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2.5 Arc length

Let f⁢(x) be a function. You’ve seen the formula (in MATH102) for the length along the curve y=f⁢(x) between x=a and x=b:

∫ab1+(d⁢yd⁢x)2⁢𝑑x

But as we saw above, sometimes we are interested in parametrized curves γ:ℝ→ℝ2. In this section we will derive a formula for the length of an arc along a parametrized curve.

Recall that the derivative of a function x:I→ℝ at a number t∈I⊆ℝ, denoted x′⁢(t), is defined as

x′⁢(t)=limδ⁢t→0⁡x⁢(t+δ⁢t)-x⁢(t)δ⁢t

(if this limit exists).

Suppose γ⁢(t)=(x⁢(t),y⁢(t)). What is the length of the arc from (x⁢(t),y⁢(t)) to (x⁢(t+δ⁢t),y⁢(t+δ⁢t))? It’s the length of the vector (x⁢(t+δ⁢t)-x⁢(t),y⁢(t+δ⁢t)-y⁢(t)).

But x⁢(t+δ⁢t)-x⁢(t)=x⁢(t+δ⁢t)-x⁢(t)δ⁢t⋅δ⁢t≈x′⁢(t)⁢δ⁢t. Similarly y⁢(t+δ⁢t)-y⁢(t)≈y′⁢(t)⁢δ⁢t.

Thus the length of the arc from (x⁢(t),y⁢(t)) to (x⁢(t+δ⁢t),y⁢(t+δ⁢t)) is approximately

|(x′⁢(t)⁢δ⁢t,y′⁢(t)⁢δ⁢t)|=|(x′⁢(t),y′⁢(t))|⁢δ⁢t=|γ′⁢(t)|⁢δ⁢t

Dividing the interval [a,b] up into segments of length δ⁢t, and taking the limit as δ⁢t tends to zero, we obtain the following formula:

L=∫t0t1|γ′⁢(t)|⁢𝑑t

Example 2.19

Determine the length of the parametrized curve γ⁢(t)=(x⁢(t),y⁢(t))=(2⁢cos⁡t-t,3⁢sin⁡t) between t=0 and t=T.

First of all, we calculate γ′⁢(t)=(-2⁢sin⁡t-1,3⁢cos⁡t). Now we try to calculate

|γ′⁢(t)|2=(-2⁢sin⁡t-1)2+(3⁢cos⁡t)2=                             

The situation seems hopeless: how could we possibly integrate the square-root of such a function? But the first thing we should do is eliminate the cos2⁡t term. Substituting in cos2⁡t=1-sin2⁡t, we obtain

|γ′⁢(t)|2=4⁢sin2⁡t+4⁢sin⁡t+1+3⁢(1-sin2⁡t)=               

Since sin⁡t≥-1 for any t, the positive root of (sin⁡t+2)2 is sin⁡t+2. Thus the length of the curve is

∫0T(sin⁡t+2)⁢𝑑t=