t) = -20t + 60\](b) What is the velocity of the ball when \(t = 4\)?
Solution:
Evaluate \(v(t)\) at \(t = 4\):
\[v(4) = -20\cdot 4 + 60 = -80 + 60 = -20\]The velocity is \(-20\) meters per second. The negative sign indicates the ball is moving downward at this time.
Practice Problems
Practice Problem 1: Rock Thrown on Mars
If a rock is thrown upward on Mars, its height (in meters) after \(t\) seconds is given by \(s(t) = -1.86t^2 + 16t\). At what time is the velocity of the rock equal to -2.6 meters per second?
Solution:
First, find the velocity function:
\[v(t) = s'(t) = \frac{d}{dt}(-1.86t^2 + 16t) = -3.72t + 16\]Set the velocity function equal to -2.6 and solve for \(t\):
\[-3.72t + 16 = -2.6\] \[-3.72t = -18.6\] \[t = \frac{-18.6}{-3.72} = 5\]The velocity of the rock is -2.6 meters per second at \(t = 5\) seconds.
Practice Problem 2: Rate of Change of Volume of a Cube
Find the rate of change of the volume of a cube with respect to the length of a side \(s\). What is the rate of change of the volume when \(s = 3\)?
Diagram Description: A three-dimensional cube with side length \(s\) marked on each visible edge. The cube is drawn in perspective to show its three-dimensional nature, with edges labeled to indicate that all sides have equal length \(s\).
Solution:
The volume of a cube with side length \(s\) is:
\[V(s) = s^3\]The rate of change of the volume with respect to the side length is:
\[V'(s) = \frac{d}{ds}(s^3) = 3s^2\]When \(s = 3\):
\[V'(3) = 3\cdot 3^2 = 3\cdot 9 = 27\]The rate of change of the volume when the side length is 3 is 27 cubic units per unit of side length.
Practice Problem 3: Population Growth Rate
The population of a town since the year 2000 can be given by \(p(t) = 200t^2 - 100t + 70000\), where \(t = 0\) corresponds to the year 2000. In which year is the population increasing at the rate of 2300 people per year?
Solution:
First, find the rate of change of the population:
\[p'(t) = \frac{d}{dt}(200t^2 - 100t + 70000) = 400t - 100\]Set the rate of change equal to 2300 and solve for \(t\):
\[400t - 100 = 2300\] \[400t = 2400\] \[t = 6\]Since \(t = 0\) corresponds to the year 2000, we have:
\[2000 + t = 2000 + 6 = 2006\]The population is increasing at a rate of 2300 people per year in the year 2006.