Find the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1.
1=5sin(theta)
theta=arcsin(1/5)
r'=5cos(theta)
I tried the integral between arcsin(1/5) and pi-arcsin(1/5) of (((5sin(theta))^2+(5cos(theta))^2))^1/2 which gives me 13.694 Webworks (a math homework website) says that my answer is wrong.
z=arcsin(1/5)=0.201358 is the angle about (0,0) where the circles intersect. So, the arc length on the outer circle is indeed
∫[z,π-z] 5 dθ = 13.6944
But now you have to add the arc on the inner circle to complete the perimeter
∫[z,π-z] 1 dθ = 2.73888
so, the whole perimeter is 16.4333
You can also do it without calculus, knowing that s=rθ, and using the same angles. But you have to note that the subtended angle in the larger circle is twice the angle about (0,0).
To find the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1, you need to use the definition of arc length in polar coordinates.
The formula for arc length in polar coordinates is given by:
L = ∫[a,b] √(r(θ)^2 + (dr(θ)/dθ)^2) dθ
Let's break down the steps to find the length of the perimeter:
Step 1: Find the intersection points of the two polar curves r=5sin(theta) and r=1.
To do this, set the two equations equal to each other:
5sin(theta) = 1
Solve for theta:
sin(theta) = 1/5
Taking the arcsin of both sides gives:
theta = arcsin(1/5)
Step 2: Calculate the derivative of r=5sin(theta) with respect to theta.
To find dr(θ)/dθ, differentiate r(θ) with respect to θ:
r'(θ) = 5cos(theta)
Step 3: Substitute the equations from Steps 1 and 2 into the arc length formula.
The arc length formula becomes:
L = ∫[arcsin(1/5), pi-arcsin(1/5)] √(r(θ)^2 + (dr(θ)/dθ)^2) dθ
Substituting the values from Steps 1 and 2 gives:
L = ∫[arcsin(1/5), pi-arcsin(1/5)] √((5sin(theta))^2 + (5cos(theta))^2) dθ
Step 4: Evaluate the integral.
Integrating the function inside the square root may be difficult to do analytically. However, you can use numerical methods or a computer algebra system (CAS) to find the value of the integral.
It seems that the value you calculated, 13.694, may be incorrect. Double-check your calculations or try using a different method to evaluate the integral.
Remember to include appropriate units (such as radians) in your final answer.
To find the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1, you need to apply the polar coordinate system.
Let's break down the problem step by step:
Step 1: Determine the intersection points of the two curves r=5sin(theta) and r=1.
To find the intersection points, set the two equations equal to each other:
5sin(theta) = 1
Solve for theta:
theta = arcsin(1/5)
The intersection points occur at theta = arcsin(1/5) and theta = pi - arcsin(1/5).
Step 2: Find the derivative of the equation r=5sin(theta) with respect to theta.
Take the derivative of r=5sin(theta) with respect to theta:
r' = 5cos(theta)
Step 3: Set up the integral for the length of the curve.
The length of a curve in polar coordinates can be calculated using the formula:
L = ∫[a, b] √(r^2 + (dr/dθ)^2) dθ
In this case, the limits of integration are from theta = arcsin(1/5) to theta = pi - arcsin(1/5).
So, the integral becomes:
L = ∫[arcsin(1/5), pi - arcsin(1/5)] √((5sin(theta))^2 + (5cos(theta))^2) dθ
Simplifying the equation inside the square root:
√(25sin^2(theta) + 25cos^2(theta)) = √25(sin^2(theta) + cos^2(theta)) = 5
L = 5 ∫[arcsin(1/5), pi - arcsin(1/5)] dθ
Step 4: Evaluate the integral.
Integrating with respect to theta:
L = 5 (theta) evaluated from theta = arcsin(1/5) to theta = pi - arcsin(1/5)
L = 5 [(pi - arcsin(1/5)) - arcsin(1/5)]
Step 5: Calculate the length of the perimeter.
Use a calculator to evaluate the expression:
L = 5 [(pi - arcsin(1/5)) - arcsin(1/5)] ≈ 15.96
Therefore, the length of the entire perimeter of the region inside r=5sin(theta) but outside r=1 is approximately 15.96 units.