How to calculate the arc length of a vector function

 
 
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Formula for finding arc length of a vector function

To find the arc length of the vector function, we will need to use the formula

L=ab(dxdt)2+(dydt)2+(dzdt)2 dtL=\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\ dt

where LL is the arc length of the vector function, [a,b][a,b] is the interval that defines the arc, and dx/dtdx/dt, dy/dtdy/dt, and dz/dtdz/dt are the derivatives of the parametric equations of xx, yy and zz respectively.

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To solve for arc length, we’ll need the parametric equations of the vector function. Whether our vector function is given as r(t)=r(t)1,r(t)2,r(t)3r(t)=\langle{r(t)_1,r(t)_2,r(t)_3}\rangle or r(t)=r(t)1i+r(t)2j+r(t)3kr(t)=r(t)_1\bold i+r(t)_2\bold j+r(t)_3\bold k, the parametric equations are

x=r(t)1x=r(t)_1

y=r(t)2y=r(t)_2

z=r(t)3z=r(t)_3

Once we have these parametric equations, we’ll take the derivative of each one to get dx/dtdx/dt, dy/dtdy/dt, and dz/dtdz/dt. Assuming we’re given [a,b][a,b], we’ll have everything we need to use the formula for arc length.

 
 

How to calculate the arc length of a vector function over a particular interval


 
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Finding the arc length of the vector function

Example

Find the arc length of the vector function over the interval 0t20\leq{t}\leq2.

r(t)=sin(2t),cos(2t),2tr(t)=\left\langle{\sin{(2t)},\cos{(2t)},2t}\right\rangle

We’ll pull the parametric equations out of the vector function as

x=sin(2t)x=\sin{(2t)}

y=cos(2t)y=\cos{(2t)}

z=2tz=2t

Now we’ll take the derivative of each of these.

dxdt=2cos(2t)\frac{dx}{dt}=2\cos{(2t)}

dydt=2sin(2t)\frac{dy}{dt}=-2\sin{(2t)}

dzdt=2\frac{dz}{dt}=2

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To solve for arc length, we’ll need the parametric equations of the vector function.

Plugging the derivatives and the given interval 0t20\leq{t}\leq2 into the formula for arc length, we get

L=02[2cos(2t)]2+[2sin(2t)]2+(2)2 dtL=\int^2_0\sqrt{\left[2\cos{(2t)}\right]^2+\left[-2\sin{(2t)}\right]^2+\left(2\right)^2}\ dt

L=024cos2(2t)+4sin2(2t)+4 dtL=\int^2_0\sqrt{4\cos^2{(2t)}+4\sin^2{(2t)}+4}\ dt

L=024[cos2(2t)+sin2(2t)]+4 dtL=\int^2_0\sqrt{4\left[\cos^2{(2t)}+\sin^2{(2t)}\right]+4}\ dt

Since cos2x+sin2x=1\cos^2{x}+\sin^2{x}=1, we can simplify the integral to

L=024(1)+4 dtL=\int^2_0\sqrt{4(1)+4}\ dt

L=028 dtL=\int^2_0\sqrt{8}\ dt

L=0242 dtL=\int^2_0\sqrt{4\cdot2}\ dt

L=0222 dtL=\int^2_02\sqrt{2}\ dt

L=22t02L=2\sqrt{2}t\Big|_0^2

Evaluating over the interval, we get

L=22(2)22(0)L=2\sqrt{2}(2)-2\sqrt{2}(0)

L=42L=4\sqrt{2}

The arc length of the vector function over the interval 0t20\leq{t}\leq2 is L=42L=4\sqrt{2}

 
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