The curve C has equation $y = 4x + 3x^{2} - 2x^{3}, \; x > 0.$ (a) Find an expression for $\frac{dy}{dx}$ - Edexcel - A-Level Maths Pure - Question 10 - 2007 - Paper 2

To check if the point P (4, 8) lies on curve C, substitute $x = 4$ into the equation:

$y = 4(4) + 3(4)^{2} - 2(4)^{3}$

Calculating:

• First, $4(4) = 16$.
• Next, $3(16) = 48$.
• Finally, $2(64) = 128$.

Hence:

$$y = 16 + 48 - 128 = 64 - 128 = -64\qquad$\textbf{(incorrect)}$.

Therefore, the point P does not lie on curve C.

To find the equation of the normal at point P, we will first determine the slope of the tangent by evaluating $\frac{dy}{dx}$ at $x = 4$:

\n Substituting $x = 4$: $\frac{dy}{dx} = 4 + 6(4) - 6(4)^{2} = 4 + 24 - 96 = -68.\n$

The slope of the normal is the negative reciprocal: $m_{normal} = -\frac{1}{-68} = \frac{1}{68}.$

Using point-slope form to find the equation of the normal: $y - 8 = \frac{1}{68}(x - 4)$

Rearranging gives: $y = \frac{1}{68}x - \frac{4}{68} + 8$ Simplifying yields: $3y = x + 20$.

To find length PQ, we determine the coordinates of Q (intersection with x-axis). Set y = 0 in the normal equation:

$0 = \frac{1}{68}x - \frac{4}{68} + 8$

Solving for x:

a) Multiply through by 68: $0 = x - 4 + 544$

b) Therefore, $x = -540$. Hence, point Q is at (-540, 0).

Now we calculate length PQ:

Using the distance formula: $PQ = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$ $PQ = \sqrt{((-540) - 4)^{2} + (0 - 8)^{2}}$

Calculating gives: $PQ = \sqrt{(-544)^{2} + (-8)^{2}} = \sqrt{295936 + 64} = \sqrt{296000} = 40\sqrt{74}.$

Thus, the length PQ is $40\sqrt{74}$.