Video Transcript
Find the length of the curve with parametric equations 𝑥 is equal to 𝑒 of the power of 𝑡 minus 𝑡 and 𝑦 is equal to four times 𝑒 to the power of 𝑡 over two, where 𝑡 is greater than or equal to zero and 𝑡 is less than or equal to two.
The question wants us to find the length of a curve defined as a pair of parametric equations. And we recall if we’re given a curve defined by a pair of parametric equations 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡. Then we can calculate the length of this curve where 𝑡 is greater than or equal to 𝛼 and 𝑡 is less than or equal to 𝛽 as 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.
Since we want to calculate the length of the curve defined by the parametric equations 𝑥 is equal to 𝑒 to the power of 𝑡 minus 𝑡 and 𝑦 is equal to four 𝑒 to the power of 𝑡 over two. We’ll set our function 𝑓 of 𝑡 to be 𝑒 to the power of 𝑡 minus 𝑡 and 𝑔 of 𝑡 to be four 𝑒 to the power of 𝑡 over two. And since we want to restrict 𝑡 to be between zero and two, we’ll set 𝛼 equal to zero and 𝛽 equal to two. So, this tells us we can calculate the length of our curve 𝐿 as the integral from zero to two of the square root of 𝑓 prime of 𝑡 squared plus 𝑔 prime of 𝑡 squared with respect to 𝑡. So, we need to calculate the derivatives of our functions 𝑓 and 𝑔.
To do this, we recall that, for any constants 𝑎 and 𝑛, the derivative of 𝑎𝑒 to the power of 𝑛𝑡 with respect to 𝑡 is just 𝑎 times 𝑛𝑒 to the power of 𝑛𝑡. Using this and the power rule for differentiation, we have that 𝑓 prime of 𝑡 is equal to 𝑒 to the power of 𝑡 minus one. And 𝑔 prime of 𝑡 is equal to two times 𝑒 to the power of 𝑡 over two. So, this gives us that the length of our curve is equal to the integral from zero to two of the square roots of 𝑒 to the 𝑡 minus one squared plus two 𝑒 to the power of 𝑡 over two squared with respect to 𝑡.
We can simplify this by noticing that 𝑒 to the power of 𝑡 minus one all squared is equal to 𝑒 to the power of two 𝑡 minus two 𝑒 to the power of 𝑡 plus one. And two times 𝑒 to the power of 𝑡 over two all squared is just equal to four 𝑒 to the power of 𝑡. This gives us the following expression.
We can simplify this further by noticing that four 𝑒 to the power of 𝑡 minus two 𝑒 the power of 𝑡 is just equal to positive two 𝑒 to the power of 𝑡. So, to calculate the length of our curve, we need to integrate from zero to two the square roots of 𝑒 to the power of two 𝑡 plus two 𝑒 to the 𝑡 plus one with respect to 𝑡.
And this is not in a standard form which we can integrate. It would be a lot easier if we could get rid of the square root. Then, we could just use our normal rules for integration to evaluate this integral. And in fact, we can do that in this case. We just need to notice that 𝑒 to the power of two 𝑡 plus two 𝑒 to the power of 𝑡 plus one is a square. It’s actually equal to 𝑒 to the power of 𝑡 plus one all squared. So, by rewriting it in this form, we can remove the square root from our integrand.
This gives us the length of our curve is equal to the integral from zero to two of 𝑒 to the power of 𝑡 plus one with respect to 𝑡. And now, this is in a form which we can integrate. We know the integral of 𝑒 to the power of 𝑡 is just equal to itself, 𝑒 to the power of 𝑡. And the integral of one is just equal to 𝑡. So, this gives us 𝑒 to the power of 𝑡 plus 𝑡 evaluated at the limits of our integral 𝑡 is equal to zero and 𝑡 is equal to two.
Evaluating this at the limits of our integral gives us 𝑒 squared plus two minus 𝑒 to the zeroth power plus zero. And we know that 𝑒 to the zeroth power is just equal to one. So, we can simplify two minus one to just give us one. Which means that we have shown that the length of the curve with parametric equations 𝑥 is equal to 𝑒 to the power of 𝑡 minus 𝑡 and 𝑦 is equal to four times 𝑒 to the power of 𝑡 over two, where 𝑡 is greater than or equal to zero and 𝑡 is less than or equal to two, is equal to 𝑒 squared plus one.
