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Unformatted text preview: SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework #2 Solutions Peter Malkin Section C.4 Question 8 The differential equation is dP dt = kP where k is some constant. We can find the general solution using the separation of variables technique. dP dt = kP ⇔ 1 P dP = kdt ⇔ integraldisplay 1 P dP = integraldisplay kdt ⇔ ln( P ) = kt + C ⇔ P = e kt + C ⇔ P = Ae kt where C and A are constants. The general solution is P = Ae kt . Let t = 0 in 1998. So, 400 , 000 = Ae ⇒ A = 400 , 000. Thus, the particular solution is P = 400 , 000 e . 015 t . In 2005, the population is P = 400 , 000 e . 015(7) ≈ 444 , 284. Question 28 Let Q be the amount of concentrate in the tank at time t . At any point in time, there is 0lbs/min of concentrate flowing into the tank, and there is 5 × Q 100 = Q 20 lbs/min flowing out of the tank. Thus, the rate of change of Q over time is dQ dt = 0- Q 20 . We can find the general solution using the separation of variables technique. We could also use the LINFODE technique....
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- Spring '08
- Calculus, Trigraph, general solution, Peter Malkin