Polar to rectangular $r = 7$

Question

I don't follow this at all. I have $r = 7$ and the formula states $x = r \cos\theta$ $y = r\sin\theta$ but my book gives $x^2 + y^2 = 49$ this is impossible. It doens't follow the formula at all. Since theta isn't present it is zero. This gives. $x = 7$ and $y = 0$ why does the formula fail?

Answer 1

The formula is fine. We have $x^2 + y^2 = r^2$ by Pythagoras theorem. Or you can use the given formulas to get $$ x^2 + y^2 = (r\cos(\theta))^2 + (r\sin(\theta))^2 = r^2(\cos^2(\theta) + \sin^2(\theta)) = r^2 $$since $\cos^2(\theta) + \sin^2(\theta) = 1$. For $r = 7$, we have $x^2 + y^2 = r^2 = 7^2 = 49$.

As you can see from the above, $\theta$ disappears because of the trigonometric identity (not because it's zero) which basically is a consequence of Pythagoras theorem.

Answer 2

If

$$x=r\cos t\;,\;y=r\sin t\implies 49=x^2+y^2=r^2\left(\cos^2t+\sin^2t\right)=r^2\implies r=7$$

since always $\,r\ge 0\,$ ...

Answer 3

This follows from the general conversion formulas from rectangular coordinates to polar coordinates. If $x = r\cos \theta, y = r\sin \theta$, then:

$$x^{2} + y^{2} = (r\cos \theta)^{2} + (r\sin\theta)^{2} = r^{2}(\cos^{2}\theta + \sin^{2}\theta) = r^{2}$$

Hence, if $r = 7$, $x^{2} + y^{2} = r^{2}$ tells us that $x^{2} + y^{2} = 49$.

N.B.: you can verify for yourself that these two equations describe the same geometric object by thinking about what they represent. In polar coordinates, the equation $r = 7$ describes the set of points in the plane that are fixed distance of $7$ from the origin. In the cartesian plane, this describes a circle of radius $7$ centered at the origin, which is described by the equation $x^{2} + y^{2} = 49$.