Study Guide

Area, Volume, and Arc Length - Arc Length

Arc Length

We've prepared our fish by cutting it into flat rectangular portions and diced our vegetables into thin slices. We've even gone off on a tangent, playing with frisbees and toilet paper tubes. We're hungry for a gourmet fish supper, but it'll be bland without any seasoning.

That's where this section comes in handy. Spices and herbs come in long strings that we need to chop. These long lines, called arcs, are written using a couple different methods, some of which we'll discuss below. We'll learn how to chop them up to find how much we have before we add them to our boiling pot.

The first thing we need to know is the length of a line. We used it for triangles, we used it for circles, and we even used it to find areas on disks. Who knew Pythagorus would be on to something when came up with the Pythagorean Theorem? Well...he probably did.

These problems are just a short review of how to use his theorem. If you have any trouble with them, you should review the Pythagorean Theorem and then come back.

Sample Problem

Find the length of the line between

1) (5,4) and (2,3)

2) (-2,-1) and (6, 3)

3) (3,5) and (-2, -6)

1) The change in x between the points (5,4) and (2,3) is 5 – 2 = 3 and the change in y is 4 – 3 = 1.

The length of the line is 2) Between the points (-2,-1) and (6, 3), Δ x = 8 and Δ y = 4.

The length of the line is 3) Between the points (3,5) and (-2, -6), Δ x = 5 and Δ y = 11.

The length of the line is When an herb is shaped like a curve (or arc) instead of a straight line, the length of the curve or arc length is the length of a piece of string that exactly covers the curve.

The idea of the string is good intuition, but it's not very useful for doing problems. Instead, like the other problems we've done in this unit to find area and volume, we are going to break our arc length problems down into parts. To find the length of the curve of a continuous function f on an interval [a,b],

• we'll chop the curve up into little pieces, • then find the length of each piece,

• and finally add all the lengths back together. We do this by letting the number of pieces go to ∞. This gives us an integral for the exact length of the curve.

Most of the assumptions we made earlier still apply if we replace "area" with "length." The exception is the first assumption, since there aren't "horizontal" or "vertical" slices when we're only working in one dimension.

We're going to use the Pythagorean Theorem a lot more in this section, albeit a little differently than we have before. For a sample function f(x) on an interval [a,b], we'll chop the curve up into tiny pieces and zoom in on the piece at position x. We want to be careful that the sun isn't shining too brightly. We aren't trying to kill a bug with a magnifying glass.

This piece is very close to being a straight line. To approximate its length, we'll pretend that it really is a straight line. The change in x over the little bit of line is Δ x. The change in y is Δ y. That's nothing new. Since we've zoomed in so much, the derivative of f at x is approximately the same as the slope of this little bit of line. In symbols, If we rearrange this equation, we get

Δ yf ' (x) Δ x.

The Pythagorean Theorem says the length of the little bit of line is We add up these quantities and let the number of pieces approach ∞ to get an integral that gives us the exact length of the curve. The integral is from a to b since those are the values that make sense for x. We can't really make these problems any harder. We just need to remember to find the length of a curve. If we were to forget the formula, it's not the zombie apocalypse. We could just draw a picture that breaks the curve up into pieces and use the Pythagorean Theorem to find the length of a piece. Then we'd just go through the steps we just used above to reconstruct the formula. We recommend practicing this a few times before the test. Memorizing a formula is one thing, but actually understanding it is another.

• Arc Length for Parametric Functions

We just learned how to chop up a simple herb like a chive. They're just simple lines that we know how to describe in terms of the variables x and y. Sometimes, we need to cut up more intricate herbs like rosemary. These lines may be described in different ways, including in parametric equations.

These parametric equations are just a different language for describing lines. They're nice because it's easier to describe the curve using this other language, and it can be easier to integrate this way too.

The process for finding the length of a curve described by parametric equations is the same. We're just using a different language. It's not even an African click language.

