An oil tanker at T is leaking oil which forms a circular oil slick - HSC - SSCE Mathematics Extension 1 - Question 7 - 2005 - Paper 1

Using the relationship established previously, we apply the derivative:

$\frac{d}{dt}(r) = \frac{\tan(\alpha)}{\frac{d\alpha}{dt}}$

Considering that (\alpha = 0.1), and using the chain rule on the radius, we set:

$\frac{dr}{dt} = \frac{1}{\sec^2(\alpha)} \cdot \frac{d\alpha}{dt}\cdot r$

After substituting the given values:

(\frac{dr}{dt} = \frac{1}{\sec^2(0.1)} \cdot 0.02 \cdot 4500)

The growth rate of the radius will be approximately given by the above formula.

To find the stationary points, we differentiate (f(x)):

$f'(x) = 3Ax^2 - A$

Setting the derivative to zero:

$3Ax^2 - A = 0 \Rightarrow A(3x^2 - 1) = 0.$

For non-zero A:

$3x^2 - 1 = 0 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm \sqrt{3} / 3.$

We need to show that (f(x)) has only one zero for the specified range of A. Analyzing (f(x)) we note it is continuous and differentiable. As our earlier findings suggest that all stationary points are defined by A, further evaluation implies A influences the intersection of the curve and the x-axis. The presence of one zero is deduced based on sign changes in calculated points.

Calculating values at endpoints gives us insights. Since (f(-1)) and (f(1)) yield the same sign across – suggesting no intersections within interval leads to a suitable conclusion that there is no zero within this domain.

Differentiating (g(θ) = 2cos(θ) + θ tan(θ) ) allows use to find (g'(θ)):

$g'(θ) = -2sin(θ) + \tan(θ) + θ\sec^2(θ)$

Analyzing this shows that an absence of stationary points arise as we follow arguments from part (ii).

Establishing that (g(θ)) possesses no stationary points implies a one-to-one function phenomenon. It essentially showcases monotonic behavior throughout the range, which leads to the conclusion that an inverse exists.