M 112 Short Course in Calculus. Chapter 2 – Rate of Change: The Derivative Sections 2.1 – Instantaneous Rate of Change V. J. Motto. Instantaneous Rate of Change. So we have the function; s(t) = -16t 2 + 100 t + 6. Let’s look over smaller and smaller intervals in the
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M 112 Short Course in Calculus
Chapter 2 – Rate of Change: The Derivative
Sections 2.1 – Instantaneous Rate of Change
V. J. Motto
Instantaneous Rate of Change
So we have the function; s(t) = -16t2 + 100 t + 6
Let’s look over smaller and smaller intervals in the
neighborhood of t = 1
Figure 2.1: Average velocities over intervals on either side of t = 1
showing successively smaller intervals
Some basic definitions
The quantity (in mg) of a drug in the blood at time t (in minutes) is given by Q = 25(0.8)t. Estimate the rate of change of the quantity at t = 3 and interpret your answer.
What kind of function is this?
What is the domain and range?
We estimate the rate of change at t = 3 by computing the average rate of change over intervals near t = 3. We can make our estimate as accurate as we like by choosing our intervals small enough.
Let’s look at the average rate of change over the interval 3 ≤ t ≤ 3.01:
A reasonable estimate for the rate of change of the quantity at t = 3 is −2.85 mg/minute.
Another Basic definition
Estimate f′ (2) if f(x) = x3.
What does the graph of f look like?
What is the domain and range of f?
Since f′(2) is the derivative, or rate of change, of f(x) = x3 at 2, we look at the average rate of change over intervals near 2. Using the interval 2 ≤ x ≤ 2.001, we see that
Visualizing the Derivative of a function
Figure 2.2: Visualizing the average
rate of change of f between a and b
Figure 2.3: Visualizing the instantaneous
rate of change of f at a
Use a graph of f(x) = x2to determine whether each of the following quantities is positive, negative, or zero:
What is the domain?
What is the range?
Figure 2.5 shows tangent line segments to the graph of f(x) = x 2
at the points x = 1, x = −1, x = 2, and x = 0. Since the derivative is the slope of the tangent line at the point, we have:
(a) f′(1) is positive.
(b) f′(−1) is negative.
(c) f′(2) is positive (and larger than f′(1)).
(d) f′(0) = 0 since the graph has a horizontal tangent at x = 0.
The graph of a function y = f(x) is shown in Figure 2.7. Indicate whether each of the following quantities is positive or negative, and illustrate your answers graphically.
Since there are no y-values, I can’t find a model for this function. I must work with the graph.
(a) Since f ′(1) is the slope of the graph at x = 1, we see in Figure 2.8 that f′(1) is positive.
(b) The difference quotient
is the slope of the secant line between x = 1 and
x = 3. We see from Figure 2.9 that this slope is
(c) Since f(4) is the value of the function at x = 4 and f(2) is the value of the function at x = 2, the expression f(4) − f(2) is the change in the function between x = 2 and x = 4. Since f(4) lies below f(2), this change is negative. See Figure 2.10.
The total acreage of farms in the US1 has decreased since 1980. See Table 2.2.
million acres per year.
Between 1980 and 2000, the amount of farm land was decreasing at an average rate of 4.7 million per year.
(b) We use the interval from 1995 to 2000 to estimate the instantaneous rate of change at 1995:
million acres per year. In 1995, the amount of farm land was decreasing at a rate of approximately 3.6 million acres per year.
Find a model for this data:
So we have the function
f(x) = 0.05x2 – 5.83x + 1039.31
Average rate of change is still the slope of the secant!
But the can actually find the instantaneous rate of change --- the first derivative by using the function f and our calculator:
That is f′(15) = -4.197.