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
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
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.
To find the polar form of the parametric equations x = 3t and y = t^2, we need to express x and y in terms of r (the distance from the origin) and θ (the angle).
Start by substituting the given parametric equations into the equation for x and y:
x = 3t
y = t^2
Next, solve for t in terms of x:
t = x / 3
Now, substitute this value of t into the equation for y:
y = (x / 3)^2
y = x^2 / 9
We can rewrite this equation as:
x^2 + y^2 = (x^2 / 9) + y^2
x^2 + y^2 = (x^2 + 9y^2) / 9
9x^2 + 9y^2 = x^2 + 9y^2
Combine like terms:
8x^2 = 0
Thus, x^2 = 0.
Now, substitute this back into the equation for y:
y = x^2 / 9
y = 0 / 9
y = 0
By substituting x and y into the polar form, we can find the answer.
One option is A. r = 9 sec θ. In polar form, r is the distance from the origin, which in this case is 9. The sec θ term represents the ratio of the hypotenuse to the adjacent side in a right triangle formed by the x-axis and the vector from the origin to the point. Since x is equal to 0, we can say that sec θ is equal to 0. Hence, r = 9 sec θ results in r = 0, which is correct given that x and y are both equal to 0.
Therefore, the polar form of the parametric equations x = 3t and y = t^2 is A. r = 9 sec θ.
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.