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12. Use the Question 12 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2022 - Paper 1

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12. Use the Question 12 Writing Booklet. (a) A direction field is to be drawn for the differential equation dy/dx = -x - 2y / x^2 + y^2 On the diagram on page 1 o... show full transcript

Worked Solution & Example Answer:12. Use the Question 12 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2022 - Paper 1

Step 1

A direction field is to be drawn for the differential equation

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Answer

To draw the direction field for the differential equation ( \frac{dy}{dx} = \frac{-x - 2y}{x^2 + y^2} ), calculate the slopes at the points P, Q, and R given in the booklet.

  1. Point P: Calculate the slope at P, for example, if P = (x1, y1), substitute these values into the equation to find ( \frac{dy}{dx} ).
  2. Point Q: Similarly, calculate the slope at Q using its coordinates.
  3. Point R: Repeat for R.

After calculating the values, draw arrows indicating the direction of the field at each point.

Step 2

Will any team be penalised? Justify your answer.

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Answer

To determine if any team will be penalized, first we need to ascertain the number of players per team:

  • Total players = 41
  • Total teams = 13
  • Average players per team = ( \frac{41}{13} \approx 3.15 )

Since any team with more than 3 players above the age limit is penalised, it’s possible for some teams to exceed the limit. For example, if one team had 4 players above the age limit, that team would be penalized. Therefore, yes, at least one team will be penalized under the condition given.

Step 3

Find the equation of the tangent to the curve y = x arctan(x) at the point with coordinates (1, π/4)

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Answer

To find the equation of the tangent, we first need to find the derivative of the function ( y = x \arctan(x) ):

  1. Differentiate using the product rule:

    • Let ( u = x ) and ( v = \arctan(x) )
    • Then ( \frac{du}{dx} = 1 ) and ( \frac{dv}{dx} = \frac{1}{1+x^2} )
    • By the product rule: [ \frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx} = \arctan(x) + x \cdot \frac{1}{1+x^2} ]
    • Substitute ( x = 1 ): [ \frac{dy}{dx} \Big|_{x=1} = \arctan(1) + 1 \cdot \frac{1}{2} = \frac{\pi}{4} + \frac{1}{2} ]
  2. Find the equation of the tangent line:

    • The slope at the point is given as ( m = \frac{\pi}{4} + \frac{1}{2} )
    • Using point-slope form ( y - y_1 = m(x - x_1) ): [ y - \frac{\pi}{4} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(x - 1) ]
  3. Convert to slope-intercept form (y = mx + c):

    • Rearranging gives the equation of the tangent.

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