Chapter2 - Section 2.1 The Tangent and Velocity Problems...

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1 Section 2.1 The Tangent and Velocity Problems • Goals – Use two problems to explain the need for the notion of limit : • The tangent problem • The velocity problem • What would it mean in general for a line to be tangent to a curve? Tangent Problem touching once touching twice Example • Let’s use a specific example to see how tangent lines in general could be defined: • We want to find an equation of the tangent line to the parabola y = x 2 at the point P (1, 1) . • Since we know the point P , the only question is that of the slope m of the line. Example (cont’d) • Our idea is to use a nearby point Q ( x , x 2 ) on the parabola and find the slope m PQ of the secant line PQ . • We choose x 1 so that Q P . Then 2 1 1 PQ x m x = Example (cont’d) • Now we allow x to approach the value 1 without ever being 1 . – Meanwhile we monitor m PQ . • The tables on the next slide give the values of m PQ as x approaches 1 both from above and below. • In both cases it seems that m PQ approaches 2 ! Example (cont’d)
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2 Example (cont’d) • We say that the slope of the tangent line is the limit of the slopes of the secant lines. • This is expressed symbolically by • For this particular case we would write lim PQ QP mm = 2 1 1 lim 2 1 x x x = Example (cont’d) • To conclude this example, we use the point-slope form to write an equation of the desired tangent line: y –1 ±=±2( x – 1) , or y = 2 x • Next we turn to another general problem whose solution involves limits: The Velocity Problem • Here our goal is to give a precise definition of “instantaneous” velocity; and again we begin with a specific example: • Suppose a ball is dropped from the top of a tower 450 m tall. We want to find the instantaneous velocity of the ball after 5 seconds. Velocity Problem (cont’d) • Our starting point is the definition of average velocity as distance per unit time. • We know that the distance s ( t ) fallen after t seconds is given by s ( t ) = 4.9 t 2 . • However, there is no time interval over which to compute an average velocity, since we want the instantaneous velocity when t = 5 . Velocity Problem (cont’d) • So we use essentially the same idea as that used for the tangent problem: • We compute the average velocity over successively smaller time intervals beginning at t = 5 . • The next slide shows this calculation for the time interval from t = 5 to t = 5.1 : Velocity Problem (cont’d) ()( ) () ( ) 22 change in position average velocity = time elapsed 5.1 5 0.1 4.9 5.1 4.9 5 0.1 49.49 m/s ss = = =
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3 Velocity Problem (cont’d) • The table on the next slide shows the results of similar calculations over successively smaller time periods. • It appears that as we shorten the time period, the average velocity is becoming closer to 49 m/s .
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This note was uploaded on 04/18/2008 for the course MATH 210 taught by Professor Zhoramanseur during the Spring '08 term at SUNY Oswego.

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Chapter2 - Section 2.1 The Tangent and Velocity Problems...

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