World record free fall On October 14, 2012, Felix Baumgartner stepped off a balloon capsule at an altitude of 127,852.4 feet and began his free fall. It is claimed that Felix reached the speed of sound 34 seconds into his fall at an altitude of 109,731 feet and that he continued to fall at supersonic speed for 30 seconds until he was at an altitude of 75,330.4 feet. Let f ( t ) equal the distance that Felix had fallen t seconds after leaving his capsule. Calculate f (0), f (34), f (64), and his average supersonic speed f ( 64 ) − f ( 34 ) 64 − 34 (in ft/s) over the time interval [34, 64] ( Source http://www.redbullstratos.com )
World record free fall On October 14, 2012, Felix Baumgartner stepped off a balloon capsule at an altitude of 127,852.4 feet and began his free fall. It is claimed that Felix reached the speed of sound 34 seconds into his fall at an altitude of 109,731 feet and that he continued to fall at supersonic speed for 30 seconds until he was at an altitude of 75,330.4 feet. Let f ( t ) equal the distance that Felix had fallen t seconds after leaving his capsule. Calculate f (0), f (34), f (64), and his average supersonic speed f ( 64 ) − f ( 34 ) 64 − 34 (in ft/s) over the time interval [34, 64] ( Source http://www.redbullstratos.com )
Solution Summary: The author calculates the functional values of f(t), and the average supersonic speed, which are 0, 18121.4 and 52522 respectively.
World record free fall On October 14, 2012, Felix Baumgartner stepped off a balloon capsule at an altitude of 127,852.4 feet and began his free fall. It is claimed that Felix reached the speed of sound 34 seconds into his fall at an altitude of 109,731 feet and that he continued to fall at supersonic speed for 30 seconds until he was at an altitude of 75,330.4 feet. Let f(t) equal the distance that Felix had fallen t seconds after leaving his capsule. Calculate f(0), f(34), f(64), and his average supersonic speed
f
(
64
)
−
f
(
34
)
64
−
34
(in ft/s) over the time interval [34, 64] (Sourcehttp://www.redbullstratos.com)
Hint: You may use the following derivative rules:
ddxsin(x)=cos(x)
ddxcos(x)=−sin(x)
ddxln(x)=1x
Find the equation of the tangent line to the curve y=4sinx at the point (π6,2).The equation of this tangent line is
Chapter 1 Solutions
Calculus, Single Variable: Early Transcendentals (3rd Edition)
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