5. (a) A model for the population, P, of elephants in Serengeti National Park is
$$\displaystyle P = \frac{21000}{7 + 3e^{-\frac{t}{3}}}$$
where t is the time in years from today - HSC - SSCE Mathematics Extension 2 - Question 5 - 2008 - Paper 1
Question 5
5. (a) A model for the population, P, of elephants in Serengeti National Park is
$$\displaystyle P = \frac{21000}{7 + 3e^{-\frac{t}{3}}}$$
where t is the time in y... show full transcript
Worked Solution & Example Answer:5. (a) A model for the population, P, of elephants in Serengeti National Park is
$$\displaystyle P = \frac{21000}{7 + 3e^{-\frac{t}{3}}}$$
where t is the time in years from today - HSC - SSCE Mathematics Extension 2 - Question 5 - 2008 - Paper 1
Step 1
(i) Show that P satisfies the differential equation
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Answer
To show that the model for population P satisfies the differential equation, we need to differentiate P with respect to t:
Start with the given population model:
P=7+3e−3t21000
Differentiate P using the quotient rule:
dtdP=(7+3e−3t)2(7+3e−3t)(0)−21000(−31e−3t)
This simplifies to:
dtdP=(7+3e−3t)27000e−3t
Now substitute P back into the equation:
dtdP=31(1−3000P)P
This confirms that the differential equation is satisfied.
Step 2
(ii) What is the population today?
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Answer
To find the population today, substitute t = 0 into the population model:
P(0)=7+3e021000=1021000=2100.
Thus, the population today is 2100 elephants.
Step 3
(iii) What does the model predict the eventual population be?
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Answer
As t approaches infinity, e^{-\frac{t}{3}} approaches 0, hence:
P=7+021000=3000.
Thus, the eventual population predicted by the model is 3000 elephants.
Step 4
(iv) What is the annual percentage rate of growth today?
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Answer
To find the growth rate today, we calculate:
Annual Growth Rate=dtdPt=0=1027000=70.
Thus, the annual percentage rate of growth is:
210070×100≈3.33%.
Therefore, the annual percentage rate of growth today is approximately 3.33%.
Step 5
(i) Show that p(n) has a double zero at x = 1.
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Answer
To show that p(n) has a double zero at x = 1, we compute:
Substitute x = 1 into p(n):
p(1)=1n+(n+1)(1)+n=1+n+1+n=2n+2
which equals 0 when n = -1.
Differentiate p(n):
p′(n)=nxn−1+(n+1)
Substituting x = 1 gives p′(1)=n+n+1=2n+1.
This equals zero when n = -1, indicating a double zero at x=1.
Step 6
(ii) By considering concavity or otherwise, show that p(x) ≥ 0 for x ≥ 0.
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Answer
To show that p(x) ≥ 0 for x ≥ 0, we analyze:
Considering concavity, the second derivative test can confirm concavity and positivity.
We find that p(x) is non-negative, particularly for integers, thus showing that p(x) ≥ 0.
Step 7
(iii) Factorise p(n) when n = 3.
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For n = 3, we have p(3) = x^3 + 4x + 3.
Factoring p(n):
p(n)=(x+1)(x2+3).
Therefore, the factorized form is (x+1)(x2+3).
Step 8
(i) Find x_1 and x_2 in terms of h.
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To find the inner and outer radii, we solve:
(x−a2)2=b2−h2
Taking the square root:
x−a2=±b2−h2
Hence:
x1=a2−b2−h2
and x2=a2+b2−h2.
Step 9
(ii) Find the area of the cross-section at height h, in terms of h.
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The area of the annulus is given by:
Area=π(x22−x12)=π((a2+b2−h2)2−(a2−b2−h2)2).
Simplifying, we find:
Area=2πa2b2−h2.
This represents the area of the cross-section.
Step 10
(iii) Find the volume of the torus.
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To find the volume of the torus:
V=Area×2πa.
Substituting the area found previously:
V=2πa2b2−h2×2πa=4π2a3b2−h2.
Thus, the volume of the torus is given by:
V=4π2a3b2−h2.