ACOW RATES OF CHANGE AND DERIVATIVES MODULE Updated 4/10/2020
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Activity B2 – Instantaneous Rate of Change – An Algebraic Approach
THIS ACTIVITY IS INCOMPLETE UNTIL THE
SECANT/TANGENT APPLET GETS FIXED
In activity B1 we used the secant/tangent applet to calculate the instantaneous rate of change. In this activity we focus on calculating the instantaneous rate of change using the formula(RS1) at right.
Now we need to add a discussion on using the secant/tangent applet to discuss how the formula for instantaneous rate of change is derived. We talked about having the “h” value shown on the applet so the students can see that as “h goes to zero” the secant line is approaching the tangent line. The secant/tangent applet also needs to have the values for the slope of the tangent and slope of secant displayed in a cell somewhere on the applet.
1. Given ( )

4 x which of the following is the first step in finding the slope of the line tangent to ( ) at x = –3? a) b) c) d)
4(4
  h lim

0 f h
(4 ) 4 x lim h

0
4 h h
 
4 h lim

0
4(4
 h h
) 4(4) h
2. Suppose ( )
 t
2
0 t 4 represents the position, in feet, of an object after t seconds.
Find the instantaneous rate of change of the object after 3 seconds by filling in the blanks below. h lim

0
 h

2
  h
2
 h lim

0
9 6 h
 h
2 
9 h

 h
 h
2 h h lim

0 lim 6

0
  h h


3. Which of the following best interprets the answer from problem 2 above? a) It took 6 seconds for the object to reach the ground.

ACOW RATES OF CHANGE AND DERIVATIVES MODULE Updated 4/10/2020
Page 2 of 2 b) At exactly 3 seconds, the object was traveling upward at a speed of 6 feet per second. c) At exactly 3 seconds, the object was traveling downward at a speed of 6 feet per second. d) The average rate of change of the object between 0 and 3 seconds was 6 feet per second.
4. Find the slope of the line tangent to ( )
 x at x = 16. Fill in the blanks with the correct numbers. h lim

0
 h lim

0
  
 

 h lim

0 h

16
16
16
4

 h lim

0 16
1
4

5. Given

1 x
, find the slope of the line tangent to ( ) at x = 5. Fill in the blanks with the correct numbers. lim h

0
1
 h
 h
1
 lim h

0
5(5
5
 h )

5
5(5

 h h ) h
 lim h

0

5(5
 h )

1
 lim h

0

1
5(5
 h )
