Lecture 18 : Wednesday May 14th

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18.1 Area and arc-length in polar co-ordinates The arc length of the curve r = f (θ) for α ≤ θ ≤ β is given by the formula Z

β

p

f 0 (θ)2 + f (θ)2 dθ.

α

We also stated the formula

Z

β

p

x0 (θ)2 + y 0 (θ)2 dθ

α

where x = f (θ) cos θ = r cos θ and y = f (θ) sin θ = r sin θ. Let’s prove that this is the same as the first formula. By the product rule of differentiation, x0 (θ) = f 0 (θ) cos θ − f (θ) sin θ y 0 (θ) = f 0 (θ) sin θ + f (θ) cos θ. So it follows that x0 (θ)2 = f 0 (θ)2 cos2 θ − 2f 0 (θ) cos θf (θ) sin θ + f (θ)2 sin2 θ y 0 (θ)2 = f 0 (θ)2 sin2 θ + 2f 0 (θ) sin θf (θ) cos θ + f (θ)2 cos2 θ. Adding these up, we get x0 (θ)2 + y 0 (θ)2 = f 0 (θ)2 cos2 θ + f 0 (θ)2 sin2 θ + f (θ)2 sin2 θ + f (θ)2 cos2 θ. Since cos2 θ + sin2 θ = 1, this is exactly f 0 (θ)2 + f (θ)2 . This proves that the formulas are actually the same. For practical purposes, the first formula is easier to use.

18.2 Density and Center of Mass The density of an object is defined by the mass of the object divided by its volume (or length, or area depending on the dimensions of the object). For example, if we assume that the earth is a sphere of radius r = 6378km and the mass of the earth is 5.9742102 4kg, then the volume of the earth is 34 πr3 ≈ 1086.78km3 which means the density of the earth is. This is not to say the density of the earth is constant, because some parts of the earth are clearly denser than others. More generally, the density of a solid object might be described by a function δ(x, y, z) at a point (x, y, z) in the object. Given this “density function” of an object, we can in theory compute the mass of the object. Since this is, in general, complicated, we will stick to the case of objects in one or two dimensions. If the object were one dimensional, for example a thin rod on the x-axis for a ≤ x ≤ b, then the 1

density might be given by a function δ(x) at point (x, 0) for a ≤ x ≤ b. The mass in this case is defined by Z b Mass = δ(x) dx. a

The center of mass is defined by Rb

xδ(x) dx . Center of mass = Ra b δ(x) dx a The center of mass of a rod is the point along the rod with the property that if the rod were balanced on a needle at that point, under the influence of gravity, the rod would be in equilibrium. For example, suppose we have a rod from (0, 0) to (0, 1) whose density at distance x feet from the origin is given by δ(x) = x pounds per foot. The mass of the rod is clearly Z 1 1 Mass = δ(x) dx = a pound. 2 0 The center of mass should be closer to (0, 1) than (0, 0), and this is certainly true: Z 1 2 Center of mass = 2 x2 dx = . 3 0 So the center of mass is at distance 2/3 of a foot from the origin. As we said, the center of mass is a point of equilibrium of an object. In two dimensions, suppose we have an object consisting of the region between a curve y = f (x) where f (x) ≥ 0 and the curve y = −f (x), for a ≤ x ≤ b. Suppose that the density at position x along the x-axis is δ(x) all along the vertical cross sectional line (which is like the example of the rod above) from y = f (x) to y = −f (x). Then the mass is Z b Mass = 2f (x)δ(x) dx. a

Then the center of mass (which is certainly on the x-axis) is given by Rb 2xf (x)δ(x) dx Center of mass = Ra b . 2f (x)δ(x) dx a It is important to note that the center of mass is a position. The center of mass formula above means that if c is the value of the right hand side, then the center of mass is at (c, 0) on the x-axis. For example, suppose we have an isosceles triangle whose base has length two and whose height is one. We place the triangle along the x-axis, so that the base of the triangle is the segment of the y-axis from y = −1 to y = 1. This triangle is drawn below: 2

Isosceles triangle

Suppose the density of each vertical line at position (x, 0) is δ(x) = x pounds per square foot. Since the triangle is the same as the region bounded by y = f (x) and y = −f (x) where f (x) = 1 − x, the mass of the triangle is Z

