Question Video: Expressing the Length of a Parametric Equation Curve as an Integral Mathematics

Video Transcript

Express the length of the curve with parametric equations 𝑥 is equal to 𝑡 plus 𝑒 to the power of negative 𝑡 and 𝑦 is equal to 𝑡 minus 𝑒 to the power of negative 𝑡, where 𝑡 is greater than or equal to zero and 𝑡 is less than or equal to two, as an integral.

We’re given a curve defined by a pair of parametric equations. We need to find an expression for the length of this curve as an integral. And we actually know how to find the length of a curve defined by a pair of parametric equations. We know if a curve is defined by a pair of parametric equations, 𝑥 is equal to some function 𝑓 of 𝑡 and 𝑦 is equal to some function 𝑔 of 𝑡. Then the length of this curve where 𝑡 is greater than or equal to 𝛼 and 𝑡 is less than or equal to 𝛽 is given by 𝑙 is equal to the integral from 𝛼 to 𝛽 of the square root of 𝑓 prime of 𝑡 squared plus 𝑔 prime of 𝑡 squared with respect to 𝑡. Provided both of our derivative functions, 𝑓 prime and 𝑔 prime, are continuous on the closed interval from 𝛼 to 𝛽.

From the question, we have our function 𝑓 of 𝑡 is 𝑡 plus 𝑒 to the power of negative 𝑡 and our function 𝑔 of 𝑡 is 𝑡 minus 𝑒 to the power of negative 𝑡. And we can also see that our value of 𝛼 is zero and our value of 𝛽 is two. The first thing we’re going to want to do is check that our derivative functions, 𝑓 prime and 𝑔 prime of 𝑡, are continuous on the closed interval from zero to two. To do this, we’ll find expressions for 𝑓 prime of 𝑡 and 𝑔 prime of 𝑡. Let’s start with 𝑓 prime of 𝑡. That’s the derivative of 𝑡 plus 𝑒 to the power of negative 𝑡 with respect to 𝑡. And we can differentiate this term by term.

First, by using the power rule for differentiation, we know the derivative of 𝑡 with respect to 𝑡 is equal to one. Next, to differentiate 𝑒 to the power of negative 𝑡, we want to multiply by the coefficient of 𝑡 in our exponent. This gives us negative 𝑒 to the power of negative 𝑡. We’ll do the same to find an expression for 𝑔 prime of 𝑡. That’s the derivative of 𝑡 minus 𝑒 to the power of negative 𝑡 with respect to 𝑡. If we then differentiate this term by term, we get one plus 𝑒 to the power of negative 𝑡. Finally, we know that one and 𝑒 to the power of negative 𝑡 are both continuous for all real values of 𝑡. So 𝑓 prime of 𝑡 and 𝑔 prime of 𝑡 are the sum or difference of two continuous functions. This means both of them are continuous for all real values of 𝑡.

Therefore, in particular, we’ve shown that 𝑓 prime and 𝑔 prime are continuous on our closed interval from zero to two. So we’re now ready to use our integral form to find the length of our curve. We get that 𝑙 is equal to the integral from zero to two of the square root of one minus 𝑒 to the power of negative 𝑡 all squared plus one plus 𝑒 to the power of negative 𝑡 all squared with respect to 𝑡. Next, we want to distribute our exponents over our parentheses. Distributing the square over our first set of parentheses, we get one minus two 𝑒 to the power of negative 𝑡 plus 𝑒 to the power of negative two 𝑡. And doing the same for our second set of parentheses, we get one plus two 𝑒 to the power of negative 𝑡 plus 𝑒 to the power of negative two 𝑡.

So by distributing the squares over both of our sets of parentheses. We now have the integral from zero to two of the square root of one minus two 𝑒 to the power of negative 𝑡 plus 𝑒 to the power of negative two 𝑡 plus one plus two 𝑒 to the power of negative 𝑡 plus 𝑒 to the power of negative two 𝑡 with respect to 𝑡. And now, we can simplify this expression. First, negative two 𝑒 to the power of negative 𝑡 plus two 𝑒 to the power of negative 𝑡 is equal to zero. Next, we have one plus one is equal to two. Finally, 𝑒 to the power of negative two 𝑡 plus 𝑒 to the power of negative two 𝑡 is equal to two 𝑒 to the power of negative two 𝑡.

So by grouping our like terms, we now have the integral from zero to two of the square root of two plus two 𝑒 to the power of negative two 𝑡 with respect to 𝑡. And this is our final answer. Therefore, we were able to show the length of the curve with parametric equations 𝑥 is equal to 𝑡 plus 𝑒 to the power of negative 𝑡 and 𝑦 is equal to 𝑡 minus 𝑒 to the power of negative 𝑡, where 𝑡 is greater than or equal to zero and 𝑡 is less than or equal to two. Can be expressed as the integral from zero to two of the square root of two plus two 𝑒 to the power of negative two 𝑡 with respect to 𝑡.