Chain Rule Calculus 1

Question

A skydiver jumps out of an airplane at an altitude of 3000 meters. Her height, as a function of time, is

h(t) = 3300 − 54t − 300e^−0.18t

where h is in meters and t is in seconds.

1. What is her velocity at the instant whent = 6

Answer 1

I am assuming there is a typo in either the problem or h(t), since at t=0 her starting height should be the same as the first first without a variable in h(t).

To find the skydivers velocity, you would need to take the derivative of position, then plug in 6 for t.

The only place where chain rule applies here is the e^-0.18t part. The derivative for e^f(x) is e^f(x) • f'(x)

h(t) = 3300 - 54t - 300e^-0.18t

h'(t) = -54 - 300e^-0.18t(-0.18) -- multiply by -0.18 for chain rule

h'(t) = -54 + 54e^-0.18t

h'(6) = -54 + 54e^-0.18(6)

h'(6) = -35.66184162 ....... -35.662 m/s