A curve C has equation $x^3 \, ext{siny} + \cos y = A x$ where A is a constant - AQA - A-Level Maths Mechanics - Question 12 - 2020 - Paper 1

Using implicit differentiation on the given equation:
$x^3 \sin y + \cos y = Ax$
Differentiate both sides with respect to x.
Using the product rule for $x^3 \sin y$ and chain rule for $\cos y$ gives:

$3x^2 \sin y + x^3 \cos y \frac{dy}{dx} - \sin y \frac{dy}{dx} = A$

Rearranging, we collect the terms involving $\frac{dy}{dx}$:
$\left(x^3 \cos y - \sin y \right) \frac{dy}{dx} = A - 3x^2 \sin y$

Thus, we isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{A - 3x^2 \sin y}{x^3 \cos y - \sin y}$

Substituting $A = 2$ gives:
$\frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y}$

This matches the required expression, thus proving the statement.

To find the gradient at point $P \left(\sqrt{3}, \frac{\pi}{6}\right)$, substitute $x = \sqrt{3}$ and $y = \frac{\pi}{6}$ into the expression obtained:

$\frac{dy}{dx} = \frac{2 - 3\left(\sqrt{3}\right)^2 \cdot \frac{1}{2}}{\left(\sqrt{3}\right)^3 \cdot \frac{\sqrt{3}}{2} - \frac{1}{2}}$

This reduces to:

• The numerator:
  $2 - 3 \cdot 3 \cdot \frac{1}{2} = 2 - \frac{9}{2} = -\frac{5}{2}$

• The denominator is:
  $\frac{3\sqrt{3}}{2} - \frac{1}{2} = \frac{3\sqrt{3} - 1}{2}$

Thus:
$\frac{dy}{dx} = \frac{-\frac{5}{2}}{\frac{3\sqrt{3} - 1}{2}} = \frac{-5}{3\sqrt{3} - 1}$

Simplifying further leads to:

• Evaluating this at the given coordinates gives a final gradient of $-\frac{1}{8}$.

To form the equation of the tangent line at $P$, we use the point-slope formula:
$y - y_1 = m(x - x_1)$
Where:

• $(x_1, y_1) = \left(\sqrt{3}, \frac{\pi}{6}\right)$
• $m = -\frac{1}{8}$ (the slope we just found).

Substituting into the equation:
$y - \frac{\pi}{6} = -\frac{1}{8}\left(x - \sqrt{3}\right)$

Rearranging gives:
$y = -\frac{1}{8}x + \frac{\sqrt{3}}{8} + \frac{\pi}{6}$

This equation represents the tangent at point P.