Respuesta :
Answer:
3. increases by 228 cats
Step-by-step explanation:
The rate of the increase in cats is given by:
[tex]3+5t[/tex]
as this is the rate, we can write it as:
[tex]\dfrac{dS}{dt} = 3 + 5t[/tex]
here t is the time in years. and S in the size of the colony
we're asked about the increase in size between the 10th and 4th year. So we need to know the sizes of the colony in those years.
We can find this out by integration. The above equation can be seen as a differential equation. Hence,
[tex]dS = (3+5t)dt\\S(t) = \displaystyle\int{3+5t}\,dt \\S(t) = 3t+\dfrac{5}{2}t^2+C[/tex]
S(t) is the function of the size of the colony at time t. We need to find the differences of sizes between t = 10 and t = 4. Basically,
[tex]\text{difference} = S(10) - S(4)\\\text{difference} = 3(10) + \dfrac{5}{2}(10)^2 + C - \left(3(4) + \dfrac{5}{2}(4)^2 + C\right) \\[/tex]
The C's will cancel out
[tex]\text{difference} = 3(10) + \dfrac{5}{2}(10)^2 -3(4) -\dfrac{5}{2}(4)^2\\\text{difference} = 228[/tex]
In between fourth and tenth year, 228 cats increases .
Since, A colony of feral cats is increasing at a rate of 3 + 5t cats per year
Let C represent colony of feral cats.
So, [tex]\frac{dC}{dt} =3+5t\\\\dC=(3+5t)dt[/tex]
Integrating on both side
We get, [tex]C(t)=3t+\frac{5}{2} t^{2}[/tex]
Number of cats in fourth year, [tex]C(4)=(3*4)+5*\frac{16}{2}=12+40=52\\[/tex]
Number of cats in tenth year, [tex]C(10)=30+\frac{5}{2}*100=30+250=280[/tex]
Number of cats increase = [tex]C(10)-C(4)=280-52=228[/tex]
Therefore, In between fourth and tenth year, 228 cats increases .
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