Circles in general Show that the polar equation
r² - 2r ...

Question

Circles in general Show that the polar equation

r² - 2r r₀ cos(θ - θ₀) = R² - r₀²

describes a circle of radius R whose center has polar coordinates (r₀, θ₀)

Accepted Answer

Welcome back, everyone. Find the R client of the polar curve R of theta equals 5 cosine of theta for theta between 0 and pi divided by 2 inclusive. A 5 pi divided by 4 units. B 5 pi divided by 2 units. C 5 units, and D 5 square root of 2 divided by 2 units. For this problem we can recall the R client formula. L is equal to the integral from alpha to beta of square root of r quad plus the derivative of R with respect to theta squared the theta. We can immediately recognize that alpha is the lower limit of integration, which is 0. And beta is the upper limit of integration which is 5 divided by 2. Now we can find R squared, which is going to be. 5 cosine theta squared, we get 25 cosine squared of theta. And now we also want to find the derivative of R with respect to theta. We can label it as our prime. The derivative is going to be the derivative of 5 cosine theta. Now we can factor out 5, and the derivative of cosine is negative sign. So we get -5 sine. Of the. We also want a square that derivative, so our prime squared is going to be. -5 sine theta squad, which gives us 25 squared of theta. So now let's set up our integral because we have all of the parameters needed. We get L equals the integral from 0 up to pi divided by 2 of square root of R2, which is 25 cosine square of theta plus R squad. So we're adding 25 sin squared of data and we're integrating with respect to. The. What we can do is factor out 25 and take square root of 25. We got 5. Integral from 0 up to pi divided by 2. Let's write it more clearly to ensure that everything is visible, so we got 5. Integral from 0 up to. I divided by 2. Of square root of. Sine squared of theta plus cosine squared of theorem the theorem. And according to the Pythagorean identity, we know that sine squared plus cosine squared is one. Square root of 1 is still one so we can just replace our integran with one multiplied by the theta. We're simply the theta, right? Well done. So we now have a relatively simple integral. The integral of the theta is just theta, and because we're multiplying by 5, we get 5 theta evaluated from 0 up to pi divided by 2. And then we can apply the evaluation theorem we get 5 imperencies. I divided by 2 minus 0, which is. 55 divided by 2 units. Which corresponds to the answer choice b. Thank you for watching.

Circles in general Show that the polar equation
r² - 2r r₀ cos(θ - θ₀) = R² - r₀²
describes a circle of radius R whose center has polar coordinates (r₀, θ₀)

Key Concepts

Polar Coordinates and Conversion to Cartesian Coordinates

Equation of a Circle in Cartesian Coordinates

Use of the Law of Cosines in Polar Form

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