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.
