Figure 1 shows a sketch of the curve C which has equation y = e^{rac{3}{3}} ext{sin } 3x, ext{ where } -rac{ ext{π}}{3} ext{ ≤ } x ext{ ≤ } rac{ ext{π}}{3} (a) Find the x coordinate of the turning point P on C, for which x > 0 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 5

To find the equation of the normal to the curve at x = 0:

1. First, compute the y value at x = 0:

   $y(0) = e^{3 \cdot 0}\text{sin}(3 \cdot 0) = 0$

   So the point is (0, 0).

2. Next, compute the derivative at this point:

   $\frac{dy}{dx}. \text{ at } x = 0 = 3e^{0}\cos(0) = 3$

3. The slope of the normal line is the negative reciprocal of the tangent slope:

   $m_{normal} = -\frac{1}{3}$

4. Using the point-slope form of the line, the equation of the normal line becomes:

   $y - 0 = -\frac{1}{3}(x - 0)\Rightarrow y = -\frac{1}{3}x$

Thus, the equation of the normal to C at the point where x = 0 is:

$y = -\frac{1}{3}x$