### Video Transcript

Find the length of the curve with
parametric equations π₯ equals π to the power of π‘ minus π‘ and π¦ equals four π
to the power of π‘ over two, where π‘ 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 π‘ greater than
or equal to πΌ and less than or equal to π½ is the definite integral between πΌ and
π½ of the square root of dπ₯ by dπ‘ squared plus dπ¦ by dπ‘ squared with respect to
π‘. Now, in this case, weβre interested
in the length of the curve, where π‘ is greater than or equal to zero and less than
or equal to π‘. So weβll let πΌ to be equal to zero
and π½ be equal to two. Then, our parametric equations are
π₯ equals π to the power of π‘ minus π‘ and π¦ equals four π to the power of π‘
over two.

Itβs quite clear that weβre going
to need to work out dπ₯ by dπ‘ and dπ¦ by dπ‘. And so, we firstly recall that the
derivative of π to the power of π‘ is π to the power of π‘. The derivative of negative π‘ is
one. So dπ₯ by dπ‘ is π to the power of
π‘ minus one. Now, weβre going to use the chain
rule to differentiate π¦ with respect to π‘. We let π’ be equal to π‘ over
two. So that dπ’ by dπ‘ is equal to
one-half. Then, dπ¦ by dπ‘ is dπ¦ by dπ’
times dπ’ by dπ‘. Now, π¦ is equal to four π to the
power of π’. So dπ¦ by dπ‘ is four π to the
power of π’ times a half, which is two π to the power of π’. But of course, we want dπ¦ by dπ‘
in terms of π‘. So we replace π’ with π‘ over
two. And we find that dπ¦ by dπ‘ equals
two π to the power of π‘ over two.

Now, in fact, for our arc length
formula, we need dπ₯ by dπ‘ squared and dπ¦ by dπ‘ squared. So weβre going to square each of
our expressions. When we do, we find that π to the
power of π‘ minus one squared is π to the power of two π‘ minus two π to the power
of π‘ plus one. And dπ¦ by dπ‘ squared is simply
four π to the power of π‘. 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 π to the power of two π‘ minus two π to
the power of π‘ plus one plus four π to the power of π‘ with respect to π‘.

We notice that negative two π to
the power of π‘ plus four π to the power of π‘ is two π to the power of π‘. And thatβs great because we see we
can factor π to the power of two π‘ plus two π to the power of π‘ plus one a
little like we would with a quadratic. We get π to the power of π‘ plus
one times π to the power of π‘ plus one or π to the power of π‘ plus one
squared. And of course, the square root of
π to the power of π‘ plus one squared is just π to the power of π‘ plus one. When we integrate π to the power
of π‘, we get π to the power of π‘. And the integral of one is π‘.

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