Homework 2 Solutions

Transcription

1 Homework Solutions 1. (a) Find the area of a regular heagon inscribed in a circle of radius 1. Then, find the area of a regular heagon circumscribed about a circle of radius 1. Use these calculations to estimate the area of a circle of radius 1. Then, use these calculations to give a decimal approimation of π. When a regular heagon is inscribed in a circle of radius 1, the heagon can be divided into 6 triangles as shown below by joining each of the vertices of the heagon to the center of the circle. Let ABO be one such triangle as below. Then angle AOB has measure equal to π/6 since each of the angles of the 6 triangles at the point O are congruent. We compute the area of one of these triangles as shown. Let M be the the midpoint of AB. Since AMO is a right triangle, we can use trigonometry to find the lengths AM and MO. Let θ denote the measure of the angle MOA. Then, since OA = 1 (being equal to the radius of the circle), AM = sin θ and MO = cos θ. Therefore, the area of triangle AOB is equal to (1/) AB MO = (1/)( sin θ)(cos θ) = 1 sin (θ), the last equality following from the double angle formula sin α cos α = sin (α). The area of the heagon is 6 times the area of the triangle AMO. Thus, if we denote the area of the heagon by A 6 and we use the observation that θ = π/6, then we see that A 6 = sin (π/6) = 3 sin (π/3) =. 1

2 This gives us a rough estimate of π. The area of the unit circle is equal to π and the area of the heagon gives us a lower bound. The computation of the area of a circumscribed heagon can be computed in a similar manner. Here is a shortcut. It is not hard to compute that the length of one of the sides of a circumscribed heagon is equal to / 3. Therefore the circumscribed heagon is similar to the inscribed heagon and the latter is the same as the former scaled by a factor of / 3. It follows that the area of the circumscribed heagon will be (/ 3) times the area of the inscribed heagon. (If lengths are scaled by a factor c > 0, then area is scaled by a factor of c.) Thus, the area of the circumscribed heagon is equal to (3/4)(3 3/) = 3. (b) Find the area A n of a regular n-gon inscribed in a circle of radius 1. What is the value of n A n? Can you justify your answer analytically (e.g. using the methods of.3)? To solve this problem, proceed eactly as in part (a) above. The only change is that here θ = π/n. Therefore, A n = n 1 sin (π/n) = n sin (π/n). The it of A n as n increases is equal to π since increasing the number of sides of the polygon will fill in the area of the circle will less and less error. If n = 96, we see that A (This is the origin of π 3.14.) The reason why n = 96 is a good choice is due to the fact that we have an eact way of computing sin π/96. The trick is to use the half angle formulas: cos (θ/) = 1 (1 + cos θ) and sin (θ/) = 1 (1 cos θ). It follows that sin (π/48) = 1 (1 1 ( (1 + ))). This simplifies to the much more appealing equation sin (π/48) =

3 . In this problem, we will study the properties of its in Theorem. of section.3. Compute each of the following its and clearly indicate which properties you are using at each step of your solution. (a) (b) ( 3 + ) 1 By Theorem.1.1, 1 =. By Theorem.1., 1 = 1. By Theorem.1.3, 1 = 1 = 1. By Theorem..1 (scalar multiple), 1 3 = 3( 1 ) = 3 1 = 3. Finally, using two applications of Theorem.. (sum), we have that 1 ( 3 + ) = = 1 3+ = Factor the numerator an denominator of the epression inside the it operator above: ( + 1)( + 1) ( + )( + 1). When we compute a it at = 1, the values of approach but never equal -1. Therefore it is valid to cancel the factors of ( + 1) and say that = (Here we are using Theorem.7 (functions which agree at all but one point).) By Theorem.1.1, 1 1 = 1 and 1 =. By Theorem.1., 1 = 1. By Theorem.. (sum), we have that ( + 1) = + 1 = = ( + ) = + = 1 + = and 3

