Tangent Lines and Rates of Change D. P. Morstad, University of North Dakota

Objectives of Assignment 1. To understand the relationship between tangent lines and secant lines. 2. To evaluate the slope of a tangent line using the limit of slopes of secant lines. 3. To evaluate average and instantaneous rates of change. 4. To recognize the relationship between rates of change and slopes of secant and tangent lines.

I.

The Relationship Between Tangent Lines And Secant Lines

A secant line intersects the graph of a function in at least two points. Using these two f ( x 2 ) − f ( x1 ) points, the slope of the secant line is easy to determine: msec = . x2 − x1 3 2.5

3

2

2.5

1.5

2

1

1.5

0.5

1 0.5

-2

-1

-0.5

1

2 -2

-1

-1

-0.5

1

2

-1

Secant line: Two points of intersection make it easy to find the slope.

Tangent line: One point of intersection makes it hard to find the slope.

A tangent line only provides one point of intersection. How can you possibly find the slope of a line if you only know one point on the line? The solution is to use secant lines that get closer and closer to the tangent line. mtan = lim msec sec→ tan

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The bold line is a tangent line. Four secant lines approach the tangent line from the left.

Each of the secant lines intersects the function, f ( x ) = x 2 , at two points. The slope is f ( xs ) − f ( xt ) msec = . xs − xt If the point of tangency is (1, 1), and xs = 0, then msec =

f ( 0) − f (1) 0 − 1 = = 1. 0 −1 0−1

If the point of tangency is (1, 1), and xs = ½, then msec =

f ( 12 ) − f (1) = 1 2 −1

1 4 1 2

−1 3 = . −1 2

If the point of tangency is (1, 1), and xs = ¾, then f ( 34 ) − f (1) 169 − 1 7 = 13 =4. 3 4 −1 4 −1 If you continue this pattern, using xs values that approach xt, the secant lines approach the tangent line and msec approaches mtan. That is, msec =

f ( x) − 1 = x→ 1 x −1 x 2 −1 lim = x →1 x − 1 ( x + 1)( x − 1) lim = x →1 x −1 lim( x + 1) = 1 + 1 = 2

mtan = lim

x →1

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II.

Using X(PLORE) To Graph Secant and Tangent Lines NOTE: If you are not working in the Computer Math Lab, you need to get a copy of the file “assign3.xpl”. You may download it via ftp from ftp.und.nodak.edu/pub/math/assign.xpl, or you make a copy from the computers in the Math Lab. Start X(PLORE), and then use * to load assign3.xpl. To examine secant lines around x = 1 for f ( x ) = x 2 , type and enter the following two

lines: f (x ) = x 2 LeftSecs(f(x), –2, 2, –1, 4, 1) This will graph f(x) in the window (–2, 2, –1, 4). Then, each time you press E, a secant line will be drawn. Finally, a yellow line will be drawn which is tangent to f(x) at x = 1. To see secant lines from the right, enter: RightSecs(f(x), –2, 2, –1, 4, 1) To see secant lines from the right and left, enter: BothSecs(f(x), –2, 2, –1, 4, 1) The magenta lines are secants to the right, red lines are secants to the left. Ex. 1: Graphically approximate the slope of the line tangent to f(x) = sin(x) at x = π/3. Solution: Enter the following three lines: f(x) = sin(x) grid(1,1) BothSecs(f(x), –8, 8, –6, 6, π/3) The green tangent line appears to rise one grid mark for every two that it goes 1 horizontally. Since slope is m = rise run , it appears that mtan ≈ 2 . Ex. 2: Graphically investigate the line tangent to f ( x ) = x 2 − 1 at x = 1. Solution: Enter f(x) = Abs(x^2 – 1) for the absolute value, and use BothSecs. Since secants from the left do not approach the same line as secants from the right, there is not a unique tangent line. Sharp corners in functions have many tangent lines, but not a unique one.

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III.

Finding Slopes Of Tangent Lines Using Limits

You have seen graphically that the tangent line is the limit of the secant lines. It would make sense then, that the slope of the tangent line is the limit of the slope of the secant lines. That is, msec approaches mtan as the other secant point approaches the point of tangency. Ex. 3: Solution:

Find the slope of the line tangent to f ( x ) = 3x 2 − 4 x + 3 at x = –2. The point of tangency is (–2, f(–2)). Let’s just use (x, f(x)) for the other secant point of intersection. We want to examine what happens to the slope as x → – 2. mtan = lim msec = x→ − 2

= lim

x→ − 2

f ( x ) − f ( −2 ) = x − ( −2 )

( 3x 2 − 4 x + 3) − ( 23) = x→ − 2 x+2 ( 3x 2 − 4 x − 20) = lim = x→ − 2 x+2 ( 3x − 10)( x + 2 ) = lim = x →−2 x+2 = lim( 3x − 10 ) = = lim

x →−2

= −16 So the slope of the line tangent to f(x) at x = –2 is –16. Ex. 4: Find an equation for the line tangent to f ( x ) = x 3 at x = 2. Solution: The point of tangency is (2, f(2)), and the equation will be y – y1 = m(x – x 1), so: mtan = lim msec = x →2

f ( x ) − f (2 ) = x →2 x−2 x3 − 8 = lim = x →2 x − 2 ( x − 2 )( x 2 + 2 x + 4) = lim = x →2 x−2 = lim( x 2 + 2 x + 4) = = lim

x→ 2

= 4+4+4 = = 12 The equation for the tangent line is: y – 8 = 12(x –2).

