Section 2.1
Example 1 (Homework #4 – minor variation)
The point P (5,3) lies on the curve
.
(a) If Q is the point (
), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for
the following values of x:
- 4.5,4.9,4.99,4.999,
- 5.5,5.1.5.01,5.001
(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(5,3).
(c) Using the results of part (b), find an equation of the tangent to the curve at P(5,3).
(d) Sketch the curve, two of the secant lines and the tangent line.
(a)
|
P |
Q |
|
|
(5,3) |
(4.5,2.915…) |
0.169048 |
|
(5,3) |
(4.9,2.983…) |
0.167132 |
|
(5,3) |
(4.99,2.998333) |
0.166713 |
|
(5,3) |
(4.999,2.999833) |
0.166671 |
|
(5,3) |
(5.5,3.082207) |
0.167714 |
|
(5,3) |
(5.1,3.016621) |
0.166206 |
|
(5,3) |
(5.01,3.001666) |
0.166620 |
|
(5,3) |
(5.001,3.000167) |
0.166662 |
(b)
Note:
Limiting value from below x = 5 is approaching 0.1666667.
Limiting value from above x = 5 is approaching 0.1666667.
Note: The average of the slopes of the two closest secant lines is
; m = 0.1666665.
Either way we can estimate the value of the slope of the tangent line to the curve is
m = 1/6.
Parts (c) and (d) will be done on the board.