Video Transcript
Calculate the arc length of the curve 𝑦 is equal to two over nine times three 𝑥 minus one to the power of three over two between the values 𝑥 is equal to a half and 𝑥 is equal to two, giving your answer as a fraction.
The question gives us a curve. And it wants us to find the length of the arc of this curve between the values 𝑥 is equal to a half and 𝑥 is equal to two. It wants us to give our answer as a fraction. And we know how to find the arc length of certain curves. We know if 𝑓 prime is a continuous function on the closed interval from 𝑎 to 𝑏, then the length of the curve 𝑦 is equal to 𝑓 of 𝑥 between the values of 𝑥 is equal to 𝑎 and 𝑥 is equal to 𝑏 is equal to the integral from 𝑎 to 𝑏 of the square root of one plus 𝑓 prime of 𝑥 squared with respect to 𝑥.
We want to find the length of the curve 𝑦 is equal to two over nine times three 𝑥 minus one to the power of three over two between 𝑥 is equal to one-half and 𝑥 is equal to two. So we’ll set 𝑓 of 𝑥 to be two over nine times three 𝑥 minus one to the power of three over two, 𝑎 to be one-half, and 𝑏 to be equal to two. Then to use our all kind formula, we just need to show that 𝑓 prime is continuous on the closed interval from 𝑎 to 𝑏. So we need to find an expression for 𝑓 prime of 𝑥. We start by noticing that 𝑓 of 𝑥 contains the composition of two functions. It contains a linear function raised to the power of three over two.
So we’ll set 𝑢 of 𝑥 to to be our inner function, three 𝑥 minus one. This means 𝑓 of 𝑥 is equal to two over nine times 𝑢 to the power of three over two. So we’ve written 𝑓 of 𝑥 to be a function of 𝑢, and 𝑢 in turn is a function of 𝑥. This means we can differentiate 𝑓 of 𝑥 by using the chain rule. And the chain rule tells us if 𝑓 is a function of 𝑢 and 𝑢 in turn is a function in 𝑥, then d𝑓 by d𝑥 is equal to d𝑓 by d𝑢 times d𝑢 by d𝑥. So by applying the chain rule, we have 𝑓 prime of 𝑥 is equal to the derivative of two over nine times 𝑢 to the power of three over two with respect to 𝑢 multiplied by the derivative of three 𝑥 minus one with respect to 𝑥.
And we can evaluate both of these derivatives by using the power rule for differentiation. We get three over two times two over nine multiplied by 𝑢 to the power of one-half times three. And we can then simplify this. We have a shared factor of two in the numerator and the denominator. And we have three times three in our numerator, which is nine. So we can divide this by nine to get one. And finally, 𝑓 is a function of 𝑥. So we should find 𝑓 prime in terms of 𝑥. We’ll do this by using our substitution 𝑢 is equal to three 𝑥 minus one. And by using our laws of exponents, this means 𝑓 prime of 𝑥 is equal to the square root of three 𝑥 minus one.
And we can see that 𝑓 prime of 𝑥 is the composition of two continuous functions. This means it’s continuous on its entire domain. And we can find the domain of this function. As long as we’re not taking the square root of a negative number, 𝑓 prime of 𝑥 exists. So we just need three 𝑥 minus one to be greater than or equal to zero. This happens when 𝑥 is greater than or equal to one-third. So 𝑓 prime of 𝑥 is continuous for all values of 𝑥 greater than or equal to one-third, which means, in particular, it’s continuous on the closed interval from one-half to two. This means we’ve justified the use of our integral formula for finding the arc length.
Substituting in 𝑎 is equal to one-half, 𝑏 is equal to two, and 𝑓 prime of 𝑥 is equal to the square root of three 𝑥 minus one. We get that our arc length is equal to the integral from one-half to two of the square root of one plus the square root of three 𝑥 minus one squared with respect to 𝑥. And we can immediately see how to simplify our integral. The square root of three 𝑥 minus one all squared is equal to three 𝑥 minus one. So the arc length simplifies to give us the integral from one-half to two of the square root of one plus three 𝑥 minus one.
And we can simplify this further since one minus one is equal to zero. So in actual fact, we just need to find the integral from one-half to two of the square root of three 𝑥 with respect to 𝑥. And to make this easier, we’ll use our laws of exponents to rewrite our integrand as root three times 𝑥 to the power of one-half. And in doing this, we see we can now evaluate this integral by using the power rule for integration. We want to add one to our exponent of 𝑥 to get three over two and then divide by this new exponent of 𝑥. This gives us root three divided by three over two times 𝑥 to the power of three over two evaluated at the limits of our integral 𝑥 is equal to one-half and 𝑥 is equal to two.
And we can simplify this slightly. Instead of dividing by three over two, we’ll multiply by the reciprocal of three over two. And the reciprocal of three over two is two over three. So this gives us a new coefficient of two root three over three. The last piece of simplification we’ll do is take out the coefficient of two root three over three from our evaluation. Now, when we evaluate this at the limits of our integral, we get two root three over three times two to the power of three over two minus one-half to the power of three over two. Remember, the question wants us to give our answer as a fraction. So we need to simplify this expression.
We’ll do this by using the fact that two to the power of three over two is two cubed raised to the power of one-half. So that’s the square root of eight, which we can simplify to give us two root two. So this tells us that two to the power of three over two is two root two. And one-half to the power of three over two is one divided by two root two. So this gives us two root three over three times two root two minus one divided by two root two. But this is still not a fraction. So we need to simplify further or start by rationalizing the denominator of the second term inside our parentheses.
We’ll multiply the numerator and the denominator by root two. And when we multiply these together and simplify, we get root two over four. So we now have two root three over three times two root two minus root two over four. We can simplify two root two minus root two divided by four to give us seven root two over four. And this gives us two root three over three times seven root two over four. And if we multiply these together and simplify, we get seven root six over six, which is our final answer.
Therefore, we’ve shown the curve 𝑦 is equal to two over nine times three 𝑥 minus one to the power of three over two between 𝑥 is equal to one-half and 𝑥 is equal to two has an arc length of seven root six over six.
