ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
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Activity B1 – Instantaneous Rate of Change – A Geometric Approach
THIS ACTIVITY IS INCOMPLETE UNTIL THE
SECANT/TANGENT APPLET GETS FIXED
I think that most of the content is o.k., but I need to make sure that the functions I
have used are going to be “applet friendly”.
The graph(RS1) shown at right represents the distance a car traveled in x hours. From the
graph we see the car traveled a total of 130 miles in 2 hours. The car’s average rate of
change, or average speed, from the beginning of the trip to the end of the trip can be
found by calculating the slope of the secant line that contains (RS2) the points  0, g (0) 
and  2, g (2)  .
g (2)  g (0) 130  0

 65
20
2
Thus, for the duration of the trip, the car traveled at an average rate of 65 miles per hour.
Notice that the car was not traveling 65 miles per hour for the whole 2 hours. Sometimes
the car was traveling more than 65 miles per hour(RS3), and other times the car was
traveling less than 65 miles per hour(RS4).
m
With a little calculus, we can use the equation of the graph to find exactly how fast the
car was traveling at a certain time and at what times the car was traveling a certain speed.
This concept is known as the instantaneous rate of change.
Suppose the car began the trip at 1:00pm and at 2:30pm the car passed a police officer
who was monitoring the speed of traffic with a radar gun. If the speed limit was 50 miles
per hour, was the car exceeding the speed limit when it passed the police officer? We can
approximate the car’s speed at 2:30pm by computing the average rate of change of the
car over smaller and smaller intervals until eventually we will get an interval so small
that we will be very, very close to x = 1.5, the value that represents 2:30pm. Let’s use the
secant/tangent applet to demonstrate the concept.
Use the secant/tangent applet(RS5) to calculate the slopes of the secant lines that contain
the following points.
1. x = 1.5 and x = 1.6
2. x = 1.5 and x = 1.55
3. x = 1.5 and x = 1.54
4. x = 1.5 and x = 1.53
ACOW RATES OF CHANGE AND DERIVATIVES MODULE
5. x = 1.5 and x = 1.52
(answer 60.55)
6. x = 1.5 and x = 1.51
(answer 60.55)
Updated 5/28/2016
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7. Approximate the instantaneous rate of change of the car at x = 1.5.
Notice as the length of the intervals decreased, the secant lines found in #1-6(RS6)
approached the tangent line. Thus, the instantaneous rate of change can be represented
by the line tangent to the curve at x = 1.5(RS7). Use the secant/tangent applet(RS8) and
type in the values shown below to find the slope of the tangent line.
The applet shows the slope of the line tangent to g ( x ) at x = 1.5 is 56.75. Thus, the car
was traveling at an instantaneous rate (or velocity) of 56.75 miles per hour at 2:30pm.
Since the speed limit was 50 mph, the car was exceeding the speed limit by 6.75 mph.
8. After 1.75 hours (105 minutes), the car described above was in a zone where the speed
limit was 70 miles per hour. Use the secant/tangent applet(RS6) to determine which of
the following is true?
a) The car was traveling the speed limit.
b) The car was exceeding the speed limit.
c) The car was traveling at a speed less than the limit.
d) The car was stopped at a stop light.
As demonstrated above, when a given function describes the position of an object at time
t, the slope of the tangent line at t = a represents the object’s velocity at t = a. For
example, g ( x ) gives the position of the car at time x, so the slope of the line tangent to
g ( x ) at x = 1.5 represents the velocity of the car at 2:30pm. If the slope of the tangent
line is negative, this means that the object was traveling in a negative direction (backward
or downward). If the slope of the tangent line is positive, then the object was traveling in
the positive direction (forward or upward). To determine how fast the object was
traveling (the speed of the object), take the absolute value of the number that represents
the velocity.
9. The graph of s(t) (RS9) represents the position of an object at time t. Determine if the
velocity of the object at points A, B, C, and D is slope (P), negative (N), zero (Z), or
undefined (U). Fill in each blank with P, N, Z, or U.
a) Point A
ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
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b) Point B
c) Point C
d) Point D
The concept of instantaneous rate of change is also useful in other content areas.
10. A company’s profit function, in millions of dollars, is given by
P( x)  0.2 x3  2.7 x 2  10.8x  10.8 , 0  x  8 , where x represents the number of
thousands of items sold. Use the secant/tangent applet(RS8) to find the company’s
instantaneous rate of change at x = 5.
11. Which of the following best interprets the answer from problem 10 above?
a) The company is losing 1.2 million dollars in profit.
b) At a production level of 5000 items, the company’s profit is increasing at
$1200 per item.
c) At a production level of 5 items, the company’s profit is decreasing at $1.20
per item.
d) At a production level of 5000 items, the company’s profit is decreasing at
$1200 per item.
e) At a production level of 5 items, the company’s profit is increasing at $1.20
per item.
12. Use the secant/tangent applet(RS8) to find the slope of the line tangent to
g ( x)  x2  2x  4 at x  4 .
Once we know the slope of the tangent line we can find the equation of the tangent line
using one of the formulas(RS10) at right.
13. Use your answer from question 12 above to find the equation of the line tangent to
g ( x)  x2  2x  4 at x  4 . Input your answer in slope-intercept form.
14. Use the secant/tangent applet(RS8) to help find the equation of the line tangent to
y  3x3  2x2  x  4 at x  1 .