SCalcET8M 3.8.017 PHYSICS HELP

Question

Suppose that a population of bacteria triples every hour and starts with 600 bacteria. Find an expression for the number n of bacteria after t hours.

n(t) =

Use it to estimate the rate of growth of the bacteria population after 1.5 hours. (Round your answer to the nearest whole number.)

n'(1.5) = 

Answer 1

The number of bacteria increases 3 times in 1 hour, 3x3=3² times in 2 hours, 3²x3=3³ times in 3 hours and thus 3^t times in t hours. Hence n(t) = n₀3^t.

n'(t) = d(n₀3^t)/dt = n₀d(e^(Ln3)t)/dt = n₀Ln3_*3^t. For n₀ = 600 and t =1.5 hours n'(1.5) = 600_*Ln3_*3^1.5 = 3425 bacteria/hour.

Answer 2

To solve this problem, start by writing out the first few terms of the sequence - let's translate the information in the problem. "triples every hour" means that each hour, the number of bacteria increases by a multiple of 3. Since the bacteria are unaware of time, each new number becomes the starting point for the next number

A table of t (hours) vs n (number) helps

t n

0 600

1 1800 = 600 *3

2 5400 = 1800 * 3 = (600*3)*3 = 600 * 3²

3 16200 = 5400 * 3 = 600 * 3³

....and so on

As you can see, n(t) = 600 * 3^t, where t is in hours

ONce you have the expression, go back and verify that it works:

when t = 0, 3^t = 1, n(0) = 600

for t = 2 , n(2) = 600 * 3² = 600 * 9 = 5400

Now that the equation has been found and verified, you can use it to make predictions.

The rate of change of a function is the first derivative of the function, in this case dn/dt

dn/dt = 600 * ∂(3^t) ⁄ ∂t

General Rule:

d (a^x) ⁄ dx = a^x ln(a)

==> dn/dt = 600 * (3^t * ln(3))

for t = 1.5, n'(t=1.5) = 600 * (3^1.5) * ln 3 = 3425 bacteria per hour CHECK: 1.5 is between 1 and 2 and 3425 is between 1800 and 5400.