Honey is poured into a glass jar at a rate of r(t)=4e^(-0.35t) ounces per minute, where t is measured in minutes and 0≤t≤5. If there is 3 ounces of honey in the jar at time t=0, how much honey is in t

Question

Honey is poured into a glass jar at a rate of r(t)=4e^-0.35t ounces per minute, where t is measured in minutes and 0≤t≤5. If there is 3 ounces of honey in the jar at time , how much honey is in the jar 4 seconds later?

Answer 1

First, we want to find the antiderivative (integrate) of r(t), which we can call h(t).

The units for h(t) would just be ounces of honey.

The antiderivative of r(t) is:

h(t) = (4 / -.35)e^(-.35t) + c

h(t) = -11.4286e^(-.35t) + c

Using initial condition, h(0) = 3, we can find c.

3 = -11.4286e^(-.35(0)) + c

3 = -11.4286 + c

c = 14.4286

Therefore,

h(t) = -11.4286e^(-.35t) + 14.4286

Because t is in minutes, we need to convert t = 4 seconds into minutes.

4 / 60 = 1/15 minutes

h(1/15) = 3.26358 ounces of honey