Question

A curve has the parametric equations.x = sin²t and y=Sin2t, 0a)Find the expression for dy/dx in terms of t.

b) Find the equation of the normal to the curve at t=π/6​

Answer 1

The solution involves finding dy/dx for the given parametric equations by differentiating both with respect to t, then finding the slope at a specific t value to determine the equation of the normal to the curve.

Step-by-step explanation:

A student asks for help with finding dy/dx for parametric equations and the equation of the normal to the curve at a specific parameter. To find dy/dx in terms of t, you differentiate both x = sin²t and y = Sin2t with respect to t and then divide the two derivatives.

1. First, find dx/dt = 2sin(t)cos(t) = sin(2t).
2. Then, find dy/dt = 2cos(2t).
3. Now, dy/dx = (dy/dt) / (dx/dt) = 2cos(2t) / sin(2t).

To find the equation of the normal to the curve at t = π/6, you first find the slope of the tangent (which is dy/dx) at this t, then use the negative reciprocal for the slope of the normal, and finally use the point-slope form to write the equation of the normal.