The Tangent and Velocity Problems Section 2.1

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

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

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)

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)

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

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

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

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

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)

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:

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 → 5.001 s 53.9 m/s 49.49 m/s 49.049 m/s 49.0049 m/s

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

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