The problem consists of two parts: (a) An aircraft flies at different speeds and bearings for certain durations. We need to find the distance from the starting point and the bearing from the airstrip. (b) Differentiate the function $\frac{(1+x)(1+2x^2)}{x}$ with respect to $x$.

1. Problem Description

The problem consists of two parts:

(a) An aircraft flies at different speeds and bearings for certain durations. We need to find the distance from the starting point and the bearing from the airstrip.

(b) Differentiate the function $\frac{(1+x)(1+2x^2)}{x}$ with respect to $x$.

2. Solution Steps

(a) i. Distance from the starting point:

First leg:

Speed = 35 km/h

Time = 2 hours

Distance, $d_1 = \text{speed} \times \text{time} = 35 \times 2 = 70$ km

Second leg:

Speed = 22 km/h

Time = 2.5 hours

Distance, $d_2 = \text{speed} \times \text{time} = 22 \times 2.5 = 55$ km

We can use the cosine rule to find the distance $d$ from the starting point. The angle between the two legs is $100^{\circ} - 15^{\circ} = 85^{\circ}$.

$d^2 = d_1^2 + d_2^2 - 2 d_1 d_2 \cos(85^{\circ})$

$d^2 = 70^2 + 55^2 - 2 \times 70 \times 55 \times \cos(85^{\circ})$

$d^2 = 4900 + 3025 - 7700 \times 0.0871557427$

$d^2 = 7925 - 670.90$

$d^2 = 7254.10$

$d = \sqrt{7254.10} \approx 85.17 \text{ km}$

ii. Bearing from the airstrip:

Let $\theta$ be the angle between the first leg and the direct distance. Using the sine rule:

$\frac{\sin{\theta}}{55} = \frac{\sin{85^{\circ}}}{85.17}$

$\sin{\theta} = \frac{55 \times \sin{85^{\circ}}}{85.17} = \frac{55 \times 0.996194698}{85.17} = \frac{54.79070839}{85.17} \approx 0.6433$

$\theta = \arcsin(0.6433) \approx 40.02^{\circ}$

Therefore, the bearing from the airstrip is $15^{\circ} + 40.02^{\circ} = 55.02^{\circ} \approx 55^{\circ}$.

(b) Differentiate $\frac{(1+x)(1+2x^2)}{x}$ with respect to $x$.

Let $y = \frac{(1+x)(1+2x^2)}{x} = \frac{1+2x^2+x+2x^3}{x} = \frac{2x^3 + 2x^2 + x + 1}{x} = 2x^2 + 2x + 1 + \frac{1}{x} = 2x^2 + 2x + 1 + x^{-1}$

$\frac{dy}{dx} = \frac{d}{dx} (2x^2 + 2x + 1 + x^{-1}) = 4x + 2 - x^{-2} = 4x + 2 - \frac{1}{x^2}$

$\frac{dy}{dx} = 4x + 2 - \frac{1}{x^2} = \frac{4x^3 + 2x^2 - 1}{x^2}$

3. Final Answer

(a) i. Distance from the starting point: 85.17 km

ii. Bearing from the airstrip: 55 degrees

(b) $\frac{dy}{dx} = 4x + 2 - \frac{1}{x^2}$ or $\frac{4x^3 + 2x^2 - 1}{x^2}$