Which is the polar form of the parametric equations x=4 cos theta and y=4 sun theta?

Answer 1

In polar form, x = r cos(theta) and y = r sin(theta), where r is the distance from the origin and theta is the angle from the positive x-axis to a point (x,y) in standard position.

To find the polar form of x = 4 cos(theta) and y = 4 sin(theta), we can use the Pythagorean identity:

r^2 = x^2 + y^2

Substituting the given equations:

r^2 = (4 cos(theta))^2 + (4 sin(theta))^2

Simplifying:

r^2 = 16(cos^2(theta) + sin^2(theta))

r^2 = 16

Taking the square root:

r = ± 4

Since x and y are both positive, we can take r = 4.

Now, we can write the polar form as:

r cos(theta) + i r sin(theta) = 4 cos(theta) + 4i sin(theta)

Or, using the exponential form of complex numbers:

4e^(i theta)

Therefore, the polar form is:

r = 4

θ = arctan(y/x) = arctan(4 sin(theta)/4 cos(theta)) = arctan(tan(theta)) = theta

So, the polar form of x=4 cos theta and y=4 sin theta is:

r = 4, θ = theta

Or, equivalently:

4e^(i theta)

Answer 2

AAAaannndd the bot gets it wrong yet again!
The polar form is just r=4
the rectangular form is x^2 + y^2 = 16
No exponentials and complex numbers are involved.

Answer 3

You are correct. Apologies for the mistake in my previous response.

The polar form of the parametric equations x = 4 cos(theta) and y = 4 sin(theta) is just r = 4, which represents a circle centered at the origin with radius 4.

The rectangular form can be found by using the Pythagorean identity as:

x^2 + y^2 = (4 cos(theta))^2 + (4 sin(theta))^2

x^2 + y^2 = 16(cos^2(theta) + sin^2(theta))

x^2 + y^2 = 16

This represents a circle centered at the origin with radius 4, as expected. Once again, apologies for any confusion caused by my earlier response.

Answer 4

Thanks oobleck !

Answer 5

You're welcome! If you have any other questions, feel free to ask.