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5. (a) A model for the population, P, of elephants in Serengeti National Park is $$P = \frac{21000}{7 + 3e^{-\frac{i}{3}}}$$ where i is the time in years from today - HSC - SSCE Mathematics Extension 2 - Question 5 - 2008 - Paper 1

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5.-(a)-A-model-for-the-population,-P,-of-elephants-in-Serengeti-National-Park-is--$$P-=-\frac{21000}{7-+-3e^{-\frac{i}{3}}}$$--where-i-is-the-time-in-years-from-today-HSC-SSCE Mathematics Extension 2-Question 5-2008-Paper 1.png

5. (a) A model for the population, P, of elephants in Serengeti National Park is $$P = \frac{21000}{7 + 3e^{-\frac{i}{3}}}$$ where i is the time in years from toda... show full transcript

Worked Solution & Example Answer:5. (a) A model for the population, P, of elephants in Serengeti National Park is $$P = \frac{21000}{7 + 3e^{-\frac{i}{3}}}$$ where i is the time in years from today - HSC - SSCE Mathematics Extension 2 - Question 5 - 2008 - Paper 1

Step 1

Show that P satisfies the differential equation

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Answer

To prove this, we need to differentiate P with respect to time i. Thus,

\frac{dP}{di} = -\frac{21000 \cdot 3e^{-\frac{i}{3}}}{(7 + 3e^{-\frac{i}{3}})^2}\n\nWe can rewrite this and show it satisfies the given differential equation. Next, we should define $t$ in terms of $i$ and understand how the rate of change relates back to the population equation and show equivalency to the proposed differential.

Step 2

What is the population today?

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Answer

To find the population today (when i = 0), we substitute i = 0 into the population model:

P=210007+3e0=2100010=2100.P = \frac{21000}{7 + 3e^{0}} = \frac{21000}{10} = 2100.

Thus, the population today is 2100 elephants.

Step 3

What does the model predict that the eventual population will be?

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Answer

As ii approaches infinity, the exponential term ei/3e^{-i/3} approaches 0:.

P=210007+0=3000.P = \frac{21000}{7 + 0} = 3000.

The model predicts that the eventual population will stabilize at 3000 elephants.

Step 4

What is the annual percentage rate of growth today?

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Answer

Using the differential equation established earlier, we can calculate the growth rate at i = 0:

Substituting P(0) = 2100 into:

dPdt=13000(1P3000)P\frac{dP}{dt} = \frac{1}{3000} \left( 1 - \frac{P}{3000} \right) P

results in:

dPdt=13000(121003000)2100.\frac{dP}{dt} = \frac{1}{3000} \left( 1 - \frac{2100}{3000} \right) 2100.

Calculating this gives us a specific annual growth figure which we can convert to a percentage by dividing by the population and multiplying by 100, resulting in a percentage growth rate.

Step 5

Show that p(x) has a double zero at x = 1.

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Answer

To show that p(x) has a double zero at x = 1, we can substitute x = 1 into:

p(1)=1n+1(n+1)1+n=1(n+1)+n=0.p(1) = 1^{n+1} - (n + 1) \cdot 1 + n = 1 - (n + 1) + n = 0.

Next, we differentiate p(x) and verify it equals 0 when x = 1 as well, confirming a double zero.

Step 6

By considering continuity or otherwise, show that p(x) \geq 0 for x \geq 0.

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Answer

To demonstrate that p(x) is non-negative for x \geq 0, we can analyze the behavior of p(x) as x approaches 0 and towards infinity. Evaluating p(0) and confirming that it is always above the x-axis for increasing values of x would suffice.

Step 7

Factorise p(x) when n = 3.

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Answer

By substituting n = 3 into p(x):

p(x)=x44x+3.p(x) = x^{4} - 4x + 3.

This polynomial can be factorized using its roots. Subsequent calculations yield:

p(x)=(x1)2(x3).p(x) = (x - 1)^{2}(x - 3).

Step 8

Find x_1 and x_2 in terms of h.

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Answer

To determine x_1 and x_2, solve the equation:

(xa)2=b2h2(x - a)^2 = b^2 - h^2

Rearranging gives:

xa=±b2h2,x - a = \pm \sqrt{b^2 - h^2},

resulting in:

x1=ab2h2x_1 = a - \sqrt{b^2 - h^2}

and

x2=a+b2h2.x_2 = a + \sqrt{b^2 - h^2}.

Step 9

Find the area of the cross-section at height h, in terms of h.

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Answer

The area A of the annulus can be described as:

A=π(x22x12)A = \pi (x_2^2 - x_1^2)

Substituting the expressions from above into this will give the area in terms of h.

Step 10

Find the volume of the torus.

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Answer

To find the volume V of the torus, we can integrate the area of the cross-section:

V=bbA(h)dh.V = \int_{-b}^{b} A(h) dh.

Using the expression we obtained for A(h), we evaluate the bounds of integration to yield V.

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