What is the arc length of sqrt(2-x^2) – Calc II question?

Use the arc length formula to find the length L of the curve below, 0 ≤ x ≤ 1. Check your answer by noting that the curve is part of a circle.

y = sqrt(2 – x^2)

• A simple way is to imagine that the equation is an equatin of a circle

y = √(2 – x²)

y² = 2 – x²

x² + y² = 2

with a center at the origin and radius of √2

when x = 0, y = √2

when x = 1, y = 1

The angle formed with center as the angle between the two points is 45 degrees, or in other words, the arc length is 45/360 of the circumference of the circle

arc length

= 45/360 x 2 π r

= (1/8)(2π)(√2)

≈ 1.1107 units

• Calc 2 Arc Length

• (a.) ??a million+(dx/dy)^2 dy=arc length dx/dy=a million/2(36-y^2)^(-a million/2)(-2y)=-y/?36… Arc length=??a million+y^2/(36-y^2) dy from 0 to 3 (b.) ignore approximately my integration calculations for now for the reason which you haven’t any longer found out this way of integration yet. only be sure my answer including your graphing utility. ?6/?36-y^2 dy 6arcsin(y/6) evaluated at limits: 6arcsin(a million/2)-6arcsin(0)=6pi/6=pi

• Hi

The formula for arc-length is:

L = (a to b) ∫ [√(1 + (dy/dx)^2)] dx

a will be x = 0 and b will be x = 1. Since y = √(2 – x^2), we can find dy/dx to be:

y = √(2 – x^2)

dy/dx = (2 – x^2)’ * (1/2)(1/√(2 – x^2)) = -x/√(2 – x^2)

Now, we just plug that into the formula. You’ll have to use a trig-substitution to solve this one.

L = (0 to 1) ∫ [√(1 + (-x/√(2 – x^2))^2)] dx

= (0 to 1) ∫ [√(1 + ((x^2)/(2 – x^2))] dx

= (0 to 1) ∫ [√(((2 – x^2)/(2 – x^2)) + ((x^2)/(2 – x^2))] dx

= (0 to 1) ∫ [√((2 – x^2 + x^2)/(2 – x^2))] dx

= (0 to 1) ∫ [√(2/(2 – x^2))] dx

Let x = √(2)sin(t), then dx = √(2)cos(t) dt and the limits change to 0 to π/4

(0 to π/4) ∫ [√(2/(2 – (√(2)sin(t))^2))] √(2)cos(t) dt

= (0 to π/4) ∫ [√(2/(2 – 2sin(t)^2))] √(2)cos(t) dt

= (0 to π/4) ∫ [√(2/(2(1 – sin(t)^2)))] √(2)cos(t) dt

= (0 to π/4) ∫ [√(1/(1 – sin(t)^2))] √(2)cos(t) dt

= (0 to π/4) ∫ [√(1/cos(t)^2)] √(2)cos(t) dt

= (0 to π/4) ∫ [√(sec(t)^2)] √(2)cos(t) dt

= (0 to π/4) ∫ [sec(t)] √(2)cos(t) dt

= (0 to π/4) ∫ [√(2)sec(t)cos(t)] dt

= (0 to π/4) ∫ [√2] dt

= [t√2] (0 to π/4)

= (π/4)√(2) – (0)√(2)

= (π/4)√(2) – 0

= √(2)π/4

≈ 1.111

So the arc-length is about 1.111.

I hope this helps!

• (3 – x^2) dx

3x – x^3 / 3 ](0,1)

3 – 1/3

8/3