Photo AI
12 marks) Use the Question 12 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2022 - Paper 1
Question 12
12 marks) 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 diagra... show full transcript
Worked Solution & Example Answer:12 marks) Use the Question 12 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2022 - Paper 1
Step 1
a) A direction field is to be drawn for the differential equation
Answer
To draw the direction field for the differential equation ( \frac{dy}{dx} = \frac{-x - 2y}{x^2 + y^2} ), we evaluate the slopes at the specified points P, Q, and R:
-
Point P (0, 0): [ \frac{dy}{dx} = \frac{-0 - 2(0)}{0^2 + 0^2} = \text{undefined} ] (The slope is undefined; vertical line)
-
Point Q (1, 1): [ \frac{dy}{dx} = \frac{-1 - 2(1)}{1^2 + 1^2} = \frac{-3}{2} ] (Slope = -1.5, indicate downward slant)
-
Point R (-1, 0): [ \frac{dy}{dx} = \frac{-(-1) - 2(0)}{(-1)^2 + 0^2} = \frac{1}{1} = 1 ] (Slope = 1, indicate upward slant)
Thus, the slopes at points P, Q, and R are vertical, downward slant, and upward slant, respectively.
Step 2
b) Will any team be penalised? Justify your answer.
Answer
To determine if any team will be penalised, we first need to calculate the number of players above the age limit per team.
Given that there are 41 players above the age limit and 13 teams, we can use the following formula to find the maximum number of players above the age limit per team:
[ \text{Average number per team} = \frac{41}{13} \approx 3.15 ]
Since any team with more than 3 players above the age limit is penalised, and we have an average of about 3.15, it indicates that at least one team is likely to have more than 3 players above the limit. Thus, at least one team will be penalised.
Step 3
c) Find the equation of the tangent to the curve y = x arctan(x) at the point with coordinates (1, π/4)
Answer
To find the equation of the tangent line, we need the derivative of the curve at the point (1, ( \frac{\pi}{4} )).
-
Differentiate the function: Let ( y = x \cdot \text{arctan}(x) ). Using the product rule: [ y' = \text{arctan}(x) + x \cdot \frac{d}{dx}(\text{arctan}(x)) ] The derivative of ( \text{arctan}(x) ) is ( \frac{1}{1+x^2} ), thus: [ y' = \text{arctan}(x) + \frac{x}{1+x^2} ]
-
Evaluate the derivative at x = 1: [ y'(1) = \text{arctan}(1) + \frac{1}{1+1^2} = \frac{\pi}{4} + \frac{1}{2} ] Since ( \text{arctan}(1) = \frac{\pi}{4} ), we exceed evaluate as ( \frac{\pi}{4} + 0.5 \approx 1.285). Thus, the slope at (1, ( \frac{\pi}{4} )) is approximately 1.285.
-
Find the equation of the tangent line using the point-slope form: [ y - \frac{\pi}{4} = m(x - 1) ] Substituting for m: [ y - \frac{\pi}{4} = 1.285(x-1) ] Finally, rearranging gives: [ y = 1.285x - 1.285 + \frac{\pi}{4} ] The tangent is in the form ( y = mx + c ).