Hovedindhold

Worked example: Arc length of polar curves

Video udskrift

- What I want to do in this video is find the arc length of one petal, I guess we could call it, of the graph of r is equal to four sine of two theta. So I want to find the length of this portion of the curve that is in red right over here. We'll do this in two phases. First of all I want to set up the definite integral for finding that arc length, and then we'll evaluate it. I'll actually use a calculator to evaluate it because it'll be a little bit more straightforward at least for this integral. Let's just start with that. I encourage you to pause the video. Knowing what we know about the formula for arc length, when we have it in polar form, see if you can apply it to figure out this arc length right over here. I'm assuming you've had a go at it, so let's remind ourselves that the arc length is going to be the integral from our starting angle to our ending angle, we'll call it from alpha to beta, of the square root of the derivative of our function. Which respect to theta squared, plus our function squared. Plus our function squared d theta. Let's figure out what r prime of theta and what r of theta actually are. I'll color code it. R prime of theta, we know what r of theta is let me just rewrite it. We know r of theta is equal to four sine of two theta. Our prime of theta is just going to be derivative of two theta with respect to theta is two times four is eight, derivative of sine is cosine, cosine of two theta. If we wanted to write the definite integral for arc length it's going to be equal to the definite integral. What's our starting bound? For the petal that we care about, we're going to start at an angle of zero radians. When we have an angle of zero radians, r is zero, that's this point right over here. We're going to start at zero radians. And we're going to go all the way up to pi over two. When sine of two times pi over two is sine of pi, which is zero, so we'll get back to this point right over here. We're going to go from zero to pi over two, to pi over two radians. It's going to be the square root of, give myself some real estate here, r prime of theta squared is going to be this thing squared. It's going to be 60-- Let me do it in that same blue color, it's going to be 64 cosine squared of two theta, then plus this squared which is going to be 16 sine squared of two theta, then of course we have our d theta. You can attempt to solve this or evaluate this analytically, it's not too straightforward so I'm going to use a calculator. It just gives us a practice with knowing that hey, there are tools out there that can help us do these types of things. I'll go to the definite integral, I'll go to second calculus on my t 85, I picked this choice, which is for the definite integral. Let me express when I'm taking the definite integral of, it's going to be the square root of 64 times cosine of two theta squared. I'm going to use x instead of theta just because it's an easier variable to use on my calculator. Cosine of two-- Actually let me make sure my parenthesis are good. Cosine of two x, alright I'll close that. Now I want to square that. I did that first part. Plus 16 times plus 16 times 16 times sine of two x. Close that, close that. Squared. I'm not sure if the calculator knows to interpret that as multiplication, so let me insert a times right over here. 64 times cosine of two x, that thing squared. Plus 16 times sine of two x, that thing squared. Let me go to the end. Let me close my radical, so that's going to close that right over there, that closes what I'm taking the square root of. Then comma, the variable that I'm integrating with respect to is x, in this case. Everywhere I saw theta here I'm just replacing with an x there. I want to do it from, instead of theta equals zero to theta equals pi over two, we'll say x is equal to zero to x is equal to pi over two. Hopefully I have not made a mistake when I've typed this in, And I get... It's chugging on a little bit. And we get... Will it actually come up? Oh finally! If we were to round it to the nearest thousandth it'll be nine point six eight eight. Approximately nine point six eight eight. Let's just see, does that make intuitive sense? It goes as far as four away, so if you just went out to four and then back down that would be about eight. Of course we have actually gone out some, so it actually makes intuitive sense that this arc length would be nine point six eight eight. Hopefully you enjoyed that.