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The point P lies on the curve with equation $$x = (4y - ext{sin}(2y))^2$$ Given that P has $(x, y)$ coordinates \( \left(p, \frac{\pi}{2}\right) \text{ where } p \text{ is a constant,} (a) find the exact value of p - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 3
Question 5
The point P lies on the curve with equation $$x = (4y - ext{sin}(2y))^2$$ Given that P has $(x, y)$ coordinates \( \left(p, \frac{\pi}{2}\right) \text{ where } p ... show full transcript
Worked Solution & Example Answer:The point P lies on the curve with equation $$x = (4y - ext{sin}(2y))^2$$ Given that P has $(x, y)$ coordinates \( \left(p, \frac{\pi}{2}\right) \text{ where } p \text{ is a constant,} (a) find the exact value of p - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 3
Step 1
find the exact value of p.
Answer
To find the value of ( p ), we first substitute ( y = \frac{\pi}{2} ) into the curve's equation:
Calculating ( y = \frac{\pi}{2} ):
[ x = (4(\frac{\pi}{2}) - \text{sin}(\pi))^2 ] [ = (2\pi - 0)^2 ] [ = (2\pi)^2 = 4\pi^2 ]
Therefore, the value of ( p ) is ( 4\pi^2 ).
Step 2
Use calculus to find the coordinates of A.
Answer
To find the coordinates of A, we need to determine the tangent line at point P. First, we differentiate the equation of the curve:
Substituting ( y = \frac{\pi}{2} ) into the derivative:
- Compute ( \text{cos}(\pi) = -1) and substitute: [ \frac{dx}{dy} = 2(4(\frac{\pi}{2}) - 0)(4 - 2(-1)) = 2(2\pi)(4 + 2) = 12\pi ]
The slope of the tangent at point P is ( 12\pi ). Using point-slope form, the equation of the tangent line can be derived:
[ y - \frac{\pi}{2} = 12\pi (x - 4\pi^2) ]
Setting ( x = 0 ) to find where it intersects the y-axis:
[ y - \frac{\pi}{2} = 12\pi (0 - 4\pi^2) ] [ y = \frac{\pi}{2} - 48\pi^3 ]
Thus, the coordinates of point A are ( (0, \frac{\pi}{2} - 48\pi^3) ).