# Question Video: Finding the Length of a Parametric Curve Mathematics • Higher Education

Find the length of the curve with parametric equations 𝑥 = 3 cos 𝑡 − cos 3𝑡 and 𝑦 = 3 sin 𝑡 − sin 3𝑡, where 0 ≤ 𝑡 ≤ 𝜋.

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### Video Transcript

Find the length of the curve with parametric equations 𝑥 equals three cos 𝑡 minus cos three 𝑡 and 𝑦 equals three sin 𝑡 minus sin three 𝑡, where 𝑡 is greater than or equal to zero and less than or equal to 𝜋.

We recall that the formula we used to find the arc length of a curve defined parametrically for values of 𝑡 from 𝛼 to 𝛽 is the definite integral between 𝛼 and 𝛽 of the square root of d𝑥 by d𝑡 squared plus d𝑦 by d𝑡 squared with respect to 𝑡. In this case, 𝑥 is equal to three cos 𝑡 minus cos three 𝑡 and 𝑦 is equal to three sin 𝑡 minus sin three 𝑡. And we’re interested in the length of the curve between 𝑡 is greater than or equal to zero and less than or equal to 𝜋.

So we’ll let 𝛼 be equal to zero and 𝛽 be equal to 𝜋. We’re also going to need to work out d𝑥 by d𝑡 and d𝑦 by d𝑡. And so, since we’re working with trigonometric expressions, we recall the derivative of cos of 𝑎𝑡 and sin of 𝑎𝑡. They are negative 𝑎 sin of 𝑎𝑡 and 𝑎 cos 𝑎𝑡, respectively, for real constant values of 𝑎. This means d𝑥 by d𝑡 is negative three sin 𝑡 minus negative three sin three 𝑡. And of course, that becomes plus three sin three 𝑡. Similarly, d𝑦 by d𝑡 is three cos 𝑡 minus three cos three 𝑡.

Before we substitute into the formula, we’re actually going to square these and find their sum. Negative three sin 𝑡 plus three sin three 𝑡 all squared is nine sin squared 𝑡 minus 18 sin 𝑡 sin three 𝑡 plus nine sin square three 𝑡. Then, three cos 𝑡 minus three cos three 𝑡 all squared is nine cos squared 𝑡 minus 18 cos 𝑡 cos three 𝑡 plus nine cos squared three 𝑡. At this stage, we recall the trigonometric identity sin squared 𝑡 plus cos squared 𝑡 equals one. And we see that we have nine sin squared 𝑡 plus nine cos squared 𝑡. Well, that must be equal to nine. Similarly, we have nine sin squared three 𝑡 plus nine cos squared three 𝑡, which is also equal to nine. And we also have negative 18 times sin 𝑡 sin three 𝑡 plus cos 𝑡 cos three 𝑡. All I’ve done here is factored the negative 18 out.

Next, we’re going to use the trigonometric identity cos of 𝐴 minus 𝐵 is equal to cos 𝐴 cos 𝐵 plus sin 𝐴 sin 𝐵. And this means that sin 𝑡 sin three 𝑡 plus cos 𝑡 cos three 𝑡 must be equal to cos of three 𝑡 minus 𝑡, which is, of course, simply cos of two 𝑡. So this becomes 18 minus 18 cos of two 𝑡. And so, we find that the arc length is equal to the definite integral between zero and 𝜋 of the square root of 18 minus 18 cos of two 𝑡 d𝑡.

Let’s clear some space and evaluate this integral. Now, in fact, the integral of the square root of 18 minus 18 cos of two 𝑡 still isn’t particularly nice to calculate. And so, we go back to the fact that cos of two 𝑡 is equal to two cos squared 𝑡 minus one. We replace cos of two 𝑡 with this expression and then distribute the parentheses. And our integrand is now equal to the square root of 36 minus 36 cos squared 𝑡. We take out the common factor of 36 and then rearrange the identity sin squared 𝑡 plus cos squared 𝑡 equals one. So that one minus cos squared 𝑡 is equal to sin squared 𝑡. So our integrand is six times the square root of sin squared 𝑡, which is, of course, simply six sin 𝑡.

When we integrate six sin 𝑡, we get negative six cos 𝑡. So the arc length is equal to negative six cos 𝑡 evaluated between those limits. That’s negative six cos of 𝜋 minus negative six cos of zero, which is equal to 12. And so, we found that the arc length of the curve that we’re interested in is 12 units. As you might expect, not only does this process work for curves defined by trigonometric equations, but also those defined by exponential and logarithmic ones.