# The point P lies on a curve with equation: x=(4y-sin2y)^2. (i) Given P has coordinates (x, pi/2) find x. (ii) The tangent to the curve at P cuts the y-axis at the point A. Use calculus to find the coordinates of the point A.

To find the x coordinate of point P, we simply substitute in the value of y at P into the equation of the curve and solve for x = 4pi^2. (ii) To start, we can differentiate x with respect to y, by using the chain rule. In other words, this is finding dx/dy. (Hint: This is the reciprocal of the gradient.) To do this, we multiply everything in the bracket by the indices value, 2. Then subtract one from the indices, to leave the whole bracket to be to the power of 1. Now we multiply all that we have left by the derivative of whats inside the bracket. All of this should come out as dx/dy = 2(4y-sin2y)(4-2cos2y). Now we can substitute the y coordinate of P into the equation to get dx/dy = 24pi. This is the reciprocal of the gradient. Therefore, the gradient of the tangent to the curve at P is 1/24pi. Now we have the gradient of the line, and a point it runs through. Therefore, we can use the standard equation to find the equation of a line: y-y1=m(x-x1) Substituting all your values in should give you y = x/24pi + 2pi/3. To find the y intercept of this line, you can set x to zero, such that y = 2pi/3.

Answered by Tobias F. Maths tutor

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