Video: Finding the Length of a Parametric Equation Curve

Video Transcript

Find the length of the curve with parametric equations 𝑥 equals 𝑒 to the power of 𝑡 minus 𝑡 and 𝑦 equals four 𝑒 to the power of 𝑡 over two, where 𝑡 is greater than or equal to zero and less than or equal to two.

We know that the formula we used to find the arc length of curves defined parametrically from values of 𝑡 greater than or equal to 𝛼 and less than or equal 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. Now, in this case, we’re interested in the length of the curve, where 𝑡 is greater than or equal to zero and less than or equal to 𝑡. So we’ll let 𝛼 to be equal to zero and 𝛽 be equal to two. Then, our parametric equations are 𝑥 equals 𝑒 to the power of 𝑡 minus 𝑡 and 𝑦 equals four 𝑒 to the power of 𝑡 over two.

It’s quite clear that we’re going to need to work out d𝑥 by d𝑡 and d𝑦 by d𝑡. And so, we firstly recall that the derivative of 𝑒 to the power of 𝑡 is 𝑒 to the power of 𝑡. The derivative of negative 𝑡 is one. So d𝑥 by d𝑡 is 𝑒 to the power of 𝑡 minus one. Now, we’re going to use the chain rule to differentiate 𝑦 with respect to 𝑡. We let 𝑢 be equal to 𝑡 over two. So that d𝑢 by d𝑡 is equal to one-half. Then, d𝑦 by d𝑡 is d𝑦 by d𝑢 times d𝑢 by d𝑡. Now, 𝑦 is equal to four 𝑒 to the power of 𝑢. So d𝑦 by d𝑡 is four 𝑒 to the power of 𝑢 times a half, which is two 𝑒 to the power of 𝑢. But of course, we want d𝑦 by d𝑡 in terms of 𝑡. So we replace 𝑢 with 𝑡 over two. And we find that d𝑦 by d𝑡 equals two 𝑒 to the power of 𝑡 over two.

Now, in fact, for our arc length formula, we need d𝑥 by d𝑡 squared and d𝑦 by d𝑡 squared. So we’re going to square each of our expressions. When we do, we find that 𝑒 to the power of 𝑡 minus one squared is 𝑒 to the power of two 𝑡 minus two 𝑒 to the power of 𝑡 plus one. And d𝑦 by d𝑡 squared is simply four 𝑒 to the power of 𝑡. Now, we substitute everything we know into our formula for the arc length. And we get the definite integral between zero and two of the square root of 𝑒 to the power of two 𝑡 minus two 𝑒 to the power of 𝑡 plus one plus four 𝑒 to the power of 𝑡 with respect to 𝑡.

We notice that negative two 𝑒 to the power of 𝑡 plus four 𝑒 to the power of 𝑡 is two 𝑒 to the power of 𝑡. And that’s great because we see we can factor 𝑒 to the power of two 𝑡 plus two 𝑒 to the power of 𝑡 plus one a little like we would with a quadratic. We get 𝑒 to the power of 𝑡 plus one times 𝑒 to the power of 𝑡 plus one or 𝑒 to the power of 𝑡 plus one squared. And of course, the square root of 𝑒 to the power of 𝑡 plus one squared is just 𝑒 to the power of 𝑡 plus one. When we integrate 𝑒 to the power of 𝑡, we get 𝑒 to the power of 𝑡. And the integral of one is 𝑡.

This means the arc length is 𝑒 squared plus two minus 𝑒 to the power of of zero plus zero. And, of course, 𝑒 to the power of zero is one. So this simplifies to 𝑒 to the power of two plus one. And the length of the curve with our parametric equations for values of 𝑡 from zero to two is 𝑒 to the power of two plus one units.