The curve C has parametric equations $x = ext{ln} \, t, \quad y = t^2 - 2, \quad t > 0$ Find (a) an equation of the normal to C at the point where $t = 3$ - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 6

To find the Cartesian equation, we need to eliminate the parameter $t$:

From the first parametric equation: $t = e^x$

Substituting $t = e^x$ into the second parametric equation:

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

Thus, the Cartesian equation of the curve C is:

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

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

The volume is given by:

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

Where $f(x) = (e^x)^2 - 2$. The bounds are from $x = \text{ln} \, 2$ to $x = \text{ln} \, 4$. Thus, we have:

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

Expanding the integrand:

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

Evaluating this integral gives:

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

Calculating the limits:

• At $x = \text{ln} \, 4$: $\frac{e^{4 \text{ln} \, 4}}{4} - 2e^{2 \text{ln} \, 4} + 4 \text{ln} \, 4 = \frac{64}{4} - 8 + 4 \text{ln} \, 4 = 16 - 8 + 4 \text{ln} \, 4$
• At $x = \text{ln} \, 2$: $\frac{e^{4 \text{ln} \, 2}}{4} - 2e^{2 \text{ln} \, 2} + 4 \text{ln} \, 2 = \frac{16}{4} - 4 + 4 \text{ln} \, 2 = 4 - 4 + 4 \text{ln} \, 2$

Combining these results:

The volume is: $V = \pi \left[ (16 - 8 + 4 \text{ln} \, 4) - (4 - 4 + 4 \text{ln} \, 2) \right] = \pi (12 + 4 \text{ln} \, 4 - 4 \text{ln} \, 2)$

Simplifying gives the final volume expression.