Find the area it encloses.
r = 5cos(θ)
3 answers
What are the limits of theta? How do you deal with negative r from theta = 90 to 270 degrees?
Isn't that just a circle with diameter from (0,0) to (5,0) ?
area would be π(5/2)^2 = 25π/4
area would be π(5/2)^2 = 25π/4
We know from converting from polar to rectangular ...
r √(x^2 + y^2) , x = rcosØ , and y = rsinØ
so
rcosØ = x
cosØ = x/r
5cosØ = 5x/r
and
√(x^2 + y^2) = 5x/(√(x^2 + y^2))
cross-multiply
x^2 + y^2 = 5x
which is a circle as described in previous reply.
r √(x^2 + y^2) , x = rcosØ , and y = rsinØ
so
rcosØ = x
cosØ = x/r
5cosØ = 5x/r
and
√(x^2 + y^2) = 5x/(√(x^2 + y^2))
cross-multiply
x^2 + y^2 = 5x
which is a circle as described in previous reply.
1 answer