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How to Find the Arc Length of a Curve using Calculus Integrals (Example 1)
In arc length questions you’ll always be given an equation for a curve as well as the domain for which the curve begins at and ends at, usually the curve’s dependent variable is r(t) and t is the independent variable used.
You just need to memorize the “arc length formula” and plug in the values stated in your problem, which include the curvature function and the domain will be the integral boundaries.
Example problem: Find the length of the curve \[r(t) = \left\langle {t,3\cos (t),3\sin (t)} \right\rangle \] over the domain \[ – 5 \le t \le 5\].
Arc length is given by the formula (memorize this! or ask your teacher/professor if it’s given to you on a “cheat sheet”):
\[L = \int_{{t_1}}^{{t_2}} {\left| {r'(t)} \right|} dt\]
We can start by finding the derivative of r(t).
Remember, the derivative of a vector-valued function is just the derivative of each component.
This means you should review and memorize concepts from finding derivatives such as the chain rule, product rule, quotient rule, and knowing how to differentiate each kind of function such as trigonometry, logarithms, exponentials, polynomials, etc.
\[r'(t) = \left\langle {1, – 3\sin (t),3\cos (t)} \right\rangle \]
Substituting this into L, using the boundaries on t, and finding the norm (the “norm” are those flat bars surrounding r'(t) in the equation for arc length two equations above, the norm is just squaring and adding each component and then taking the square root of that, see below), we will obtain:
\[L = \int_{ – 5}^5 {\sqrt {{{(1)}^2} + {{( – 3\sin (t))}^2} + {{(3\cos (t))}^2}} } dt\]
\[L = \int_{ – 5}^5 {\sqrt {1 + 9{{\sin }^2}(t) + 9{{\cos }^2}(t)} } dt\]
And since we have a bunch of trig, we can use the well-known identity \[{\sin ^2}(t) + {\cos ^2}(t) = 1\]. Keep simplifying your result.
\[L = \int_{ – 5}^5 {\sqrt {1 + 9} } dt\]
\[L = \int_{ – 5}^5 {\sqrt {10} } dt\]
\[L = \sqrt {10} (5 – ( – 5))\]
\[L = 10\sqrt {10} \]
How to plot parametric curves in MATLAB
Are you curious what the curve in the question looks like? It’s from here onward that your mathematics courses are probably starting to look more and more abstract. BUT, it doesn’t have to be that way!
If you wanted to plot it, you’d need to make a 3-D plot, and substitute in values of t from -5 to 5. Each time you substitute t into the three equations, you will get three results, each result is a dimension, either the x, y, or z dimension.
It’s usually painstaking to plot these by hand because it takes a significant amount of time to do computations to gather the points. But with computational software such as MATLAB, you can easily make a script to sketch the curve very easily.
Use the following MATLAB script. You can use it for other parametric curves as well, just replace the three equations with what you were given, and also replace the domain in the last line of code.
syms t
xt = t;
yt = 3*cos(t);
zt = 3*sin(t);
fplot3(xt,yt,zt,[-5 5])
xlabel('x')
ylabel('y')
zlabel('z')You will acquire the following plot:
So that’s what we found the arc length of! Play around with our script and see what kinds of crazy curves you can make!
Click here to see another arc length example problem involving re-parameterization of a curve.
Try another topic, such as finding the potential and work of a conservative vector field.