It will be a circle.
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To apply the transformation ( w = z + \frac{1}{z} ) to the circle defined by ( |z| = 2 ), we can express ( z ) in polar form as ( z = 2e^{i\theta} ), where ( \theta ) ranges from ( 0 ) to ( 2\pi ). Substituting this into the equation for ( w ), we get ( w = 2e^{i\theta} + \frac{1}{2e^{i\theta}} = 2e^{i\theta} + \frac{1}{2} e^{-i\theta} ). This simplifies to ( w = 2e^{i\theta} + \frac{1}{2}(\cos \theta - i \sin \theta) ), which describes a new curve in the ( w )-plane. The resulting curve can be analyzed further to understand its geometric properties.
To convert the curve (x^3 + y^3 = 3axy) into polar form, we use the substitutions (x = r\cos\theta) and (y = r\sin\theta). This gives us the polar equation (r^3(\cos^3\theta + \sin^3\theta) = 3ar^2\cos\theta\sin\theta), which simplifies to (r = \frac{3a\cos\theta\sin\theta}{\cos^3\theta + \sin^3\theta}). To find the area encircled by the loop, we can use the formula for the area in polar coordinates, (A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta). Evaluating this integral over one loop (typically from (0) to (\frac{\pi}{2}) for the symmetric shape) yields the area (A = \frac{3\pi a^2}{8}).
Yes. (Theta in radians, and then approximately, not exactly.)
theta = arcsin(0.0138) is the principal value.
I regret that the browser provided by answers.com is incapable of displaying even simple graphics.
To apply the transformation ( w = z + \frac{1}{z} ) to the circle defined by ( |z| = 2 ), we can express ( z ) in polar form as ( z = 2e^{i\theta} ), where ( \theta ) ranges from ( 0 ) to ( 2\pi ). Substituting this into the equation for ( w ), we get ( w = 2e^{i\theta} + \frac{1}{2e^{i\theta}} = 2e^{i\theta} + \frac{1}{2} e^{-i\theta} ). This simplifies to ( w = 2e^{i\theta} + \frac{1}{2}(\cos \theta - i \sin \theta) ), which describes a new curve in the ( w )-plane. The resulting curve can be analyzed further to understand its geometric properties.
To convert the curve (x^3 + y^3 = 3axy) into polar form, we use the substitutions (x = r\cos\theta) and (y = r\sin\theta). This gives us the polar equation (r^3(\cos^3\theta + \sin^3\theta) = 3ar^2\cos\theta\sin\theta), which simplifies to (r = \frac{3a\cos\theta\sin\theta}{\cos^3\theta + \sin^3\theta}). To find the area encircled by the loop, we can use the formula for the area in polar coordinates, (A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta). Evaluating this integral over one loop (typically from (0) to (\frac{\pi}{2}) for the symmetric shape) yields the area (A = \frac{3\pi a^2}{8}).
It's possible
Yes. (Theta in radians, and then approximately, not exactly.)
theta = arcsin(0.0138) is the principal value.
Yes, it is.
x2+y2=2y into polar coordinates When converting Cartesian coordinates to polar coordinates, three standard converstion factors must be memorized: r2=x2+y2 r*cos(theta)=x r*sin(theta)=y From these conversions, you can easily get the above Cartesian equation into polar coordinates: r2=2rsin(theta), which reduces down (by dividing out 1 r on both sides) to: r=2sin(theta)
Theta equals 0 or pi.
It also equals 13 12.
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
The answer depends on what theta is and the units of its measurement.