Question

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

A.r=9 sec theta
B.r=9 tan theta sec theta
C.r=9 tan^2 theta
D.r=9 sec^2 theta

Answer 1

AAAaannndd the bot gets it wrong yet again!
You lost track of the 9, and otherwise just got lost ...
r sinθ = (r cosθ)^2/9
9r sinθ = r^2 cos^2θ
r = 9 secθ tanθ

so, B

Answer 2

We can start by eliminating the parameter t using the equation x = 3t or t = x/3. Substituting this into the equation y = t^2 gives y = (x/3)^2 = x^2/9.

Now, we can convert the Cartesian equation y = x^2/9 into polar form. Using the identity x = r cos(theta) and y = r sin(theta), we get:

r sin(theta) = (r cos(theta))^2/9

Multiplying both sides by 9 and simplifying, we get:

r^2 sin(theta) = r cos^2(theta)

Dividing both sides by cos^2(theta), we get:

r^2 tan^2(theta) + r^2 = r sec^2(theta)

Simplifying further, we get:

r^2 (tan^2(theta) + 1) = r^2 sec^2(theta)

r^2 = r^2 sec^2(theta) / (tan^2(theta) + 1)

r^2 = (r^2 sec^2(theta)) / (sec^2(theta) - 1)

Taking the square root of both sides, and using the fact that r is always positive, we get:

r = sqrt(r^2 sec^2(theta) / (sec^2(theta) - 1))

Simplifying further, we get:

r = sqrt[(r^2 sec^2(theta)) / (sec^2(theta) - 1)] * (sec(theta) / sec(theta))

r = sqrt[(r^2) / (1 - sin^2(theta))] * sec(theta)

r = r sec(theta) / cos(theta)

r = 3sec(theta)

Therefore, the polar form of the parametric equations x = 3t and y = t^2 is r = 3sec(theta), which is option A.

Answer 3

I apologize for the mistake in my previous response. You are correct that I lost track of the 9 in the equation. The correct way to solve the problem is as follows:

x = 3t and y = t^2 imply t = x/3. Substituting this into y = t^2, we get y = x^2/9.

To convert this equation into polar form, we can use the trigonometric identities:

x = r cos(theta)

y = r sin(theta)

Therefore, we have r sin(theta) = (r cos(theta))^2/9, which simplifies to 9 r sin(theta) = r^2 cos^2(theta). Solving for r, we get:

r = 9 sec(theta) tan(theta)

Thus, the polar form of the parametric equations x = 3t and y = t^2 is r = 9 sec(theta) tan(theta), which is option B.

Thank you for bringing this to my attention.