Rate of Change: Bees in Wildflower Meadow (a-c)

[chwala]

Homework Statement

See attached.

Relevant Equations

differentiation

part (a)

The number of Bees per Wildflower plant.

part (b)

$\dfrac{dB}{dF}= \dfrac{dB}{dt} ⋅\dfrac{dt}{dF}$$\dfrac{dB}{dF}=\left[\dfrac{2-3\sin 3t}{5e^{0.1t}}\right]$

$\dfrac{dB}{dF} (t=4)= 0.4839$part (c)

For values of $t>12$ The number of Bees per wildflower plants reduces drastically at 3 bees per 10 plants (number of bees are becoming insignificant)...that may not be a true representation of the model.Insight welcome...

 

[Mark44]

I haven't checked your numbers, but I don't see anything wrong with your work, otherwise.

 

[Steve4Physics]

part (b)
.
$\dfrac{dB}{dF} (t=4)= 0.4839$

You can’t justify giving the answer to four significant figures. The parameters in the equations are only precise to one sig. fig. I'd round to two sig. figs. as a compromise.

part (c)

For values of $t>12$ The number of Bees per wildflower plants reduces drastically at 3 bees per 10 plants (number of bees are becoming insignificant)...that may not be a true representation of the model.

I think what they are getting at is this...

The question states that the data are acquired during a number of weeks over summer. During summer the number of wildflowers can reasonably be expected to steadily increase. But after 12 weeks (t>12) we will have entered autumn and the number of wildflowers will be decreasing. This is not correctly modelled by $F(t) = e^{0.1t}$.

An improved version of $F(t)$ might include seasonal variations over a complete year.

 

[mjc123]

Part a: the rate of change of number of bees with number of plants. Note that dB/dt can be negative, but you don't have negative bees per plant.