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.
