Word problem with growth rates.

Question

I have this word problem involving growth rates. Mind checking my work?

A bacteria culture initially contains $100$ cells and grows at a rate proportional to its size. After an hour the population has increased to $420$.

(a) Find an expression for the number of bacteria after $t$ hours.
(b) Find the number of bacteria after $3$ hours.
(c) Find the rate of growth after $3$ hours.
(d) When will the population reach $10,000$?

So...

a) $P(0) = 100$ and $P(1) = 420$

$$P(t) = 100 \cdot e^{kt}$$ $$e^k = 4.2$$ $$k = \ln(4.2)$$ So model is: $$P(t) = 100 \cdot e^{\ln(4.2) \cdot t}$$

Is this right?

b) $p(3) = 100 \cdot e^{\ln(4.2) \cdot 3} = 7408.8$

c) What is the derivative at $p(3)$? Having trouble with c and d here. I assume for c I can just do

$$7408.8 \cdot 4.2$$ because I guess this is the definition? But why?

Answer 1

Note that

$$P(t) = 100 \cdot e^{ln(4.2) \cdot t}\implies P'(t) = 100 \cdot e^{ln(4.2) \cdot t}\cdot \ln4.2=k\cdot P(t)$$

thus

$$P'(3) = 100 \cdot e^{ln(4.2) \cdot 3}\cdot \ln4.2$$

indeed the following derivative rule applies

$$f(t)=e^{g(t)}\implies f'(t)=e^{g(t)}\cdot g'(t)$$

Answer 2

You were able to turn the question 1. into the correct formula $$p(t)=100\cdot e^{kt}$$ in a snap of fingers, don't try to tell me you have problems with (c).

The derivative of the above is $p'(t)=k\cdot p(t)$ because that was the requirement remember:

grows at a rate proportional to its size

Threfore you don't multiply $p(3)$ by $4.2$ but by $k=\ln(4.2)$ to get the growth rate.