Z

1

1

2x(1 − x) dx =

2f (x)δ(x) dx =

Mass =

0

0

1 of a pound. 3

The center of mass is Z

Z

1

Center of mass = 3

1

2xf (x)δ(x) dx = 6 0

0

1 x2 (1 − x) dx = . 2

So in this case the center of mass of the triangle is actually at (1/2, 0). This makes sense because although the density is increasing to the right, the area is decreasing. Here is an entertaining example. Suppose we stack cubic blocks one on top of the other, but so that each successive block in the stack overlaps the previous one by some amount. Each block will have constant density, and each block is one foot in length, height and width. We would like to know how much overhang we can create with n blocks stacked one after the other and so that the system of blocks does not collapse. According to physics, the blocks will not collapse provided the center of mass of the whole system is above the first block. Now suppose that the ith block overlaps the (i − 1)th block by 1/(2i) feet for i = 1, 2, . . . , n − 1 (the first block is the zeroth block). Let’s place the blocks in the pattern shown below along the x-axis and suppose that the first i blocks have center of mass at (xi , 0). Suppose that the center of mass of just the ith block alone 3

is at (ci , 0). Since the blocks have uniform density, the center of mass of the zeroth block is at x0 = c0 = 21 . The center of mass (x1 , 0) of the first two blocks is given by 1 3 x1 = (x0 + c1 ) = 2 4 since c1 = 1. This is still above the first block, so the system is stable so far. For the first three blocks, the center of mass is at (x2 , 0) where 1 (2x1 + c2 ) 3 1 3 5 = ( + ) 3 2 4 11 = . 12

x2 =

Again this is above the first block so we’re okay. How can we work out the center of mass for the first i blocks? Well suppose we know the center of mass xi−1 of the first i − 1 blocks. Then 1 xi = (ixi−1 + ci ) i+1 as we did for the first few blocks. We’re going to show that xi < 1 − 2−2i for all i – which means the whole system is stable, and at the same time we’ll show that 1 1 1 ln(i + 1) < ci + < ln i. 2 2 2 Let’s begin with the second task first. We know that for i ≥ 1, ci +

1 1 1 1 = 1 + + + ··· + 2 2 4 2i

since the overhang in the ith block is 1/2i. This is exactly a left Riemann sum for the function y = 1/2x when 1 ≤ x ≤ (i + 1). This is an overestimate of the area under the curve y = 1/2x, whose precise value is Z i+1 1 1 dx = ln(i + 1). 2x 2 1 Therefore ci +

1 1 > ln(i + 1). 2 2

On the other hand,

1 1 1 1 = + + ··· + 2 2 4 2i and this is a right Riemann sum for 1 ≤ x ≤ i. This underestimates the area under y = 1/2x, so Z i 1 1 1 ci + < dx = ln i. 2 2 1 2x ci +

4

So we have shown what we wanted about ci . To show that the system is stable, we need xi < 1 for all i. Remember 1 xi = (ixi−1 + ci ). i+1 But now we know ci < 21 ln i, We could take a good guess at what xi should be: we already had x0 = 12 , x1 = 43 = 1 − 14 , 11 1 1 x2 = 12 = 1 − 12 . So how about guessing xi = 1 − (i+1)2 i ? It turns out that this is the right guess, and we can prove it by mathematical induction using the formula for xi in terms of xi−1 . Define the harmonic sum n X 1 1 Hn = − . r 2 r=1 In general, the center of mass of the n blocks is given by 1 1 1 1 1 (H1 − + H2 − + H3 − + · · · + Hn − ). n 2 2 2 2 If this is less than one, then the center of mass of all the blocks is above the first block, and the system is stable. There is a very simple inequality that we can use here to see what Hn looks like. Remember that Z n 1 dx = ln n. 1 x This can be interpreted as the area under the curve y = 1/x for 1 ≤ x ≤ n. This area can be approximated by a left Riemann sum. If we use a left Riemann sum to estimate this area, splitting the range [1, n] into unit intervals, we get precisely Apx Area =

n−1 X 1 r=1

r

= Hn−1 .

