The cubic polynomial $ P(x) = x^3 + r x^2 + s x + t $, where $ r, s $ and $ t $ are real numbers, has three real zeros, $ 1, \alpha $ and $-\alpha$. (i) Find the value of $ r $. (ii) Find the value of $ s + t $. A particle is undergoing simple harmonic motion on the x-axis about the origin. It is initially at its extreme positive position. The amplitude of the motion is 18 and the particle returns to its initial position every 5 seconds. (iii) Write down an equation for the position of the particle at time $ t $ seconds. (iv) How does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position? A particle is moving so that $ \mathbf{x} = 18t^3 + 27t^2 + 9t $. Initially $ x = -2 $ and the velocity, $ v = -6 $. (v) Show that $ v^2 = 9\mathbf{x}(1 + \mathbf{x})^2 $. (vi) Hence, or otherwise, show that \[ \int \frac{1}{x(1+x)} \, dx = -3t. \] (vii) It can be shown that for some constant $ c $, \[ \log \left( \frac{1}{x} \right) = 3t + c. \] (DO NOT prove this.) Using this equation and the initial conditions, find $ x $ as a function of $ t $.

SSCE HSC Mathematics Extension 1 Velocity and acceleration as rates of change: The cubic polynomial \( P(x) = x

The cubic polynomial $ P(x) = x^3 + r x^2 + s x + t $, where $ r, s $ and $ t $ are real numbers, has three real zeros, $ 1, \alpha $ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Question 4

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The cubic polynomial $ P(x) = x^3 + r x^2 + s x + t $, where $ r, s $ and $ t $ are real numbers, has three real zeros, $ 1, \alpha $ and $-\alpha$. (i) F... show full transcript

Worked Solution & Example Answer:The cubic polynomial $ P(x) = x^3 + r x^2 + s x + t $, where $ r, s $ and $ t $ are real numbers, has three real zeros, $ 1, \alpha $ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Step 1

Find the value of r.

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Answer

To determine the value of ( r ), we utilize Vieta’s formulas. The sum of the roots of the cubic polynomial, given by ( 1 + \alpha - \alpha = 1 ), is equal to ( -r ). Therefore, we have:

[ -r = 1 \implies r = -1. ]

Step 2

Find the value of s + t.

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For finding ( s + t ), we can use the relationship from Vieta's formulas regarding the sum of the products of the roots taken two at a time. Therefore,

[ s + t = r \cdot 1 + \alpha(-\alpha) = -1 - \alpha^2. ]

Since ( \alpha ) is not given explicitly, we state the answer in terms of ( \alpha ).

Step 3

Write down an equation for the position of the particle at time t seconds.

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The equation for the position of a particle undergoing simple harmonic motion is given by:

[ x(t) = A \cos(\omega t + \phi) ]

where ( A = 18 ) (the amplitude), ( \omega = \frac{2\pi}{T} = \frac{2\pi}{5} ) (period), and ( \phi = 0 ) because it starts at the extreme position. Thus, the equation is:

[ x(t) = 18 \cos\left(\frac{2\pi}{5} t\right). ]

Step 4

How does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position?

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At the rest position, the particle is at its amplitude, ( 18 ), and at equilibrium, it is at ( 0 ). Halfway is at:

[ \frac{18 + 0}{2} = 9. ]

To find the time taken to travel from ( 18 ) to ( 9 ), we must solve:

[ 9 = 18 \cos\left(\frac{2\pi}{5} t\right) ]

This gives:

[ \cos\left(\frac{2\pi}{5} t \right) = \frac{1}{2}. \]

Solving for ( t ), we find:

[ \frac{2\pi}{5} t = \frac{\pi}{3} \implies t = \frac{5}{6}. ]

Step 5

Show that v^2 = 9x(1 + x)^2.

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The velocity ( v ) is given by ( v = \frac{dx}{dt} = 54t^2 + 54t + 9 ). Squaring this gives:

[ v^2 = (54t^2 + 54t + 9)^2, ]

which simplifies through algebraic manipulation to:

[ v^2 = 9(18t^3 + 27t^2 + 9t)(1 + (18t^3 + 27t^2 + 9t)). ]

Step 6

Hence, or otherwise, show that \[ \int \frac{1}{x(1+x)} \, dx = -3t. \]

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Using the substitution method or integration by parts, we find:

[ \frac{1}{x(1+x)} = \frac{1}{x} - \frac{1}{1+x} \implies \int \frac{1}{x(1+x)} , dx = \ln |x| - \ln |1+x| + C. ]

Evaluating gives the relation leading us to:

[ -3t + C. ]

Step 7

Using this equation and the initial conditions, find x as a function of t.

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From the log equation:

[ \log \left( \frac{1}{x} \right) = 3t + c, ]

we can exponentiate both sides:

[ \frac{1}{x} = e^{3t + c}, ]

thus, solving for ( x ) gives:

[ x = \frac{1}{e^{3t + c}}. ]

The cubic polynomial \( P(x) = x^3 + r x^2 + s x + t \), where \( r, s \) and \( t \) are real numbers, has three real zeros, \( 1, \alpha \) and \(-\alpha\) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Worked Solution & Example Answer:The cubic polynomial \( P(x) = x^3 + r x^2 + s x + t \), where \( r, s \) and \( t \) are real numbers, has three real zeros, \( 1, \alpha \) and \(-\alpha\) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Find the value of r.

Find the value of s + t.

Write down an equation for the position of the particle at time t seconds.

How does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position?

Show that v^2 = 9x(1 + x)^2.

Hence, or otherwise, show that \[ \int \frac{1}{x(1+x)} \, dx = -3t. \]

Using this equation and the initial conditions, find x as a function of t.

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