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A curve C has equation $$x^3 \sin y + \cos y = Ax$$ where A is a constant - AQA - A-Level Maths Pure - Question 12 - 2020 - Paper 1
Question 12
A curve C has equation $$x^3 \sin y + \cos y = Ax$$ where A is a constant. C passes through the point P(\sqrt{3}, \frac{\pi}{6}) 12 (a) Show that A = 2 12 (b) ... show full transcript
Worked Solution & Example Answer:A curve C has equation $$x^3 \sin y + \cos y = Ax$$ where A is a constant - AQA - A-Level Maths Pure - Question 12 - 2020 - Paper 1
Step 1
Show that A = 2
Answer
To show that A = 2, we start by substituting the values of x and y from the point P(\sqrt{3}, \frac{\pi}{6}) into the equation:
-
Substitute:
Let x = \sqrt{3} Let y = \frac{\pi}{6} -
Compute (\sin \frac{\pi}{6}) and (\cos \frac{\pi}{6}):
(\sin \frac{\pi}{6} = \frac{1}{2}) and (\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}) -
Substitute these values into the equation:
[ \left(\sqrt{3}\right)^3 \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) = A\left(\sqrt{3}\right) ] This simplifies to:
[ \frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = A\sqrt{3} ]
[ \frac{4\sqrt{3}}{2} = A\sqrt{3} ]
[ 2\sqrt{3} = A\sqrt{3} ] -
Divide both sides by (\sqrt{3}):
[ A = 2 ]
Thus, we have shown that A = 2.
Step 2
Show that \(\frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y}\)
Answer
To find (\frac{dy}{dx}) using implicit differentiation, we start with the equation:
(x^3 \sin y + \cos y = Ax)
Differentiate both sides with respect to x:
[ 3x^2 \sin y + x^3 \cos y \frac{dy}{dx} - \sin y \frac{dy}{dx} = A ]
Rearranging gives:
[ (x^3 \cos y - \sin y) \frac{dy}{dx} = A - 3x^2 \sin y ]
Isolating (\frac{dy}{dx}) yields:
[ \frac{dy}{dx} = \frac{A - 3x^2 \sin y}{x^3 \cos y - \sin y} ]
Substituting in A = 2, we have:
[ \frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y} ]
Thus, the expression is verified as (\frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y}).