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.
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???
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}???.