Photo AI

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 icon

Question 12

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.png

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

96%

114 rated

Answer

To find the value of A, we start by substituting the coordinates of point P (\sqrt{3}, \frac{\pi}{6}) into the equation of the curve.

  1. Substitute x = \sqrt{3} and y = \frac{\pi}{6} into the equation: x3siny+cosy=Axx^3 \sin y + \cos y = Ax

    This gives:

    (3)3sin(π6)+cos(π6)=A3\left(\sqrt{3}\right)^3 \sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right) = A \sqrt{3}

  2. Calculate the sine and cosine values:

    • (\sin\left(\frac{\pi}{6}\right) = \frac{1}{2})
    • (\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2})
  3. Substitute these values back into the equation: (3)312+32=A3\left(\sqrt{3}\right)^3 \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} = A \sqrt{3}

  4. Calculate the left-hand side: 332+32=A3\frac{3\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = A \sqrt{3} 432=A3\frac{4\sqrt{3}}{2} = A \sqrt{3}

  5. Simplifying gives: 23=A32\sqrt{3} = A \sqrt{3}

  6. Dividing both sides by (\sqrt{3}) leads to: A=2A = 2.

Step 2

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

99%

104 rated

Answer

To show the required differentiation, we start with the equation:

x3siny+cosy=2xx^3 \sin y + \cos y = 2x

  1. Differentiate both sides with respect to x using implicit differentiation:

    • For the left side: (3x^2 \sin y + x^3 \cos y \frac{dy}{dx} - \sin y \frac{dy}{dx}
    • For the right side: 2
  2. This gives: 3x2siny+(x3cosysiny)dydx=23x^2 \sin y + (x^3 \cos y - \sin y) \frac{dy}{dx} = 2

  3. Rearranging for (\frac{dy}{dx}) we have: (x3cosysiny)dydx=23x2siny(x^3 \cos y - \sin y) \frac{dy}{dx} = 2 - 3x^2 \sin y

  4. Isolating (\frac{dy}{dx}): dydx=23x2sinyx3cosysiny\frac{dy}{dx} = \frac{2 - 3x^2 \sin y}{x^3 \cos y - \sin y}

  5. This confirms the required expression.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;