The curve C has parametric equations $x = ext{ln} \, t, \, y = t^2 - 2, \, t > 0$ Find (a) an equation of the normal to C at the point where $t = 3$ - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 6

To find the cartesian equation of C, we eliminate the parameter $t$:

1. From $x = \text{ln} \, t$, we have: $t = e^{x}$

2. Substitute $t$ into the equation for $y$: $y = (e^{x})^2 - 2$

Thus, the cartesian equation of C is: $y = e^{2x} - 2$

To find the volume of the solid generated by rotating area $R$ around the x-axis, we use the disk method:

1. The volume $V$ is given by: $V = \pi \int_{a}^{b} [f(x)]^2 \, dx$ where $f(x) = e^{2x} - 2$.

2. The boundaries of integration are determined from the x-values at which $y = 0$: $e^{2x} - 2 = 0 \Rightarrow e^{2x} = 2 \Rightarrow 2x = \ln 2 \Rightarrow x = \frac{1}{2} \ln 2$ The exact values are at $x = \ln 2$ and $x = \ln 4$.

3. Set up the integral: $V = \pi \int_{\ln 2}^{\ln 4} (e^{2x} - 2)^2 \, dx$

4. Evaluating the integral:

   • Expand $(e^{2x} - 2)^2 = e^{4x} - 4e^{2x} + 4$ and integrate term by term: $= \pi \left[ \frac{e^{4x}}{4} - 2e^{2x} + 4x \right]_{\ln 2}^{\ln 4}$

5. Calculate the definite integral:

   • Substitute the limits and simplify to find the volume.