How to find K in Q(t) = 15(1 - e^-kt)

Question

A 50-gallon barrel is filled completely with pure water. Salt water with a concentration of 0.3 pounds/gallon is then pumped into the barrel, and the resulting mixture overflows at the same rate. The amount of salt (in pounds) in the barrel at time t (in minutes) is given by Q(t) = 15(1 − e^−kt) where k > 0.

(a) Find k if there are 5.5 pounds of salt in the barrel after 10 minutes. Round your answer to 4 decimal places.
k ≈
so I manipulated it and got -.0312 but it said I was wrong so I was wondering how to solve this portion.
(b) What happens to the amount of salt in the barrel as t → ∞?
Q → 0.3

Q → 1
Q → 15
Q → 5.5
Q → 50

For this I don't even have a guess because I would graph it to figure it out but I still don't know k so.... Would there be any other possible way to solve this?

please help me figure these out, I am stuck.

Answer 1

(a). 5.5 = 15(1 - e^-10k)

 

      5.5 = 15 - 15e^-10k

 

     -9.5 = -15e^-10k

 

      0.6333333 = e^-10k

 

     ln(0.6333333) = -10k

 

       k ≈ 0.0457

 

(b). Q(t) = 15(1 - 1/e^kt)  

 

      Since k > 0, 1/e^kt → 0 as t → ∞

 

      So, Q(t) → 15 as t → ∞

 

      As t→∞, the amount of salt in the tank approaches 15 lb.  

Answer 2

Q(t) = 5.5

t = 10

 

Therefore,

 

5.5 = 15(1 - e^-10k)

 

 

Divide both sides of the equation by 15.  This will isolate the k terms.

 

0.37 = 1 - e^-10k

 

 

Subtract 1 on both sides of the equation.

 

-0.63 = -e^-10k

 

0.63 = e^-10k

 

 

Natural log both sides of the equation and bring any exponents down as the coefficient of the log.

 

ln(0.63) = -10kln(e)

 

 

Knowing that    ln(e)=1

 

ln(0.63) = -10k

 

 

Divide both sides of the equation by -10.

 

ln(0.63) / -10 = k

 

0.0462 = k

 

 

I think when you did your working, you did not watch your signs.

 

 

b)

 

Take the parent part of the equation, which is   f(t)=-e^-kt.  This a graph of an exponential decay flipped upside down.  So as t increases infinity, the amount of salt slowly increases, until it reaches certain limit.

Answer 3

(a)

5.5 = 15(1 - e^10k)

0.3666... ≈ 1 - e^10k

-0.6333... ≈ -e^10k

0.6333 ≈ e^10k

ln 0.6333 ≈ 10k

-0.4568 ≈ 10k

-0.04568 ≈ k