Question Video: Calculating the Arc Length of a Curve Defined by a Rational Function | Nagwa Question Video: Calculating the Arc Length of a Curve Defined by a Rational Function | Nagwa

# Question Video: Calculating the Arc Length of a Curve Defined by a Rational Function Mathematics • Higher Education

Calculate the arc length of the curve π¦ = (π₯β΄/32) + (1/π₯Β²) between π₯ = 1 and π₯ = 2, giving your answer as a fraction.

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### Video Transcript

Calculate the arc length of the curve π¦ is equal to π₯ to the fourth power divided by 32 plus one over π₯ squared between π₯ is equal to one and π₯ is equal to two, giving your answer as a fraction.

The question wants us to find the arc length of a curve between the values of π₯ is equal to one and π₯ is equal to two. It wants us to give our answer as a fraction. And we know how to calculate the arc length for certain curves. We know if π prime is a continuous function on a closed interval from π to π. Then we can find the length of the curve π¦ is equal to π of π₯ between the values of π₯ is equal to π and π₯ is equal to π. By evaluating the integral from π to π of the square root of one plus π prime of π₯ squared with respect to π₯.

In our case, weβre trying to find the length of the curve π¦ is equal to π₯ to the fourth power over 32 plus one over π₯ squared between π₯ is equal to one and π₯ is equal to two. So, weβll set π of π₯ to be π₯ to the fourth power over 32 plus one over π₯ squared, π to be equal to one, and π to be equal to two. Now, to use this arc length formula, we need to show that our function π prime is continuous on the closed interval from π to π.

To do this, weβre going to need to find an expression for π prime of π₯. Weβll start by using our laws of exponents to rewrite π of π₯ as π₯ to the fourth power over 32 plus π₯ to the power of negative two. We can now differentiate this term by term by using the power rule for differentiation. We get π prime of π₯ is equal to four π₯ cubed over 32 minus two π₯ to the power of negative three. And weβll simplify and rewrite this to give us π₯ cubed over eight minus two over π₯ cubed.

And we can see that π prime of π₯ is the difference between a polynomial and a rational function. We know polynomials are continuous for all real values of π₯ and rational functions are continuous on their domain. In other words, our rational function is continuous everywhere except where π₯ is equal to zero. This means, in particular, our rational function is continuous on the closed interval from one to two. And we know our polynomial is also continuous on the closed interval from one to two. This means that their difference must also be continuous on this interval.

What this means is weβve justified our use of our integral formula to find the arc length of our curve. So, weβll substitute π is equal to one, π is equal to two, and π prime of π₯ is equal to π₯ cubed over eight minus two over π₯ cubed into our arc length formula. This gives us the length is equal to the integral from one to two of the square root of one plus π₯ cubed over eight minus two over π₯ cubed squared with respect to π₯.

Next, weβll want to simplify our integrand. Weβll start by distributing the square over our parentheses. We can do this by using the binomial formula or by using the FOIL method. Weβll get π₯ to the sixth power over 64 minus two times two π₯ cubed over eight π₯ cubed plus four over π₯ to the sixth power. And we can simplify our second term. We can cancel the shared factor of π₯ cubed in the numerator and the denominator. And this leaves us with negative two times two divided by eight, which simplifies to give us negative one-half.

This gives us the integral from one to two of the square root of one plus π₯ to the sixth power over 64 minus one-half plus four divided by π₯ to the sixth power with respect to π₯. We can then simplify our integrand even more, since one minus one-half is equal to positive one-half. And this leaves us with the following integral. This is a very-tough-looking integral. Itβs hard to see how we could evaluate this.

We might be tempted to start trying to use a substitution. But remember, the first thing we should always try is to simplify our integrand. It would be nice if we could write our inner function as a square, since then this would cancel with the square root which weβre taking. So, letβs try doing this. We know one square root of π₯ to the sixth power over 64 is π₯ cubed over eight. And one square root of four divided by π₯ to the sixth power is two divided by π₯ cubed.

We then want to think what would happen if we squared the sum or the difference of these two terms. Letβs start with squaring the sum of the two terms. We know we would get π₯ cubed over eight all squared, which would give us π₯ to the sixth power over 64. We would get two over π₯ cubed all squared, which would give us four divided by π₯ to the sixth power. But we would then also get two times π₯ cubed over eight times two over π₯ cubed. And if we calculate this, this simplifies to give us one-half.

So, in actual fact, our inner function was a square; itβs equal to π₯ cubed over eight plus two over π₯ cubed all squared. And we can then take the square roots of both sides of this equation. Remember, we want the positive square roots. So, this will be π₯ cubed over eight plus two over π₯ cubed. So, we can now instead evaluate the integral from one to two of this function with respect to π₯.

So, the length for our arc is equal to the integral from one to two of π₯ cubed over eight plus two over π₯ cubed with respect to π₯. And to evaluate this integral, weβll start by rewriting two over π₯ cubed as two π₯ to the power of negative three. We can now evaluate this integral term by term by using the power rule for integration. Applying this and simplifying, we get π₯ to the fourth power over 32 minus one over π₯ squared evaluated at the limits of our integral, π₯ is equal to one and π₯ is equal to two.

Now, we evaluate this at the limits of our integral. We get two to the fourth power over 32 minus one over two squared minus one to the fourth power over 32 minus one over one squared. And then, if we calculate this expression, we see we get 39 divided by 32. And this is a fraction, so this is our final answer.

Therefore, weβve shown the curve π¦ is equal to π₯ to the fourth power over 32 plus one over π₯ squared between π₯ is equal to one and π₯ is equal to two has an arc length of 39 divided by 32.

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