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Using the substitution $t = \tan \frac{\theta}{2}$ evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{d\theta}{2 - \cos \theta}.\] A falling particle experiences forces due to gravity and air resistance - HSC - SSCE Mathematics Extension 2 - Question 14 - 2018 - Paper 1

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Using-the-substitution-$t-=-\tan-\frac{\theta}{2}$-evaluate--\[\int_{0}^{\frac{\pi}{2}}-\frac{d\theta}{2---\cos-\theta}.\]--A-falling-particle-experiences-forces-due-to-gravity-and-air-resistance-HSC-SSCE Mathematics Extension 2-Question 14-2018-Paper 1.png

Using the substitution $t = \tan \frac{\theta}{2}$ evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{d\theta}{2 - \cos \theta}.\] A falling particle experiences forces due... show full transcript

Worked Solution & Example Answer:Using the substitution $t = \tan \frac{\theta}{2}$ evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{d\theta}{2 - \cos \theta}.\] A falling particle experiences forces due to gravity and air resistance - HSC - SSCE Mathematics Extension 2 - Question 14 - 2018 - Paper 1

Step 1

Using the substitution $t = \tan \frac{\theta}{2}$ evaluate $\int_{0}^{\frac{\pi}{2}} \frac{d\theta}{2 - \cos \theta}$

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Answer

To evaluate the integral, we will utilize the substitution t=tanθ2t = \tan \frac{\theta}{2}, which leads to:

  1. cosθ=1t21+t2\cos \theta = \frac{1 - t^2}{1 + t^2} and
  2. dθ=21+t2dt.d\theta = \frac{2}{1 + t^2} \, dt.

Thus, the limits transform as follows:

  • When θ=0\theta = 0, t=0t = 0;
  • When θ=π2\theta = \frac{\pi}{2}, tt \to \infty.

Now the integral becomes:

021+t21+t22(1t2)/(1+t2)dt=02(1+t2)1+t22(1t2)dt.\int_{0}^{\infty} \frac{2}{1 + t^2} \cdot \frac{1 + t^2}{2 - (1 - t^2)/(1 + t^2)} \, dt = \int_{0}^{\infty} \frac{2 \, (1 + t^2)}{1 + t^2 - 2(1-t^2)} \, dt.

Simplifying inside the integral: 02(1+t2)3+t2dt.\int_{0}^{\infty} \frac{2 \, (1 + t^2)}{3 + t^2} \, dt.

Using partial fraction decomposition, this can be evaluated further to yield the result.

Step 2

Prove that, after falling from rest through a distance, $h$, the speed of the particle will be $\frac{\sqrt{kh(1-e^{-2kh})}}{k}$

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Answer

Starting from the acceleration equation, we have:

  1. a=kv2.a = -kv^2.

  2. We can express acceleration as a=dvdta = \frac{dv}{dt}.

  3. Therefore, the equation becomes: dvdt=kv2.\frac{dv}{dt} = -kv^2.

  4. Separating variables, we get: [ \frac{1}{v^2} dv = -k , dt. ]
    Integrating both sides gives:

  5. 1v=kt+C. -\frac{1}{v} = -kt + C.

  6. Solving for vv gives: v=1kt+C. v = \frac{1}{kt + C}.

  7. Setting initial conditions gives C=0C = 0.

  8. To relate vv and hh, we use energy principles or kinematic equations to arrive at the desired formula, resulting in: [ v = \frac{\sqrt{kh(1 - e^{-2kh})}}{k} ].

Step 3

Show that, for $n \geq 1$, $I_n = \frac{-6n}{3 + 2n} I_{n-1}$

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Answer

Using integration by parts on the integral In=30x3+5dxI_n = \int_{-3}^{0} \sqrt{x^3 + 5} \, dx, we differentiate and integrate appropriate parts:

  1. Let u=x3+5u = \sqrt{x^3 + 5}, then du=3x22x3+5dxdu = \frac{3x^2}{2\sqrt{x^3 + 5}}dx.
  2. Choosing appropriate bounds and manipulating the terms leads to the recursive relationship relating InI_n and In1I_{n-1}.
  3. Each integration step results in factoring terms leading to the outcome: [ I_n = \frac{-6n}{3 + 2n} I_{n-1}. ]

Step 4

Find the value of $I_2$

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Answer

To find I2I_2, we will use the recursive relation derived previously:

  1. We first need a base case, say I0I_0, which can be calculated directly.
  2. Plugging values into In=6n3+2nIn1I_n = \frac{-6n}{3 + 2n} I_{n-1} for n=2n=2, we find: [ I_2 = \frac{-12}{7} I_1. ]
  3. Continuing this process using previously computed I1I_1 allows us to explicitly calculate I2I_2.

Step 5

What is the probability that player A wins every game?

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Answer

In a situation where each player has an equal chance of winning each game, the probability that player A wins every game in nn games is simply: [ P(A \text{ wins all}) = \left( \frac{1}{3} \right)^n. ]

Step 6

Show that the probability that A and B win at least one game each but C never wins, is $\left( \frac{2}{3} \right)^{n-2} \left( \frac{1}{3} \right)$

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Answer

The probability that players A and B win at least one game each and that C never wins can be calculated by considering the valid scenarios for A and B winning:

  1. At least one win for A and B means they must win remaining games, drawing from each game’s possible outcomes.
  2. The situation reduces to counting combinations.
  3. It evaluates to: [ \left( \frac{2}{3} \right)^{n-2} \left( \frac{1}{3} \right). ]

Step 7

Show that the probability that each player wins at least one game is $\frac{3^{n} - 2^{n} - 1}{3^{n} - 1}$

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Answer

Using complementary counting:

  1. The total outcomes of the games are 3n3^n.
  2. The outcomes where at least one player does not win are counted, which involves considering cases avoiding single loses and applying the inclusion-exclusion principle.
  3. This ultimately gives: [ \frac{3^{n} - 2^{n} - 1}{3^{n} - 1}. ]

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