Math in the Real World

I pulled these two problems from shelovesmath.com

Problem 1: To solve this problem, you need to use the differential equation: (dT/dt)=k(T-60) where t is in minutes and T is temperature in degrees. You must solve this equations first for T by isolating it on one side of the equation. (dT/dt)=k(T-60) → (dT/T-60)=kdr → ഽ(dT/T-60)=ഽkdt → ln|T-60|=kt+C → e^(ln|T-60|)=e^(kt)+C → |T-60|=Ce^(kt) → T=Ce^(kt)+60 → Initial Condition: (0,100), 100=Ce^(0•k)+60 → C=40, T=40e^(kt)+60. So if the temperature is 80° after 10 minutes: 80=40e^(k•10)+60 → e^(k•10)=ln(5) → k=(ln(5)/10) → k≈ -.0693147. Now we can find the time it takes to cool to 70°, so plug 70 in for T and -.0693147 for k in T=40e^(kt)+60. 70=40e^(-.0693147t)+60 → e^(-.0693147t)=(1/4) → -.0693147t=ln(.25) → t=(ln(.25)/-.0693147) → t≈ 20 minutes to cool to 70°. I chose this problem because it wasn't too complicated and it deals with things that most people deal with everyday, hot drinks, and how seemingly fast they cool down.

Problem 2: To solve this problem we will use the equation y=Ce^(kt), where C is the starting amount and k is the constant of proportionality. The data points (t,y) can be filled in from the question, (2,200) and (5,800). First, we solve for C by plugging in the two points separately into the equation for C, then setting them equal to each other and solving for k. Then we plug the k value into one of the initial equations for C: (2,200) → 200=Ce^(2k) → C= (200/e^(2k)), (5,800) → 800=Ce^(5t) → C=(800/e^(5t)), (200/e^(2k))=(800/e^(5t)) → 200e^(5k)=800e^(2k) → e^(5k)=4e^(2k) → ln(e^(5k))=ln(4e^(2k)) → ln(e^(5k))=ln(4)+ln(e^(2k)), the ln's and e's cancel out → 5k=ln(4) =2k → k=(ln(4)/3) → k≈ .462 → C= (200/e^[2(ln(4)/3)] → C≈ 79.37 bacteria initially. This means y=79.37e^(.462t). To get how many hours it'll take to get 50,000 bacteria, plug in 50,000 for y in y=79.37e^(.462t): 50,000=79.37e^(.462t) → (50,000/79.37)=e^(.462t) → ln(50,000/79.37)=.462t → t≈ 13.95 hours until there are 50,000 bacteria. I chose this problem because it was familiar. I had done problems similar to this one in biology, so I was interested to see how it applied to calculus.