Suppose we want to find the length of the curve described by parametric equations x(t) and y(t), on the interval atb.

We break up the curve into little pieces, as usual: The Pythagorean Theorem says that the length of a little piece is (approximately) We need to get Δ x and Δ y in more useful terms, preferably in terms of t, so we can end up with an integral with respect to t.

The amount x changes over a little time interval is approximately multiplied by the length of the time interval. Similarly, for y, This means the length of a little piece is approximately This simplifies to When we integrate over the values that make sense for t, we get the length of the curve: This equation is just like the one we derived in the previous section, but now the equation is parameterized with variable t. We're using the notation instead of the notation y'(t) to make it clear that t is the independent variable. Since there are so many variables floating around, using prime notation here isn't the best idea. It's best not to get confused with dots and quotation or prime marks floating around a page of mathematical soup.

Here is a recap video of Pythagorean Theorem, just in case you need it.

Sample Problem

Find the length of the parametric curve described by

x(t) = t2, y(t) = t3

for 0 ≤ t ≤ 2.

You may use a calculator to evaluate the integral.

Explain why the answer is reasonable.

First, we find the derivatives Using our new formula, the length of the curve is If we graph the curve, we get This curve is close enough to the straight line between (0,0) and (4,8), so we would expect the length of the curve to be close to the length of that line.

The length of the line is When we're asked to explain why an answer is reasonable, we should compare the curve to a line or circle whose length we know how to find. Even when we're not explicitly asked to do this, it's a good idea as a way to check ourselves. If a curve is as long as a green bean, we don't want to trust an answer that says it's as long as Route 66.

With parametric functions we have to take a little extra care with the limits of integration. The limits of integration are values of t. If we aren't given the values of t, we have to find them ourselves.

Fun fact: "normal" functions are just a special case of parametric functions. The function f(x) can be parametrized as

x(t) = t

y(t) = f(t).

Since x and t are the same thing, the curve f(x) for axb is the same as the parametrized curve for atb. We can find the length of the curve by working with the parametrized version.

The derivative of

x(t) = t is and the derivative of y(t) = f(t) is Then the length of the curve is Since x and t are the same, we could just as well write this as which is the same thing the formula we had earlier.

This means we don't need to remember multiple formulas for arc length. Having to memorize less is better. It'll keep us from making mistakes between formulas, or from mistaking a panda for a polar bear.

If we remember the version for parametric equations, we can find the arc length of a normal function by using the parametrization x = t.

We prefer the parametric version. It looks just a bit more complicated, but it's more fun. The integral Looks like it's using the Pythagorean Theorem, which makes it easier to remember.

• Arc Length for Polar Functions

Polar functions are a special case of parametric functions. These are just equations disguised as nasty trigonometric functions. It's sort of like they're wearing intricate Halloween costumes, like a stick figure man costume made of black cloth and glow sticks. We can't tell who it is, but we can do our best to use our knowledge about a person to figure it out.

The polar function r = f(t) can be parametrized as

x(t) = r cos t = f(t) cos t

y(t) = r sin t = f(t) sin t

We're using t instead of θ so we can talk about instead of , to be consistent with the earlier discussion of parametric functions. To find and , we have to apply the product rule carefully.

Sample Problem

Find the arc length for one petal of the polar function r = sin(3t).

The function looks like The first petal corresponds to the interval . We get

x(t) = r cos t = sin(3t)cos t

y(t) = r sin t = sin(3t)sin t.

Using the product rule gives us We use the arc length formula and get It looks complicated and drawn out, but we can always just use a calculator to solve the integral. If our teacher asks us to solve this one, they're looking to torture us.

Sometimes, a slightly modified costume simplifies the calculations. We can use the identity

sin2 t + cos2 t = 1

to get a nicer formula for the arc length of a polar function.

If we want to make the formula look more like it goes with polar functions, we can put θ in place of t: This formula is nice to keep handy, but we probably don't need to memorize it unless we need to do a lot of polar arc length problems. 