# Math Help - DIfferential equations

1. ## DIfferential equations

Hey i have some revision questions that i'm having trouble with.

1. A koala population N is subject to a birth rate a and a deah rate of b+y p per head of population per annum. Ecologists have modelled the rate of change of this population over time using the following:
dN/dt = (a-b)N-yN^2

It is knowln that a = 0.2, b=0.04, y= 2x10^-4 and the initial population was 200.
Determine if and when the population will have doubled in size.

2. The department of civil aviation have strict rules concerning the heights of structures/airborne articles near an airport. A weather balloon is to be tethered to the ground by a 250m rope. Because of the air traffic regulations the balloons height must not exceed 200m. Throughout the day a clinometer is used to measure the angle of elevation to the nearest degree. At 1pm the angle was measure at 53 degrees. Can the weather people be sure the balloon's height is within regulations.

Verify your decision using a calculus method mentioning any pertinent assumptions.

2. Originally Posted by phillipvdw2001
1. A koala population N is subject to a birth rate a and a deah rate of b+y p per head of population per annum. Ecologists have modelled the rate of change of this population over time using the following:
dN/dt = (a-b)N-yN^2
This is a Bernoulli differential equation. Or, it is separable, if you prefer:
$\frac{dN}{dt} = (a - b)N - yN^2$

$\frac{dN}{(a - b)N - yN^2} = dt$

So
$\int_{N_0}^N \frac{dN}{(a - b)N - yN^2} = \int_0^t dt$

and you can go from there.

-Dan

3. does trigonometry come under 'calculus'? if not, ignore the rest of this post ..

2) You have been told that the length of the diagonal side is 250m.
The angle of this line, to the ground is 53 $^{o}$.

From this, assuming that the 'ground' is perfectly level, you are then able to draw a perpendicular, upright, line from here (as shown by the dotted line in the diagram below.)

This the creates a right-angled triangle, from which you can use trigonometry to calculate the length of the vertical side.

You will see that I have already labelled the sides for you - this is so that you can match it to :

s\o/h c\a/h t\o/a

the adj[acent] side is not needed, so we will use:

s\o/h = (sine)\(opposite)/(hypotenuse)

$\text {(sine)} = \frac {\text {(opposite)}}{\text {(hypotenuse)}}$

now rearrange so that the the opposite is the subject:
$\text {hyp}(\text {sin} A^{o}) = \text {opp}$

$250sin53^{o} = \text {opp}$

$\text {opp} = 199.65887751182321157100020101723$

$\text {vertical height of the balloon} = 199.66m$

Because of the air traffic regulations the balloons height must not exceed 200m.
The vertical height of the balloon is 199.66m; therefore, it does NOT exceed the 200m and is within regulation.