Question

Find the exact length of the curve (parametrics)?

Answer 1

In this section we will look at the arc length of the parametric curve given by,

x=f(t)y=g(t)α≤t≤βx=f(t)y=g(t)α≤t≤β

We will also be assuming that the curve is traced out exactly once as tt increases from aa to bb. We will also need to assume that the curve is traced out from left to right as tt increases. This is equivalent to saying,

dxdt≥0for α≤t≤βdxdt≥0for α≤t≤β

So, let’s start out the derivation by recalling the arc length formula as we first derived it in the arc length section of the Applications of Integrals chapter.

L=∫dsL=∫ds

where,

ds=√1+(dydx)2dxif y=f(x),a≤x≤bds=√1+(dxdy)2dyif x=h(y),c≤y≤dds=1+(dydx)2dxif y=f(x),a≤x≤bds=1+(dxdy)2dyif x=h(y),c≤y≤d

We will use the first dsds above because we have a nice formula for the derivative in terms of the parametric equations (see the Tangents with Parametric Equations section). To use this we’ll also need to know that,

dx=f′(t)dt=dxdtdtdx=f′(t)dt=dxdtdt

The arc length formula then becomes,

L=∫βα  ⎷1+⎛⎝dydtdxdt⎞⎠2dxdtdt=∫βα    ⎷1+(d

Example 1 Determine the length of the parametric curve given by the following parametric equations.x=3sin(t)y=3cos(t)0≤t≤2πx=3sin⁡(t)y=3cos⁡(t)0≤t≤2π

We know that this is a circle of radius 3 centered at the origin from our prior discussion about graphing parametric curves. We also know from this discussion that it will be traced out exactly once in this range.

So, we can use the formula we derived above. We’ll first need the following,

dxdt=3cos(t)dydt=−3sin(t)dxdt=3cos⁡(t)dydt=−3sin⁡(t)

The length is then,

L=∫2π0√9sin2(t)+9cos2(t)dt=∫2π03√sin2(t)+cos2(t)dt=3∫2π0dt=6π