1. The Tangent and Velocity Problems
Section 2.1

2. The Tangent Problems
The word tangent is derived from the Latin word tangens, which means “touching.”
The tangent to a curve is a line that touches the curve

3. Estimating Slopes at a Given point…
#1) Graph f(x).

4. Estimating Slopes at a Given point…
#2) Estimate m at x=3.
x =… to x =…
Points (x, y)
m between x points
x=3→x=4
x=3→x=3.5
x=3→x=3.25
x=2.75→x=3
x=2.5→x=3
x=2→x=3
𝟑,𝟐 ; 𝟒, 𝟓
.236
𝟑,𝟐 ; 𝟑.𝟓, 𝟒.𝟓
.242
𝟑,𝟐 ; 𝟑.𝟐𝟓, 𝟒.𝟐𝟓
.246
𝟐.𝟕𝟓, 𝟑.𝟕𝟓 ; 𝟑, 𝟐
.254
𝟐.𝟓, 𝟑.𝟓 ; 𝟑, 𝟐
.258
𝟐, 𝟑 ; 𝟑, 𝟐
.268
Thus, m=1/4 at (3,2)

5. Estimating Slopes at a Given point…
#3) Estimate m at x=8.
x =… to x =…
Points (x, y)
m between x points
x=8→x=9
x=8→x=8.5
x=8→x=8.25
x=7.75→x=8
x=7.5→x=8
x=7→x=8
.162
.164
.166
.168
.169
.172
Thus, m=1/6 at (8,3)

6. Estimating Slopes at a Given point…
#4) Estimate m at x=a for a>-1.
x=3
x=8
x=a
f(3)=2
f(8)=3

7. Example
Find an equation of the tangent line to the parabola y = x2 at the point P (1, 1).
Solution:
To find an equation of the tangent line t find m

8. Example – Solution .9 .99 .999 1.001 1.01 1.1 2 x =… mPQ 1 1.9 1.99
Find an estimate ofm by choosing a nearby point Q (x, x2) on the parabola and find the slope of the secant line PQ.
x =…
mPQ .9
.99
.999
1.001
1.01
1.1 2 1
1.9
1.99
1.999
We choose x  1 so that Q  P .
2.001
2.01
2.1 3

9. Example – Solution It seems the slope of the tangent line t ism = 2.
Slope of the tangent line is the limit of the slopes of the secant lines,
and
Use point-slope form to find equation through (1, 1)
y – 1 = 2(x – 1)
y = 2x – 1

10. Example
Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds.
Solution: If the distance fallen after t seconds is denoted by s (t) and measured in meters,
s (t) = 4.9t 2
Finding the velocity after 5s, a single instant of time (t = 5)

11. Example – Solution
Approximate the quantity by computing the average velocity over the brief time interval of a tenth of a second from t = 5 to t = 5.1:

12. Example – Solution 5 s → 6 s 5 s → 5.1 s 5 s → 5.01 s 5 s → 5.001 s
Make a table showing the results of similar calculations of the average velocity over successively smaller time periods.
Time Interval
Ave. Velocity
5 s → 6 s
5 s → 5.1 s
5 s → 5.01 s
5 s → s
53.9 m/s
49.49 m/s
m/s
m/s

13. Example – Solution It seems shortening the time period,
the average velocity is becoming closer to 49 m/s.
Thus, instantaneous velocity when t = 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t = 5.
Therefore, the (instantaneous) velocity after 5 s is v = 49 m/s

14. 2.1 The Tangent and Velocity Problems
Summarize Notes
Read Section 2.1
Homework
Pg.86 #1,3,5-8