Question

How do i solve this?
A curve is defined parametrically by x=3sint, y=cos 3t, 0 lesser than or equal to t lesser than or equal to 2pi Find the equation of the line tangent to the curve at the point defined at t= 2pi/9

Answer 1

What point do you get if you plug in \(t=\dfrac{2\pi}{9}\)? This will be the "point" in the point-slope formula.

As for the slope, that's given by the derivative \(\dfrac{dy}{dx}\) evaluated at said point. To find the derivative, you must first take the derivatives with respect to the parameter \(t\):
\[\frac{dy}{dx}=\frac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}\]
Then plug in the point you found earlier to find the slope.