MHB Convert r=7cos(theta) into a rectangular equation

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The polar equation r = 7cos(θ) can be converted into a rectangular equation by using the relationships between polar and rectangular coordinates. Starting with r = 6(cos(θ)), we substitute cos(θ) with x/r, leading to the equation r^2 = 6x. By substituting r^2 with x^2 + y^2, we arrive at x^2 + y^2 = 6x. Rearranging this into standard form, we complete the square to find that the equation represents a circle with a radius of 3 units, centered at the point (3, 0). This conversion illustrates the relationship between polar coordinates and their rectangular counterparts effectively.
Elissa89
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So we're learning to plot polar equations, which easy enough. But I got a question in the homework that wasn't covered in class:

Convert r=7cos(theta) into a rectangular equation. Use x and y values. I know how to convert when it's x=r*cos(theta) or y=r*sin(theta) and r and theta is given. But this is different and I don't know how to do it.
 
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Okay, we are given the polar equation:

$$r=6\cos(\theta)$$

Now, from:

$$x=r\cos(\theta)\implies \cos(\theta)=\frac{x}{r}$$

We may write:

$$r=6\left(\frac{x}{r}\right)$$

Multiply through by \(r\):

$$r^2=6x$$

We know:

$$r^2=x^2+y^2$$

Hence, we have:

$$x^2+y^2=6x$$

This would technically suffice, but I would prefer to continue and put into standard form:

$$x^2-6x+y^2=0$$

Complete the square on \(x\):

$$(x-6x+9)+y^2=9$$

$$(x-3)^2+y^2=3^2$$

Now it's easy to see we have a circle of radius 3 units centered at (3,0).
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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