Point P on the curve, x = 2tan( y+ π/12), has a y-coordinate of π/4. Find an equation for the normal to the curve at P.
Step 1 - Firstly, we need to find the x-coordinate of point P. This will allow us to find the normal equation required in the question. To find the x-coordinate, we plug in the value of y given in the question, π/4, in the equation also given in the question. This gives us an x-coordinate of 2√3. Now we know point P has coordinates (2√3, π/4).Step 2 - Now we must differentiate the equation in the question to give dx/dy. This will help us find the gradient of the normal at P. To differentiate this, we must know that the derivative of tan(x) is just sec2(x), and thus, the derivative of 2tan(y+π/12) is 2sec2(y+π/12).Step 3 - Now we know that dx/dy = 2sec2(y+π/12), but we want to find dy/dx, as this form always give us the gradient, not dx/dy. To find dy/dx, we simply invert dx/dy by 'flipping it'. This means dy/dx = 0.5cos2(y+π/12).Step 4 - To find the gradient at P, we plug in the y value given in the question into our dy/dx. This gives us a value of 1/8.Step 5 - As the question asks for the gradient of the normal, and not the tangent, we must use the negative inverse of 1/8, which is -8. Examiners expect you to know that gradient of a normal to any tangent is just the negative inverse, so make sure you read the question carefully!Step 6 - Using our equation for a straight line, y-y1 = m(x-x1), we plug in our y-coordinate (given in question), x-coordinate (calculated in step 1) and gradient (calculated in step 5). This gives us the answer of:
y= -8x +16√3 + π/4
