a circle with a center of radius 5 has a chord placed horizontally 4 units above the center of the circle. find the length of the arc of the circle above the chord.
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Superimpose a coordinate system on the circle so that the center of the circle is the origin. Then, the equation of the circle is:
x^2 + y^2 = 5^2 ==> x^2 + y^2 = 25.
A chord placed 4 units above the circle intersects the circle when:
x^2 + 4^2 = 25 ==> x = ±3.
So, we will be integrating from -3 to 3.
By solving x^2 + y^2 = 25 for the positive solution for y, we have:
y = √(25 - x^2).
By differentiating, we have:
dy/dx = -x/√(25 - x^2).
So:
1 + (dy/dx)^2 = 1 + [-x/√(25 - x^2)]^2
= 1 + x^2/(25 - x^2)
= [(25 - x^2) + x^2]/(25 - x^2)
= 25/(25 - x^2)
==> √[1 + (dy/dx)^2] = 5/√(25 - x^2).
Therefore, the length of the arc above the chord is:
L = ∫ √[1 + (dy/dx)^2] dx (from x=a to b)
= ∫ 5/√(25 - x^2) dx (from x=-3 to 3)
= 5arcsin(x/5) (evaluated from x=-3 to 3)
= 10arcsin(3/5)
≈ 6.44.
I hope this helps!
x^2 + y^2 = 5^2 ==> x^2 + y^2 = 25.
A chord placed 4 units above the circle intersects the circle when:
x^2 + 4^2 = 25 ==> x = ±3.
So, we will be integrating from -3 to 3.
By solving x^2 + y^2 = 25 for the positive solution for y, we have:
y = √(25 - x^2).
By differentiating, we have:
dy/dx = -x/√(25 - x^2).
So:
1 + (dy/dx)^2 = 1 + [-x/√(25 - x^2)]^2
= 1 + x^2/(25 - x^2)
= [(25 - x^2) + x^2]/(25 - x^2)
= 25/(25 - x^2)
==> √[1 + (dy/dx)^2] = 5/√(25 - x^2).
Therefore, the length of the arc above the chord is:
L = ∫ √[1 + (dy/dx)^2] dx (from x=a to b)
= ∫ 5/√(25 - x^2) dx (from x=-3 to 3)
= 5arcsin(x/5) (evaluated from x=-3 to 3)
= 10arcsin(3/5)
≈ 6.44.
I hope this helps!