Since y = 1/x is decreasing, this is an overestimate of the actual area, which means Hn−1 ≥ log n. On the other hand, the right Riemann sum is Apx Area =

n X 1 r=2

r

= Hn − 1

and this is an underestimate of the actual area. It follows that Hn − 1 ≤ log n. Our conclusion is Hn − 1 ≤ log n ≤ Hn−1 . Going back to the example of the blocks, this means the center of mass of the n blocks is at most ! Ã n−1 1 n X 1− + ln i . n 2 i=1 5

Finally we have to deal with Consider the problem of finding the area between the given curve and the x-axis in each of the diagrams below. 1 1

0.8 0.8 1

0.6 0.6

0.8

0.6

0.4

0.4 0.4

0.2

0.2 0.2

0

0 -1

-0.5

y=

0

√

0.5

1

0 -1

-0.5

0

0.5

x

1

-1

-0.5

0

1 − x2

y = 1 − |x|

0.5

1

x

x

y = 1 − x2

It is not hard to find the area between the x-axis and the curve √ in the first two graphs, since the curve y = 1−|x| defines a triangle and the curve y = 1 − x2 defines a semicircle, and we have formulas for the areas of these shapes: 1 Area(b) = π. 2

Area(4) = 1

Now in the third picture, we have the parabola f (x) = 1−x2 . We don’t have a formula for the area of such a shape, so let’s spend some time seeing how we could get an approximate area for this shape.

1.2 Approximating area How could we get a good estimate of the area? Since we know how to find the area of a rectangle, we could approximate the region by summing up the areas of rectangles as shown below. These sums are called Riemann sums.

Approximate area 6

The rectangles in the picture under the curve f (x) = 1 − x2 are defined precisely as follows. First cut interval under the curve on the x-axis into ten intervals of equal length. We could write out all the intervals, starting with [−1, −0.8] followed by [−0.8, −0.6] and so on ending with [0.8, 1]. Rather than write them numerically, lets list the rectangles symbolically by ¤i for i = 1, 2, . . . , 10. So, for example, the base of ¤1 is [−1, −0.8], and the base of ¤7 is [0.4, 0.6]. We can also give the height of the each rectangle ¤i . Ht(¤1 ) = f (−1) = 0 Ht(¤2 ) = f (−0.8) = 0.36

···

Ht(¤10 ) = f (0.8) = 0.36.

Notice that the height of ¤i is just the value of the function f at the left endpoint of the ith interval. We call this a left Riemann sum. To find the approximate area, we just add up the areas of the rectangles ¤i to get Apx Area = Area(¤1 ) + Area(¤2 ) + Area(¤3 ) + · · · + Area(¤10 ) = 0.2 · (0 + 0.36 + 0.64 + 0.84 + 0.96 + 1 + 0.96 + 0.84 + 0.64 + 0.36) = 1.32 So the approximate area found using these rectangles is 1.32. It turns out that the actual area is 34 , and we will see how to work this out in a flash later on using integration. There isn’t any reason to use the rectangles we gave above, where the height of ¤i was the value of f at the left endpoint of the ith interval. We could have chosen the height of ¤i to be f at the right endpoint. If we do that, then the height of ¤1 is now f (−0.8) = 0.36, the height of ¤2 is f (−0.6) = 0.64, and so on. The area is called a right Riemann sum. Alternatively, we could pick the midpoint of the ith interval and let the height of ¤i be the value of f at this midpoint to get another Riemann sum. The pictures of these rectangles are shown below:

Right Midpoint endpoint Let’s work out the approximate areas by finding the values of the Riemann sums: in the first picture Apx Area = 0.2 · (0.36 + 0.64 + 0.84 + 0.96 + 1 + 0.96 + 0.84 + 0.64 + 0.36 + 0) = 1.32 7

as expected since this is just a reflection of the pictures with the blue rectangles. In the second picture, where the height of the ith rectangle is f (0.2i − 1.1), we get Apx Area = 0.2 · (0.36 + 0.64 + 0.84 + 0.96 + 1 + 0.96 + 0.84 + 0.64 + 0.36 + 0) = 1.34. Now we have overestimated the area. There is no reason to stick to ten rectangles. In fact, a key observation is that the more rectangles we use, the better approximation we will have for the areas. For instance, consider increasing the number of blue rectangles to twenty. The picture looks as follows:

Finer rectangular partitions Now the intervals are [−0.1i, −0.1i + 0.1] for i = −10 . . . 9 and the height of the ith rectangle is f (−0.1i). Let’s see that we get a better gurss of the true area, which is 34 : Apx Area = 0.1 · (0 + 0.19 + 0.36 + · · · + 1 + 0.91 + · · · + 0.19) = 1.33. This is already accurate to within two decimal places of the true area. Using a computer, we can get more and more accurate estimates using more and more rectangles. For example, the table below shows the better and better accuracy using left endpoint rectangles: Number of rectangles Apx Area

10 20 1.32 1.33

8

100 1.3332

1000 1.333332