Figure 2 shows a sketch of the curve C with parametric equations $x = 27 \sec t$, $y = 3 \tan t$, $0 \leq t \leq \frac{\pi}{3}$ (a) Find the gradient of the curve C at the point where $t = \frac{\pi}{6}$ - Edexcel - A-Level Maths Pure - Question 12 - 2013 - Paper 1

To find the Cartesian equation, we eliminate the parameter t:

1. From $x = 27 \sec t$, we can express $\sec t$ as $\frac{x}{27}$.
2. From $y = 3 \tan t$, we can use the identity $\tan^2 t + 1 = \sec^2 t$: $\tan t = \sqrt{\sec^2 t - 1} = \sqrt{\left(\frac{x}{27}\right)^2 - 1}.$
3. Thus, $y = 3 \sqrt{\left(\frac{x}{27}\right)^2 - 1}.$
4. Manipulating gives: $\frac{y^2}{9} = \frac{x^2}{729} - 1 \rightarrow x^2 = \left(\frac{y^2}{9} + 1\right)729.$
5. Solving leads to: $y = \left(x - 9\right)^{\frac{1}{3}}$.
6. Therefore, the values of a and b are:
   • $a = 9$
   • $b = 216.$

To determine the volume of the solid formed by rotating the region R around the x-axis, we use the formula: $V = \int_{a}^{b} \pi y^2 \, dx.$

1. We know that the limits are from $x = 27$ to $x = 125$ and replacing y gives: $V = \pi \int_{27}^{125} \left(3 \tan(t)\right)^2 \cdot \frac{dx}{dt} \, dt,$ where we express dx in terms of dt.
2. Substituting the values into integral:
   • After integration yields: $V = \frac{4236}{5}.$
3. Thus, the volume of the solid of revolution is: $\boxed{\frac{4236}{5}}.$