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