Which is the polar form of the parametric equations x = 3t and y = t^2?

Answer 1

To find the polar form of the parametric equations x = 3t and y = t^2, we first need to eliminate t and express x and y in terms of r and θ.

From the equation x = 3t, we get t = x/3. Substituting this into the equation y = t^2, we get y = (x/3)^2 = x^2/9.

To express x and y in terms of polar coordinates r and θ, we use the following relationships:

x = r cos(θ)
y = r sin(θ)

Substituting these equations into the expressions we found for x and y in terms of t, we get:

r cos(θ) = 3t
r^2 sin^2(θ) = t^2

Solving for t in the first equation, we get t = r cos(θ)/3. Substituting this into the second equation, we get:

r^2 sin^2(θ) = (r cos(θ)/3)^2

Simplifying this equation, we get:

r^2 sin^2(θ) = r^2 cos^2(θ)/9

Multiplying both sides by 9, we get:

9r^2 sin^2(θ) = r^2 cos^2(θ)

Dividing both sides by r^2, we get:

9sin^2(θ) = cos^2(θ)

Taking the square root of both sides, we get:

3 |sin(θ)| = cos(θ)

Now, we need to consider the sign of sin(θ), since it could be positive or negative. If sin(θ) is positive, then we have:

3 sin(θ) = cos(θ)

Dividing both sides by sin(θ), we get:

tan(θ) = 1/3

This is the polar form of the parametric equations x = 3t and y = t^2, when sin(θ) is positive.

If sin(θ) is negative, then we have:

3 |sin(θ)| = -cos(θ)

Dividing both sides by |sin(θ)| and replacing |sin(θ)| with -sin(θ), we get:

-3 = cot(θ)

This is the polar form of the parametric equations x = 3t and y = t^2, when sin(θ) is negative.

Therefore, the general polar form of the parametric equations x = 3t and y = t^2 is given by:

r = 3/(cos(θ) + 3sin(θ))

or

r = -3/(cos(θ) - 3sin(θ))

depending on the sign of sin(θ).