Question

1. The biomass B(t) of a fish population is the total mass of the members of the population at time t. It is the product of the number of individuals N(t) in the population and the average mass M(t) of a fish at time t.
   
   In the case of guppies, breeding occurs continually. Suppose that at time t = 4 weeks the population is 820 guppies and is growing at a rate of 50 guppies per week. The average mass is 1.2 grams and is increasing at a rate of 0.14 grams per week. At what rate is the biomass increasing when t = 4?
   
2. If g is differentiable the reciprocal rule says
   

Use the quotient rule to prove the reciprocal rule.

Answer 1

Solution :- The biomass B(t) of a fish population is the total mass of the members of the population at time t. It is the product of the number of individuals N(t) in the population and the average mass M(t) of a fish at time t.

That is B(t) = N(t) · M(t), { Using the product rule of derivative }

Also B′(t) = N(t) · M′ (t) + M(t) · N ′ (t)

Further Given that  at t =4 we have

B′(4) = N(4) · M′ (4) + M(4) · N ′ (4)

= 820 · 0.14 + 1.2 · 50 = 174.8

Thus, the biomass is increasing at the rate of 174.8 g/week.

b) Quotient rule: [f(x)/g(x)]' = [f'(x)g(x)-f(x)g'(x)] / [g(x)]^2

let f(x)=1, then

[1/g(x)]' = [g(x)(1)'-1*g'(x)]/[g(x)]^2

=-g'(x)/[g(x)]^2

Hence the proof