Find the area it encloses.
r = 5cos(θ)
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
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.
To find the area enclosed by the curve defined by the equation r = 5cos(θ), you can use the formula for finding the area of a polar curve.
The general formula for finding the area enclosed by a polar curve is given by:
A = (1/2)∫[a, b] (r(θ))^2 dθ
Where r(θ) is the polar function and a, b are the angles of rotation that enclose the desired region.
For the given equation r = 5cos(θ), we need to determine the angle values that enclose the desired region. To do this, we find the points of intersection of the curve with itself.
To find the points of intersection, we set the equation equal to itself and solve for θ:
5cos(θ) = 5cos(θ)
This equation is true for any value of θ. Therefore, the curve intersects itself at all angles.
Since the curve intersects itself at all angles, the desired region is the entire curve.
Now we can calculate the area using the formula:
A = (1/2)∫[0, 2π] (5cos(θ))^2 dθ
Simplifying the equation:
A = (1/2)∫[0, 2π] 25cos^2(θ) dθ
Applying the half-angle identity for cosine:
A = (1/2)∫[0, 2π] 25(1 + cos(2θ))/2 dθ
Simplifying further:
A = (25/4)∫[0, 2π] (1 + cos(2θ)) dθ
The integral of (1 + cos(2θ)) with respect to θ is:
A = (25/4) [θ + (1/2)sin(2θ)] |[0, 2π]
Evaluating the integral at the limits:
A = (25/4) [(2π + (1/2)sin(4π)) - (0 + (1/2)sin(0))]
Simplifying:
A = (25/4) [2π + 0 - 0 - 0]
A = (25/4) * 2π
A = (25/2)π
Therefore, the area enclosed by the curve r = 5cos(θ) is (25/2)π square units.