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|=C e^ (kt) → T=C e^ (kt)+60 → Initial Condition: (0,100), 100=C e ^(0•k)+60 → C=40, T=40 e^ (kt)+60 . So if the temperature is 80° after 10 minutes: 80=40 e ^(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=40 e^ (kt)+60. 70=40 e ^(-.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 ...