in the equation (x-4)^2+y^2=16 the letters x and y represent rectangular coordinates. Write the equivalent equation using polar coordinates. solve for r
from your previous question, recall
sinØ = y/r --> y = rsinØ
cosØ = x/r --> x = rcosØ
(x-4)^2+y^2=16
(rcosØ -4)^2 + r^2sinØ = 16
r^2 cos^2 - 8rcosØ + 16 + r^2sin^2 Ø = 16
r^2 (cos^2 Ø + sin^2 Ø) - rcosØ + 16 = 16
r^2 -rcosØ = 0
r - cosØ = 0
check:
http://www.wolframalpha.com/input/?i=polar+plot+r+%3D+cos%C3%98
http://www.wolframalpha.com/input/?i=plot+(x-4)%5E2%2By%5E2%3D16
Clearly the graph is a circle of radius 4 with center at (4,0).
You know that r = cosθ is a circle with radius 1/2 and center at (0,1/2).
so, we have r = 8cosθ
Steve has the correct equation.
I don't know why my 8 was dropped in my 3rd last line, just careless.
To write the given equation in polar coordinates, we need to substitute the rectangular coordinates (x, y) with their equivalent in polar coordinates (r, θ). In polar coordinates, r represents the distance from the origin, and θ represents the angle with respect to the positive x-axis.
To convert from rectangular to polar coordinates, we use the following equations:
r = √(x^2 + y^2)
θ = arctan(y / x)
Applying these conversions, we can rewrite the equation (x-4)^2 + y^2 = 16 as:
(r cos θ - 4)^2 + (r sin θ)^2 = 16
Now, let's solve this equation for r.
Expanding the equation, we get:
(r^2 cos^2 θ - 8r cos θ + 16) + (r^2 sin^2 θ) = 16
Combining like terms:
r^2 cos^2 θ + r^2 sin^2 θ - 8r cos θ = 0
Using the trigonometric identity cos^2 θ + sin^2 θ = 1, we have:
r^2 - 8r cos θ = 0
Factoring out r, we obtain:
r(r - 8 cos θ) = 0
To solve for r, we have two possibilities:
1. r = 0
2. r - 8 cos θ = 0
If r = 0, it represents the origin in polar coordinates. So, let's solve the second equation for r:
r - 8 cos θ = 0
r = 8 cos θ
Therefore, the equivalent equation in polar coordinates is r = 8 cos θ.