Question

Question #3e769

Answer 1

`rho=(-sintheta+root2(sin^2theta+80cos^2theta))/(2cos^2theta)`
for `0< theta < 2pi`

Explanation:

Starting from the equations of passage from cartesian (rectangular) to polar coordinates
`x=rhocostheta`, `y=rhosintheta` and `rho=root2(x^2+y^2)`

we can replace them inside the given equation and get

`rho^2cos^2theta+rhosintheta-20=0` this can be seen as a second degree equation in the unknown `rho` that can be solved like this

`rho=(-sintheta+-root2(sin^2theta+80cos^2theta))/(2cos^2theta)`
that describes our function for `0< theta < 2pi`.

As long as the sign in front of the root is concerned, we have to consider that, according to its defintion `rho` must be positive for any `0 < theta < 2pi`.
As a matter of fact the polar function describing the given function is the one with the plus sign linking the two terms at the numerator

`rho=(-sintheta+root2(sin^2theta+80cos^2theta))/(2cos^2theta)`
for `0< theta < 2pi`

Answer 2

Assuming that intended equation is `x^2+(y-4)^2=4^2`, `r =8 sin theta, theta in [0, pi]`. .

Explanation:

`x^2+(y-4)^2=4^2` represents the circle , with center C(0, 4) and

radius 4.

Let O be the pole and `P(r, theta)` be any point on the circle.

OC=CP=4 and, easily, #anglePOC=angleCOP-pi/2-theta.angle

OCP=2theta#.

It is immediate from the isosceles `triangle`,

`r=OP=(CP+CO)cos(pi/2-theta)=8 sin theta.`

However, the following method befits any circle, with given center

and radius.

`OP^2=OC^2+CP^2-2(OC)(CP)cos angle OCP`

So,

`r^2=4^2+4^2-2(4)(4)(2 cos^2theta-1))=64(1-cos^2theta)=64sin^2theta`

This is simply, `r = 8 sin theta, theta in [0, pi]`, for the whole circle.

Perhaps, some readers might not like this answer, because I have

not used `(x, y)=r(cos theta, sin theta)` at all. If you use this, it is

relatively a short method, for this problem. I agree. Yet, what I did

was for vector orientation

Answer 3

Assuming the equation was `x^2+(y-4)^4=16`

`r=8sintheta`

Explanation:

`x=rcostheta`
`y=rsintheta`

`r^2cos^2theta+(rsintheta-4)^2=16`
`r^2cos^2theta+r^2sin^2theta-8rsintheta+cancel(16)=cancel(16)`
`r^2(cos^2theta+sin^2theta)=8rsintheta`
`r^cancel(2)= 8cancel(r)sintheta`
`r=8sintheta`