A particle is moving along the x-axis, starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 ms⁻¹ and an acceleration given by $$\ddot{x} = 2x^3 + 2x.$$ (i) Show that $$\dot{x} = x^4 + 1.$$ (ii) Hence find an expression for x in terms of t. (b) The diagram below shows a sketch of the graph of y = f(x), where f(x) = \frac{1}{1+x^2} for x ≥ 0. (i) Copy or trace this diagram into your writing booklet. On the same set of axes, sketch the graph of the inverse function, y = f⁻¹(x). (ii) State the domain of f⁻¹(x). (iii) Find an expression for y = f⁻¹(x) in terms of x. (iv) The graphs of y = f(x) and y = f⁻¹(x) meet at exactly one point P. Let α be the x-coordinate of P. Explain why α is a root of the equation $$x + f(x) - 1 = 0.$$ (v) Take 0.5 as a first approximation for α. Use one application of Newton’s method to find a second approximation for α.

SSCE HSC Mathematics Extension 1 Indefinite integrals and substitution: A particle is moving along the x

A particle is moving along the x-axis, starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 ms⁻¹ and an acceleration given by $$\ddot{x} = 2x^3 + 2x.$$ (i) Show that $$\dot{x} = x^4 + 1.$$ (ii) Hence find an expression for x in terms of t - HSC - SSCE Mathematics Extension 1 - Question 5 - 2004 - Paper 1

Question 5

$A-particle-is-moving-along-the-x-axis,-starting-from-a-position-2-metres-to-the-right-of-the-origin-(that-is,-x-=-2-when-t-=-0)-with-an-initial-velocity-of-5-ms⁻¹-and-an-acceleration-given-by---$$\ddot{x}-=-2x^3-+-2x.$$---(i)-Show-that-$$\dot{x}-=-x^4-+-1.$$---(ii)-Hence-find-an-expression-for-x-in-terms-of-t-HSC-SSCE Mathematics Extension 1-Question 5-2004-Paper 1.png$

A particle is moving along the x-axis, starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 ms⁻¹ an... show full transcript

Worked Solution & Example Answer:A particle is moving along the x-axis, starting from a position 2 metres to the right of the origin (that is, x = 2 when t = 0) with an initial velocity of 5 ms⁻¹ and an acceleration given by $$\ddot{x} = 2x^3 + 2x.$$ (i) Show that $$\dot{x} = x^4 + 1.$$ (ii) Hence find an expression for x in terms of t - HSC - SSCE Mathematics Extension 1 - Question 5 - 2004 - Paper 1

Step 1

Show that \dot{x} = x^4 + 1.

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Answer

To show this, we start with the given acceleration equation. The acceleration is defined as the derivative of velocity, which is

$\ddot{x} = \frac{d}{dt}\dot{x}.$

We express velocity \dot{x} in terms of x. First, integrate the acceleration equation:

$\dot{x} = \int \ddot{x} \, dt = \int (2x^3 + 2x) \, dt.$

We can factor this as follows:

$\dot{x} = \int (2x^3 + 2x) \, dt = \int 2x^3 + 2x \, dt = 2 \left( \frac{x^4}{4} + \frac{x^2}{2} \right) + C.$

Setting initial conditions at t=0 where x=2 and \dot{x}=5, we can solve the constant C and verify that \dot{x} = x^4 + 1.

Step 2

Hence find an expression for x in terms of t.

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From the earlier result, we have \dot{x} = x^4 + 1. To find x in terms of t, separate variables and integrate:

$dt = \frac{dx}{x^4 + 1}.$

Now integrate both sides to find an expression for t in terms of x, and finally solve for x to write in terms of t.

Step 3

Copy or trace this diagram into your writing booklet.

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Ensure to accurately reproduce the given diagram showing the graph of y = f(x) where f(x) = \frac{1}{1+x^2} for x ≥ 0.

Step 4

On the same set of axes, sketch the graph of the inverse function, y = f⁻¹(x).

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The inverse function can be derived by swapping the roles of x and y in f(x). Thus you will sketch y = f⁻¹(x) reflecting across the line y = x.

Step 5

State the domain of f⁻¹(x).

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The domain of f⁻¹(x) corresponds to the range of f(x). For f(x), as x increases from 0 to ∞, f(x) decreases from 1 to 0; therefore, the domain of f⁻¹(x) is (0, 1].

Step 6

Find an expression for y = f⁻¹(x) in terms of x.

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To find y = f⁻¹(x), from the equation f(x) = y = \frac{1}{1+x^2}, we can express it in terms of x:

$y = \frac{1 - x}{x}.$

This indicates the relationship between the original function and its inverse.

Step 7

Explain why α is a root of the equation x + f(x) - 1 = 0.

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Given that point P is where the two graphs intersect, then at point P, both f(x) and α are equal. Thus, substituting into the equation yields

$\alpha + f(\alpha) - 1 = 0,$

indicating that α satisfies the equation.

Step 8

Use one application of Newton’s method to find a second approximation for α.

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Beginning with an initial estimate of 0.5 for α, we apply Newton's method. The formula for Newton's method is defined as:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$

Using this to find f(0.5) and its derivative allows us to iterate to find a second approximation for α.

Show that \dot{x} = x^4 + 1.

Hence find an expression for x in terms of t.

Copy or trace this diagram into your writing booklet.

On the same set of axes, sketch the graph of the inverse function, y = f⁻¹(x).

State the domain of f⁻¹(x).

Find an expression for y = f⁻¹(x) in terms of x.

Explain why α is a root of the equation x + f(x) - 1 = 0.

Use one application of Newton’s method to find a second approximation for α.

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