4 Finally, by Theorem..4 (quotient), we have that = 1( + 1) 1 ( + ) = 0 1 = 0. (c) 1 cos 0 sin If is close to zero (more precisely, if < π/), then (1 + cos ) 0. Therefore, 1 cos sin The above simplifies to = 1 cos sin 1 + cos 1 + cos. 1 cos sin (1 + cos ) = sin sin (1 + cos ). When we compute the it as approaches zero, the value of is never equal to zero. Therefore, it is valid to cancel the factors of sin in the above and say (by Theorem.7) that 1 cos sin = 0 sin cos. By Theorems.6.1 and.6., we have that 0 sin = sin 0 = 0 and 0 cos = cos 0 = 1. By Theorem.1.1, we have that 0 1 = 1. By Theorem.. (sum), we have that (1 + cos ) = 1 + cos = = Finally, by Theorem..4 (quotien), we have that 0 sin 1 + cos = 0 sin 0 (1 + cos ) = 0 = 0. The approach taken in these eercises in problem # is what might be called an aiomatic approach. The point of this eercise was to try to understand how often we implicitly use properties and theorems about its when we make computations. The upshot is that one these properties and theorems are established, computation becomes as easy as doing arithmetic in many cases. 4

5 3. This is an eample of a challenging problem. Only a few eam or quiz problems will be challenging, but here is a chance to see if you re ready! Solve eercise 70 in section.. Your solution should begin with a clear statement the problem. Eplain how to simplify algebraically if 1 < 0 or 0 < 1. To simplify the epression algebraically, we use the fact that = if 0 and = if 0. From this it follows that + 1 = + 1 if 1 and + 1 = ( + 1) if 1. Similarly, 1 = 1 if 1 and 1 = ( 1) if 1. If we compute a it as approaches 0, then we only need to consider -values close, but not equal to 0. Therefore, we may restrict our attention to -values such that 1 1 and 0. Using this, we have that = ( + 1) ( 1) = = provided that 1 1 and 0. Therefore, the above it is equal to. This can also be observed by testing values close to 0. Be careful to test both values to the right and the left when computing its of epressions which involve the absolute value. You can also find this it by sketching the graph, although this is more difficult for this problem. The graph is quite interesting however. Notice that if 1, then = ( + 1) ( 1) =. Similarly, if 1, then = ( + 1) ( 1) =. You can now sketch the graph by using the fact that the curve y = 1/ is a hyperbola. 4. Eercise 14 in section.: create a table of values for the function and use this to estimate the it. Then use a graphing utility to graph the function and confirm your result. 5

6 Let f() = ( 3 + 8)/( + ). We are to consider the it of f() as approaches. To get a good estimate, you most likely needed to use a calculator. And similarly, it seems like the graph would be difficult to sketch by hand. There is a way to proceed, however. We can factor ( 3 + 8) = ( + )( + 4). This is a special case of the following factorization: ( n+1 y n+1 ) = ( y)( n + n 1 y+ n y + + y n +y n 1 +y n ). The above can be directly verified by first distributing the and then the y and then look for cancellation in the resulting sum. Two special cases of the above are when y = 1, ( n+1 1) = ( 1)( n + n 1 + n ), and when y = 1, n+1 +1 = (+1)( n n 1 + n + +( 1) n +( 1) n 1 +( 1) n ). In our case, we have = ( + )( + 4) = provided that. Since in the it as approaches, we never let equal the value of, we can say that = + 4 = ( ) ( ) + 4 = Eercise 4 in section.. Find the it L of the epression below. Then find δ > 0 such that f() L < 0.01 whenever 0 < c < δ. ( 4 ) 4 It is clear that the it will be equal to 4 4 =. So L =. Since the it is as approaches 4, we see that c = 4. To solve this eercise, work backwards. We want to be able to conclude that f() L < Re-writing this, we see that we want to be able to conclude that 4 = <

7 This is what we would like to conclude based on a choice of value of a positive number δ. The role of δ is to specify how close to c = 4 the values of should be in order for this conclusion to be valid. For instance if δ = 1, then can be any number within 1 of c = 4. If, for instance, = 4.5, then f(4.5) L = = 0.5, so δ = 1 is not sufficient. We could continue to guess... or we can look for a way to see how the epression c < δ is related to the epression f() L. Since c = 4, I will try to factor out this epression: f() L = = 1 4. Since we want 1 4 < 0.01 I see that if 4 < (0.01) = 0.0, then this will be true. So, choose δ =