Photo AI
Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point \((0, 1)\) - HSC - SSCE Mathematics Extension 1 - Question 14 - 2022 - Paper 1
Question 14
Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point \((0, 1)\). The vectors \(\vec{u}\) and \(\ve... show full transcript
Worked Solution & Example Answer:Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point \((0, 1)\) - HSC - SSCE Mathematics Extension 1 - Question 14 - 2022 - Paper 1
Step 1
Find the particular solution to the differential equation \((x - 2) \frac{dy}{dx} = xy\) that passes through the point \((0, 1)\).
Answer
To find the particular solution, we start with the given differential equation:
Rearranging gives:
This is a separable equation, which we can write as:
Integrating both sides yields:
Exponentiating, we obtain:
where (k = e^C).
Now, we use the initial condition (y(0) = 1):
\Rightarrow k = -\frac{1}{2}$$ Thus, the particular solution is: $$y = -\frac{1}{2}(x - 2) = -\frac{1}{2}x + 1.$$Step 2
The vectors \(\vec{u}\) and \(\vec{v}\) are not parallel. Prove that \(|\vec{u} - \lambda \vec{v}|\) is smallest when \(\lambda = \lambda_0\).
Answer
Let (\lambda = \lambda_0). We can express the distance as:
Using the property of projections, we find that:
This holds due to the projection theorem, confirming that the minimum distance occurs when (\lambda = \lambda_0).
Step 3
Show that \(d\) must be less than 37% of the maximum possible range of the projectile.
Answer
The maximum range (R) of a projectile launched with speed (u) at angle (\theta) is given by:
The speed of the target is (u) and the projectile has an initial speed of (2u). Since the target moves away at half the projectile’s speed, we need to consider how far the projectile travels while the target moves.
Calculating the time (t) until the projectile reaches maximum height:
In this time, the target travels a distance of:
Thus,
which leads to the result that for a hit, (d < 0.37 \cdot R).
Step 4
Given management's decision and using a suitable approximation, find the maximum number of tickets that can be sold for a flight which has 350 seats.
Answer
Using the binomial approximation, let (n = 350) and the probability (p = 0.05). We need the number of passengers (k) such that:
Where (X \sim Binomial(n, p)). Using normal approximation:
\sigma = \sqrt{np(1-p)} = \sqrt{350 \cdot 0.05 \cdot 0.95}$$ To determine the maximum possible number of tickets that can be sold, we need: $$Z = \frac{k - \mu}{\sigma}$$ Setting this less than 1 for a 20% threshold provides the number of additional tickets that can be sold without exceeding seat limits.