y = Cekt
Next, we plug in y = 4,000 and t = 0 to solve for C.
4,000 = Ce^0
C = 4,000
Now, we have y = 4,000ekt. Next, we plug in y = 6,500 and t = 3 to solve for k.
6,500 = 4,000e^3 k
Therefore, our equation for the population of bacteria, y, at time, t, is y ≈ 4,000e0.162t, or if we
want an exact solution, it is y = 4,000 . Finally, we can solve for the population at time t
= 10: y ≈ 4,000e0.162(10) ≈ 20,212 or y = 4,000 ≈ 20,179. (Notice how even with an
“exact” solution, we still have to round the answer. And, if we are concerned with significant
figures, the population can be written as 20,000.)
- 45 m
Because acceleration is the derivative of velocity, we can write = −9. We are also told that
v = 18 and h = 45 when t = 0. We solve this differential equation by separation of variables.
We want to get all of the v variables on one side of the equals sign and all of the t variables on
the other side. We can do this easily by multiplying both sides by dt: dv = −9 dt. Next, we
integrate both sides:
∫^ dv = ∫ −9 dt
v = −9 + C 0
Now we can solve for C 0 by plugging in v = 18 and t = 0: 18 = −9 (0) + C 0 so C 0 = 18. Thus,
