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 = 5 weeks
 the population is 829 guppies and is growing at a rate of 50 guppies per week, while the average mass is 1.3 g and is increasing at a rate of 0.14 g/week. At what rate is the biomass increasing when 
t = 5?
 (Round your answer to one decimal place.)

The biomass \( B(t) \) is defined as the total mass of the fish population, which is the product of the number of individuals \( N(t) \) and the average mass of each individual fish \( M(t) \). Mathematically, we can express the biomass as:

\[
B(t) = N(t) \cdot M(t)
\]

We are asked to find the rate of change of the biomass at \( t = 5 \). To do this, we'll differentiate \( B(t) \) with respect to time \( t \) using the product rule.

The product rule states that:

\[
\frac{dB(t)}{dt} = \frac{dN(t)}{dt} \cdot M(t) + N(t) \cdot \frac{dM(t)}{dt}
\]

We are given the following values at \( t = 5 \):

- \( N(5) = 829 \) guppies
- \( \frac{dN(t)}{dt} = 50 \) guppies per week (rate of change of the population)
- \( M(5) = 1.3 \) g (average mass per fish)
- \( \frac{dM(t)}{dt} = 0.14 \) g/week (rate of change of the average mass)

Now, we substitute these values into the product rule formula:

\[
\frac{dB(t)}{dt} = 50 \cdot 1.3 + 829 \cdot 0.14
\]

Let’s compute this step by step:

\[
\frac{dB(t)}{dt} = 50 \cdot 1.3 + 829 \cdot 0.14 = 65 + 116.06 = 181.06
\]

So, the rate at which the biomass is increasing at \( t = 5 \) weeks is approximately:

\[
\boxed{181.1 \text{ grams per week}}
\]

Let me know if you'd like any further clarifications!

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**Related Questions**:
1. How does the product rule apply to other biological models involving growth?
2. How would the biomass change if the population growth rate were to decrease?
3. What would happen to the biomass rate if the average mass growth slowed down?
4. How could we model population dynamics if the fish population has a carrying capacity?
5. How might external factors (like food supply) affect the rates \( \frac{dN(t)}{dt} \) and \( \frac{dM(t)}{dt} \)?

**Tip**: When dealing with product functions like \( B(t) = N(t) \cdot M(t) \), remember to always apply the product rule when differentiating with respect to time or any other variable.

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 = 5 weeks the population is 829 guppies and is growing at a rate of 50 guppies per week, while the average mass is 1.3 g and is increasing at a rate of 0.14 g/week. At what rate is the biomass increasing when t = 5? (Round your answer to one decimal place.)

Learn how to calculate the rate of change of biomass in a fish population using the product rule in calculus. This detailed example involves population growth, differentiation, and the rates of change of individuals and average mass. Suitable for advanced high school students or introductory college calculus learners.

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