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IV.

Average And Instantaneous Rates Of Change

The average rate of change of a function, f(x), over an interval, a ≤ x ≤ b, is defined to be the change in the function from a to b, divided by the change in x from a to b. If one is compelled to always be formulaic about everything, it is: average rate of change of f(x) over [a, b] is:

f ( b) − f ( a ) . b−a

You might recognize this as being the slope of the secant line which intersects f(x) at (a, f(a)) and (b f(b)). If you can evaluate f(x) at a and b, this is simple to evaluate. If you only have a graph of the function, it’s still pretty simple. You will still need to have Assign3.xpl loaded in X(PLORE) for these last three examples. The grid is markings are 1 unit × 1 unit. Ex. 5:

Type and enter “example5”. You should see a decreasing curve on a grid. What is the average rate of change of f(x) over [1, 4]? What is the average rate of change of f(x) over [2, 8]?

Solution: Looking at the graph, notice that at x = 1, y = 4 and at x = 4, y = 1. So the average f (1) − f ( 4) 4 − 1 rate of change is: = = −1 . 1− 4 1− 4 From x = 2 to x = 8, you can see the function goes from (2, 2) to (8, ½), so the f ( 8) − f ( 2) 12 − 2 average rate of change here is = = − 14 . 8− 2 8− 2 The instantaneous rate of change of a function at a given point is the limit of the average rates of change at that point. That is, f ( x ) − f (a ) instantaneous rate of change of f(x) at x = a is: lim . x→ a x−a Hopefully, you recognize this as being the slope of the line tangent to f(x) at x = a. Ex. 6:

Type and enter “example6”. You should see the same decreasing curve on the grid. What is the instantaneous rate of change of f(x) at x = 2?

Solution: Press E once to see the right secant lines, press it again to see the tangent line. Pressing E two more times shows the left secant and tangent lines. Looking at the green tangent line, notice that it goes down one unit for every unit it goes to the −1 right. Consequently, the instantaneous rate of change is: = −1. 1

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Ex. 7:

Type and enter “example7”. You should see the same decreasing curve on the grid. What is the instantaneous rate of change of f(x) at x = 1? Solution: Press E once to see the secant and tangent lines. Looking at the graph of the green tangent line, notice that it goes down four units for every unit it goes to the −4 right. Consequently, the instantaneous rate of change is: = −4 . 1 V.

Practice Problems

1.

Let a) b) c) d)

f ( x ) = 3x 2 − 6x + 7 . Find the slope of the secant line that intersects f(x) at x = 2 and x = 4. Find the average rate of change of f(x) between x = 2 and x = 4. Find the slope of the line tangent to f(x) at x = 2. Find the instantaneous rate of change of f(x) at x = 2.

2.

Let a) b) c) d)

f ( x ) = 3x + 4 . Find the slope of the secant line that intersects f(x) at x = –1 and x = 3. Find the average rate of change of f(x) between x = –1 and x = 3. Find the slope of the line tangent to f(x) at x = –1. Find the instantaneous rate of change of f(x) at x = –1.

3.

Let a) b) c) d)

f ( x ) = −2 x 2 + 5 x − 2 . Find the slope of the secant line that intersects f(x) at x = –1 and x = 1. Find the average rate of change of f(x) between x = –1 and x = 1 Find the slope of the line tangent to f(x) at x = 1. Find the instantaneous rate of change of f(x) at x = 1.

4.

Let f ( x ) =

5.

Let f ( x ) = x 2 − 4 .

x +1 . x −1 a) Find the slope of the secant line that intersects f(x) at x = 2 and x = 4. b) Find the average rate of change of f(x) between x = 2 and x = 4. c) Find the slope of the line tangent to f(x) at x = 2. d) Find the instantaneous rate of change of f(x) at x = 2.

a) b) c) d)

Find the slope of the secant line that intersects f(x) at x = 2 and x = 3. Find the average rate of change of f(x) between x = 2 and x = 3. Find the slope of the “right” line tangent to f(x) at x = 2. Find the instantaneous rate of change of f(x) at x = 2 “from the left.”

Solutions: 1. 12,12,6,6; 2. 3,3,3,3; 3. 5,5,1,1; 4. − 23 , − 23 , –2 –2; 5. 5,5, 4,–4. 16