Precalc0110to0111-page19

Precalc0110to0111-page19 - s 1 ( ) 0.01 = 3.0301 mph 1,...

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(Section 1.11: Limits and Derivatives in Calculus) 1.11.8 Using Form 4 of a difference quotient, The velocity function v is defined by: vt () = lim h ± 0 st + h () ² st () h Re Example 1: Numerical approaches; Approximating derivatives Let’s say the position function s is defined by: st () = t 3 on 0, 2 ± ² ³ ´ . We want to find v 1 () , the instantaneous velocity of the car at t = 1 or a = 1 . We will first consider average velocities on intervals of the form 1, 1 + h ± ² ³ ´ . Here, we let h ± 0 + . Interval Value of h (in hours) Average velocity, s 1 + h () ± s 1 () h 1, 2 ± ² ³ ´ 1 s 2 () ± s 1 () 1 = 7 mph 1, 1.1 ± ² ³ ´ 0.1 s 1.1 () ± s 1 () 0.1 = 3.31 mph 1, 1.01 ± ² ³ ´ 0.01 s 1.01 ()
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Unformatted text preview: s 1 ( ) 0.01 = 3.0301 mph 1, 1.001 0.001 s 1.001 ( ) s 1 ( ) 0.001 = 3.003001 mph + 3 mph These average velocities approach 3 mph, which is v 1 ( ) . The reader can verify that v 1 ( ) = 3 mph in the Exercises. Part D will help. WARNING 1 : Tables can sometimes be misleading. The table here does not represent a rigorous evaluation of v 1 ( ) . Answers are not always integer-valued. If h is close to 0, we may be able to use the corresponding average velocity as an approximation for v 1 ( ) ....
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This document was uploaded on 12/29/2011.

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