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 2 - 2010 - Paper 7

To find the Cartesian equation of the curve C, we can express y in terms of x by eliminating the parameter t:

1. From the parametric equation $x = \text{ln} t$, we can solve for t:

   $t = e^x$

2. Substitute $t$ into the equation for $y$:

   $y = t^2 - 2 = (e^x)^2 - 2 = e^{2x} - 2$

Thus, the Cartesian equation of C is:

$y = e^{2x} - 2$

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

1. The volume V is given by:

   $V = \pi \int_{a}^{b} [f(x)]^2 \, dx$

   Here, $f(x) = e^{2x} - 2$. The bounds are from $x = \ln 2$ to $x = \ln 4$.

2. Set up the integral:

   $V = \pi \int_{\ln 2}^{\ln 4} (e^{2x} - 2)^2 \, dx$

   Expanding the integrand:

   $(e^{2x} - 2)^2 = e^{4x} - 4e^{2x} + 4$

   So the volume becomes:

   $V = \pi \int_{\ln 2}^{\ln 4} (e^{4x} - 4e^{2x} + 4) \, dx$

3. Calculate the integral:

   $V = \pi \left[ \frac{e^{4x}}{4} - 2e^{2x} + 4x \right]_{\ln 2}^{\ln 4}$

4. Evaluate at the limits:

   For $x = \ln 4$:

   $= \frac{e^{4\ln 4}}{4} - 2e^{2\ln 4} + 4(\ln 4)$

   For $x = \ln 2$:

   $= \frac{e^{4\ln 2}}{4} - 2e^{2\ln 2} + 4(\ln 2)$

5. Compute the difference and simplify:

   After calculations and simplification, the exact volume is:

   $V = \pi(36 + 4\ln 2)$