Expert: Ahmed Salami - 11/7/2008

Question
I'm having trouble with word problems in my calculus class, hope you can help me understand better.  Here are my questions:

1)  A water bucket containing 10 gallons of water develops a leak.  The volume v of water in the bucket t seconds later is given by v(t)=10(1-(t/100))^2 until the bucket is empty 100 seconds later.

a) At what rate is the water leaking after exactly 1  minute?

b) When is the instantaneous rate of change of volume equal to the average rate of change of volume from t=0 to t=10?

c) At the instant described in part "b", how much water is in the tank?

Thanks for you help, I really appreciate it.

Answer
Hi Robert,
V(t) = 10[1 - (t/100)]^2
The rate of change of volume at any time t is,
dV/dt = 10.2[1 - (t/100)].(-1/100)
     = (-20/100)[1 - (t/100)]
     = (-1/5)[1 - (t/100)]
     = [(t/100) - 1]/5
a)At t = 1 minute = 60s
dV/dt = [(60/100) - 1]/5
     = [(0.6 - 1]/5
     = -0.4/5
     = -0.08 gallons/s

b)The average rate of change of volume from t=0 to t=10 is
V(ave) = [V(10) - V(0)]/(10 - 0)
V(10) = 10[1 - (10/100)]^2
     = 10[1 - 0.1]^2
     = 10(0.9)^2
     = 10(0.81)
     = 8.1 gallons
V(0) = 10 gallons
V(ave) = (8.1 - 10)/(10 - 0)
      = -1.9/10
      = -0.19
The instantaneous rate of change of volume at any time is
dV/dt = [(t/100) - 1]/5
So,
[(t/100) - 1]/5 = -0.19
(t/100) - 1 = -0.95
t/100 = 0.05
t = 5s

c)At t = 5s,
V(5) = 10[1 - (5/100)]^2
     = 10[1 - 0.05]^2
     = 10(0.95)^2
     = 10(0.9025)
     = 9.025 gallons

Regards