Project #1 Chapter 4.1 Q#11.
Group members: Raisa Ratri and Richard Li
A candy maker makes 500 pounds of candy per week, while his large family eats the candy at a rate equal to Q(t)/10 pounds per week, where Q(t) is the amount of candy present at the time.
- Find Q(t) for t > 0 if the candy maker has 250 pounds of candy at t = 0.
- Find limt–> ∞ Q(t).
Q’ = 500 – Q(t) / 10
= (5000 – Q(t)) / 10
= -dQ(t) / (5000 – Q(t)) = -dt / 10 We find the derivative
= Ln(5000 – Q(t)) = -t / 10 + k K is a constant
= 5000 – Q(t) = Ce^(-t / 10) C is any constant, but in this problem we need to find C
Plug in when t = 0 Q(0) = 250
250 = 500 – Ce^(0/10)
250 = 500 – C
C = 4750
Q(t) = 5000 – 4750e^(-t/10)
Limt–> ∞ Q(t) 5000 – 4750e^(-t/10) = 5000
It will go to 5000 because as t gets larger, the whole number gets smaller because it is a negative exponent, so 5000 – 0 is 5000.
I am 20 years old. I am majoring in mechanical engineering. This is my 3rd year at City Tech. The only thing I am good at is gaming. I play League and Valorant. When I graduate I think I hope to come out with ideas that will help with everyday life.