Question Video: Arc Length of Planar Curves Mathematics

Video Transcript

Calculate the arc length of the curve of 𝑦 equals the square root of four minus 𝑥 squared between 𝑥 equals zero and 𝑥 equals two, giving your answer to five decimal places.

Using Libenie’s notation, the formula for the arc length of a curve is given by the definite integral evaluated between 𝑎 and 𝑏 of the square root of one plus d𝑦 by d𝑥 squared with respect to 𝑥. We know that 𝑦 is equal to the square root of four minus 𝑥 squared, so we’ll begin by working out d𝑦 by d𝑥. If we write 𝑦 as four minus 𝑥 squared to the power of one-half, then we can use the general power rule to find the derivative of this function with respect to 𝑥. It’s a half times four minus 𝑥 squared to the power of negative half multiplied by the derivative of the function that sits inside the brackets. That’s negative two 𝑥.

Dividing through by two and rewriting d𝑦 by d𝑥, we obtain negative 𝑥 over the square root of four minus 𝑥 squared. We let 𝑎 be equal to zero and 𝑏 be equal to two. And this means the length of the curve that we’re interested in is equal to the definite integral between zero and two of the square root of one plus negative 𝑥 over the square of four minus 𝑥 squared squared with respect to 𝑥. And this simplifies quite nicely. The integrand becomes the square root of one plus 𝑥 squared over four minus 𝑥 squared. We can simplify the expression inside the square root by multiplying both the numerator and denominator of the number one by four minus 𝑥 squared. And when we add that, we get four minus 𝑥 squared plus 𝑥 squared over four minus 𝑥 squared. Negative 𝑥 squared plus 𝑥 squared is zero. And so our integrand becomes the square root of four over four minus 𝑥 squared.

This can be simplified even further. If we take out our factor of four, we see that the integrand could be written as the square root of four times one over the square of four minus 𝑥 squared. The square to four is two. And we’re allowed state constant factors outside of the integral. So we have that 𝐿 is equal to two times the definite integral evaluated between zero and two of one over the square of four minus 𝑥 squared. Now, this might look really nasty. But we can evaluate this integral by using a substitution. We recall that the derivative of arc sine of 𝑥 is one over the square root of one minus 𝑥 squared. So we rewrite our integrand slowly.

This time we take out our factor four on the denominator. And we see that two times one over the square of four cancels. We can also write 𝑥 squared over four as 𝑥 over two squared. We choose 𝑢 then for our substitution to be equal to 𝑥 over two. Then d𝑢 by d𝑥 is equal to one-half. And we can say this is the equivalent to saying two d𝑢 equals d𝑥. We’re also going to need to change the limits on our integral. When 𝑥 is equal to two, 𝑢 is equal to two divided by two, which is one. And when 𝑥 is equal to zero, 𝑢 is zero divided by two, which is zero.

So we now see that 𝐿 is equal to the definite integral between zero and one of one over the square root of one minus 𝑢 squared times two d𝑢. Once again, we’ll take out our common factor of two. And we can now evaluate the integral of one over the square root of one minus 𝑢 squared. It’s simply arc sin of 𝑢. And we’re looking to evaluate two times arc sine of 𝑢 between one and zero. That’s two times arc sin of one minus arc sin of zero, which is simply 𝜋. And we see that correct to five decimal places, the arc length of the curve 𝐿 is 3.14159.