The Tangent and Velocity Problems Section 2 1
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 =… x=3→x=4 x=3→x=3. 5 x=3→x=3. 25 x=2. 75→x=3 x=2→x=3 Points (x, y) m between x points. 236 . 242. 246. 254. 258. 268 Thus, m=1/4 at (3, 2)
Estimating Slopes at a Given point… #3) Estimate m at x=8. x =… to x =… x=8→x=9 x=8→x=8. 5 x=8→x=8. 25 x=7. 75→x=8 x=7→x=8 Points (x, y) m between x points. 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 f(3)=2 f(8)=3 x=a
Example Find an equation of the tangent line to the parabola y = x 2 at the point P (1, 1). Solution: To find an equation of the tangent line t find m
Example – Solution Find an estimate ofm by choosing a nearby point Q (x, x 2) on the parabola and find the slope of the secant line PQ. x =… We choose x 1 so that Q P. 0. 9. 999 1. 001 1. 1 2 1 m. PQ 1. 999 2. 001 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 = 2 x – 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. 9 t 2 Finding the velocity after 5 s, 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 Make a table showing the results of similar calculations of the average velocity over successively smaller time periods. 5 5 Time Interval s s → → Ave. Velocity 6 s 53. 9 m/s 5. 1 s 49. 49 m/s 5. 01 s 49. 049 m/s 5. 001 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 Ø Summarize Notes Ø Read Section 2. 1 Ø Homework Ø Pg. 86 #1, 3, 5 -8
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