Arc Length


The arc length, \( s \) along the graph of a parametric function \( (x,y) = \left(x(t),y(t)\right) \) from \( t=a \) to \( t=b \) is given by:

s=ab(dxdt)2+(dydt)2dt s = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right )^2 }\, dt

We can see why by examining a linear approximation of the arclength:

Δs2=Δx2+Δy2 \Delta s^2 = \Delta x^2 + \Delta y^2 by the Pythagorean Theorem, which means that for very small Δs \Delta s , dsΔs=Δx2+Δy2 ds \approx \Delta s = \sqrt{\Delta x^2 + \Delta y^2} .

Also, notice that in the the case where x(t)=t x(t)=t , we have (x,y)=(t,y(t)) (x,y) = (t,y(t)) , and the arc length becomes:

s=ab1+(dydt)2dt s = \int_a^b \sqrt{1+\left( \frac{dy}{dt} \right )^2 }\, dt

which we can use for non-parametrized functions.


Show that the arc length of a semi-circle of radius 1 is π \pi .

Let (x,y)=(cosθ,sinθ) (x,y)=(\cos \theta, \sin \theta) , which describes the unit-circle. Then the arc length around the the semi-circle is given by:

s=0π(sinθ)2+(cosθ)2dθ s = \int_0^\pi \sqrt{\left(\sin \theta\right)^2 + \left(- \cos \theta\right)^2}\, d\theta

But sin2θ+cos2θ=1 sin^2 \theta + \cos^2 \theta = 1 by the Pythagorean Identity, so s=0πdθ=π. s = \int_0^\pi\, d\theta = \pi. \, _\square

Application and Extensions

Gabriel's Horn

Consider the surface area and volume of of the solid formed by rotating the region bounded by the x x-axis, y=1 y=1 , and y=1x y = \frac{1}{x} , around the x x -axis. This solid is called Gabriel's Horn.


The volume of the solid of revolution can be found using the disk method:

V=π0adxx2=π(11a) V = \pi \int_0^a \frac{dx}{x^2} = \pi \left( 1 - \frac{1}{a} \right)

Now consider what happens as we allow a a to approach infinity:

limaπ(11a)=π \displaystyle \lim_{a \to \infty} \pi \left( 1 - \frac{1}{a} \right) = \pi

Surface Area

The surface area of a solid of revolution is given by the formula

S=2πabr(x)1+(dydx)2dx S = 2\pi \int_a^b r(x)\sqrt{1+\left( \frac{dy}{dx} \right )^2 } \, dx

In this case, since ddx(1x)=1x2 \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} that gives us:

S=2π1a1x1+(1x2)2dx=2π1a1x1+1x4dx \begin{aligned} S &= 2\pi \int_1^a \frac{1}{x} \sqrt{1+\left( -\frac{1}{x^2} \right )^2 } \, dx \\ &= 2\pi \int_1^a \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } \, dx \end{aligned}

This integral is hard to evaluate, but since in our interval 1+1x41 \sqrt{1+ \frac{1}{x^4}} \geq 1 and 1x>0 \frac{1}{x} > 0 :

2π1a1x1+1x4dx2π1a1xdx 2\pi \int_1^a \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } \, dx \geq 2\pi \int_1^a \frac{1}{x} \, dx

It follows that:

S2π1a1xdxS2πlna \begin{aligned} S &\geq 2\pi \int_1^a \frac{1}{x} \, dx \\ S &\geq 2\pi \ln a \end{aligned}

But lima2πlna= \displaystyle \lim_{a \to \infty} 2\pi \ln a = \infty , which means that Gabriel's Horn has infinite surface area, but a volume of only π \pi !

Note by Arron Kau
7 years, 4 months ago

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