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Question 12 (16 marks) Use the Question 12 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2022 - Paper 1

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Question 12 (16 marks) Use the Question 12 Writing Booklet. (a) A direction field is to be drawn for the differential equation $$ \frac{dy}{dx} = \frac{-x - 2y}{x... show full transcript

Worked Solution & Example Answer:Question 12 (16 marks) 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 equation dydx=x2yx2+y2\frac{dy}{dx} = \frac{-x - 2y}{x^2 + y^2}, we first evaluate the slopes at specific points, such as P, Q, and R.

  1. Select points: Choose coordinates for P, Q, and R from the Writing Booklet.
  2. Calculate slopes: For example, at point P with coordinates (x_p, y_p):
    slope=xp2ypxp2+yp2slope = \frac{-x_p - 2y_p}{x_p^2 + y_p^2} Repeat for Q and R.
  3. Draw the direction field: Use the calculated slopes to sketch the lines representing the direction field.

Step 2

Will any team be penalised? Justify your answer.

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Answer

To determine if any team will be penalised, we first calculate the total number of players exceeding the age limit.

  1. Total players above age limit: There are 41 players found above the age limit.
  2. Teams' total players: Each team has an average of 41 / 13 = 3.15 players above the age limit.
  3. Apply the threshold: Since any team with more than 3 penalised players will be penalised, at least one team will be penalised as the average indicates there is at least one team with more than 3.

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 tangent line to the curve at the given point, we first need to determine the slope of the curve using differentiation.

  1. Differentiate the function:
    y=xarctan(x)y = x \cdot \arctan(x)
    Using the product rule:
    dydx=arctan(x)+x11+x2\frac{dy}{dx} = \arctan(x) + x \cdot \frac{1}{1+x^2}
    Evaluate at x=1x = 1:
    dydxx=1=arctan(1)+112=π4+12\frac{dy}{dx} \bigg|_{x=1} = \arctan(1) + 1 \cdot \frac{1}{2} = \frac{\pi}{4} + \frac{1}{2}
  2. Find the slope: Calculate the value to get the slope mm.
  3. Use point-slope form: With slope mm and point (1,π4)(1, \frac{\pi}{4}), the equation of the tangent is:
    yπ4=m(x1)y - \frac{\pi}{4} = m(x - 1)
    Rearranging gives the required equation in the form y=mx+cy = mx + c.

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