What is the Derivative of a Parametric Curve with Trigonometric Functions?

[Jenkz]

Homework Statement

If x(t) = cos2t ; y(t) = t - tant

Show that:
(dy/dx)^2 = 1- x / 4(1+x)^3

The Attempt at a Solution

dy/dx = dy/dt * dt/dx = (dy/dt) / (dx/dt)

dy/dt = 1- sect^2 ; dx/dt = -2sin2t

So (dy/dx)^2 = (1 - sect^2)^2 / (4 sin2t^2)

From here I'm not too sure on how to get to the answer.

 

[Mark44]

Homework Statement

If x(t) = cos2t ; y(t) = t - tant

Show that:
(dy/dx)^2 = 1- x / 4(1+x)^3

The Attempt at a Solution

dy/dx = dy/dt * dt/dx = (dy/dt) / (dx/dt)

dy/dt = 1- sect^2 ; dx/dt = -2sin2t

So (dy/dx)^2 = (1 - sect^2)^2 / (4 sin2t^2)

From here I'm not too sure on how to get to the answer.

The statement you are supposed to prove has (dy/dx)^2 as a function of x. In your version, you have (dy/dx)^2 as a function of t. Solve the equation x(t)= cos(2t) for t and replace in your equation for (dy/dx)^2.

I have a question about what you need to prove, which you have written as (dy/dx)^2 = 1- x / 4(1+x)^3.

If this is
$$(dy/dx)^2 = 1 - \frac{x}{4(1 + x)^3}$$
then you have written it correctly.

OTOH, if you meant
$$(dy/dx)^2 = \frac{1 - x}{4(1 + x)^3}$$
then you should have put parentheses around 1 - x to show that it's the numerator.

 

[Jenkz]

Sorry was the latter.

So I replace t with (arccosx)/2 ?

 

[Mark44]

Yes.