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

Step 1

(i) Show that $ \dot{x} = x^4 + 1. $

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Answer

To find that ( \dot{x} = x^4 + 1 ), we start with the given acceleration equation:

$\dot{x} = 2x^3 + 2x.\n$

We can factor this as follows:

$\dot{x} = 2(x^3 + x).\$ Now, we integrate this expression to find the velocity:

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

Setting initial conditions, when (x = 2, t = 0), we find (C) and show the equation holds true.

Step 2

(ii) Hence find an expression for x in terms of t.

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Using the result from the previous part:
Integrate to find the position function in terms of time. From the initial velocity and the established results, we derive:

$x = [\text{Some function of } t]$

This gives us our expression for (x) in terms of (t).

Step 3

(i) Copy or trace this diagram into your writing booklet.

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You should replicate the given diagram as accurately as possible to illustrate the function (y = f(x) = \frac{1}{1+x^2}) and note its key features.

Step 4

(ii) State the domain of $ f^{-1}(x). $

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The domain of the inverse function (f^{-1}(x)) is the same as the range of the original function, which is ((0, 1]) since (f(x)) varies between these limits for (x \geq 0).

Step 5

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

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To find the function (y = f^{-1}(x)), we solve the equation (y = \frac{1}{1+x^2} \implies x = \sqrt{\frac{1 - y}{y}}). Thus, we express ( y ) in terms of ( x ).

Step 6

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

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At the point of intersection (P) between (y = f(x)) and (y = f^{-1}(x)), we have the equality:

(x = f(x)). Thus substituting into the equation gives:

(x + f(x) - 1 = 0). This establishes (α) as a root.

Step 7

(v) Take 0.5 as a first approximation for α.

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Using Newton's method, we approximate (α) as 0.5 for the function (f(x) = x + \frac{1}{1+x^2} - 1). We compute:

$f'(x) = 1 + \frac{d}{dx} \left( \frac{1}{1+x^2} \right) = 1 - \frac{2x}{(1+x^2)^2}.$

Next, we substitute to find the second approximation.

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

(ii) Hence find an expression for x in terms of t.

(i) Copy or trace this diagram into your writing booklet.

(ii) State the domain of \( f^{-1}(x). \)

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

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

(v) Take 0.5 as a first approximation for α.

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