Calculus/Differential equation of decay.
Expert: Paul Klarreich - 5/27/2008
QuestionLet v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>0. After her chute opens, her velocity satisfies the differential equation
dv/dt=-2(v+17), with initial condition v(0)=-47.
Use separation of variables to finda an expression for v in terms of t, where t is measured in seconds.
Terminal velocity is defined as limv(t) as t approaches infinity. Find the terminal velocity of the skydiver to the nearest foot per second.
It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
AnswerQuestioner: harman
Category: Calculus
Private: No
Subject: ap calculus
Question: Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>0. After her chute opens, her velocity satisfies the differential equation
dv/dt=-2(v+17), with initial condition v(0)=-47.
Use separation of variables to finda an expression for v in terms of t, where t is measured in seconds.
Terminal velocity is defined as limv(t) as t approaches infinity. Find the terminal velocity of the skydiver to the nearest foot per second.
It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
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Hi, Harman,
If dv/dt = -2(v + 17), write:
dv
------ = -2dt
v + 17
and integrate:
ln(v + 17) = -2t + (small) c
Solve:
v + 17 = e^(-2t+c) = e^-2t e^2 = (big) C e^-2t
v = C e^-2t - 17
Now if v(0) = -47,
-47 = C e^-2(0) - 17
-30 = C
v = -30 e^-2t - 17
....................
Terminal velocity is defined as limv(t) as t approaches infinity. Find the terminal velocity of the skydiver to the nearest foot per second.
As t --> infinity, e^-2t --> 0, so your limiting speed is -17.
..................
It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
Set v = -20 (not 20, because she is going down.
-20 = -30 e^-2t - 17
Solve for t:
-3 = -30 e^-2t
1/10 = e^-2t
Take ln of both sides:
ln(1/10) = - ln 10 = - 2t
and you can do the rest.