How to calculate the arc length of a vector function
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=\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}\ dt???
where ???L??? is the arc length of the vector function, ???[a,b]??? is the interval that defines the arc, and ???dx/dt???, ???dy/dt???, and ???dz/dt??? are the derivatives of the parametric equations of ???x???, ???y??? and ???z??? 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)=\langle{r(t)_1,r(t)_2,r(t)_3}\rangle??? or ???r(t)=r(t)_1\bold i+r(t)_2\bold j+r(t)_3\bold k???, the parametric equations are
???x=r(t)_1???
???y=r(t)_2???
???z=r(t)_3???
Once we have these parametric equations, we’ll take the derivative of each one to get ???dx/dt???, ???dy/dt???, and ???dz/dt???. Assuming we’re given ???[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 ???0\leq{t}\leq2???.
???r(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)}???
???y=\cos{(2t)}???
???z=2t???
Now we’ll take the derivative of each of these.
???\frac{dx}{dt}=2\cos{(2t)}???
???\frac{dy}{dt}=-2\sin{(2t)}???
???\frac{dz}{dt}=2???
To solve for arc length, we’ll need the parametric equations of the vector function.
Plugging the derivatives and the given interval ???0\leq{t}\leq2??? into the formula for arc length, we get
???L=\int^2_0\sqrt{\left[2\cos{(2t)}\right]^2+\left[-2\sin{(2t)}\right]^2+\left(2\right)^2}\ dt???
???L=\int^2_0\sqrt{4\cos^2{(2t)}+4\sin^2{(2t)}+4}\ dt???
???L=\int^2_0\sqrt{4\left[\cos^2{(2t)}+\sin^2{(2t)}\right]+4}\ dt???
Since ???\cos^2{x}+\sin^2{x}=1???, we can simplify the integral to
???L=\int^2_0\sqrt{4(1)+4}\ dt???
???L=\int^2_0\sqrt{8}\ dt???
???L=\int^2_0\sqrt{4\cdot2}\ dt???
???L=\int^2_02\sqrt{2}\ dt???
???L=2\sqrt{2}t\Big|_0^2???
Evaluating over the interval, we get
???L=2\sqrt{2}(2)-2\sqrt{2}(0)???
???L=4\sqrt{2}???
The arc length of the vector function over the interval ???0\leq{t}\leq2??? is ???L=4\sqrt{2}???.
