#$&* course Mth 174 5/25 3:25 pm query 12
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 + y^3 + y^4 + y^5 + … is a geometric series. We prove this by first factoring out a y^2 and getting y^2(1+y + y^2 + y^3 + … now we use the formula a/(1-y) and here a=y^2 so we have a/(1-y) = y^2/(1-y) cross multiplying we get a(1-y) = y^2(1-y) and a=y^2 so we can confirm that a indeed = y^2. Now we need to find our common ratio y^3/y^2 = y^4/y^3 by simplifying we get y=y so we know that y is what each term is multiplied by so y is the common ratio. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The common ratio is y^3 / y^2 = y^4 / y^3 = y. If we factor out y^2 we get y^2 ( 1 + y + y^2 + … ), which is in the standard form a ( 1 + r + r^2 + …) with a = y^2 and r = y. For | y | < 1 this series would converge to sum y^2 * Sum ( 1 + y + y^2 + . . . = y^2 * 1 / (1 - y). ** I understood the series to begin with y^2. However, if the series is understood to begin with a constant, then factoring out y^2 results in the first term becoming 1. Now the series is 1 + y + y^2 + ... Upon rereading the section, I couldn't find anything that says the first term a must be a constant. Must it be? The sum of a geometric series can be expressed in a number of different ways. Your text says that a + a x + a x^2 + ... sums to a / (1 - x). If you use this form then the series must start with constant number a. An alternative statement is even more restrictive: 1 + x + x^2 + ... = 1 / (1 - x). Of course this is the same as the preceding form, just dividing through by the non-zero constant a. You could express this series as y^2 ( 1 + y + y^2 + ...), in which case you would use a = y^2 and get the result a / (1 - y) = y^2 / (1 - y). Alternatively you could just say that since the expression 1 + y + y^2 + ... is 1 / 1 - y, then y^2 ( 1 + y + y^2 + ...) = y^2 / (1 - y). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###ok I think I understand correctly??? ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q problem 9.2.16 (4th edition 9.2.28) (3d edition 9.1.23) (formerly 9.4.24) bouncing ball 3/4 ht ratio YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are told a ball dropped from a height of 10 ft hits the floor and bounces. Each bounce is ¾ of the height of the previous bounce. After hitting the ball for the first time, we are told the ball rises to a height of 10*(3/4) or 7.5 ft. Part A: find an expression for the height to which the ball rises after it hits the floor for the nth time. We know that bounce 1= 10(3/4) Bounce 2 = 10(3/4)^2 Bounce 3 = 10(3/4)^3 So an expression for the nth bounce would be 10*(3/4)^n Part b: find an expression for the total vertical distance the ball has travelled when it hits the floor for the first, second, third, and fourth times. 1st bounce = 10+10(3/4) 10 is the initial vertical drop. = 17.5ft 2nd bounce = 10+10(3/4) + 10(3/4)^2 = 23.125 ft 3rd bounce = 10+10(3/4) + 10(3/4)^2+ 10(3/4)^3 = 27.344 ft 4th bounce = 10+10(3/4) + 10(3/4)^2+ 10(3/4)^3+ 10(3/4)^4 = 30.508 ft Part c. find an expression for the total vertical distance the ball has travelled for the nth time it hits the floor, express in closed form. Sn= a + a(x) + a(x^2) + . . . + a(x^(n-1)) with a = 10 and x = ¾ If the ball bounces from height h bounces to ¾ of the original height so 1st bounce = h+h(3/4) = 2h(3/4) = 3/2h 2nd bounce = 3/2 h(3/4) 3rd bounce = 3/2 h * (3/4)^2 4th bounce = 3/2 h * (3/4)^3 So the total sum of the first 4 bounces is 3/2h+ [3/2 h(3/4)+ 3/2 h * (3/4)^2+ 3/2 h * (3/4)^3+. . . ] Factoring out common terms we get 3/2h + (3/2h*(3/4)) * [1+(3/4) + (3/4)^2] The total sum of n bounces found by using 1/(1-x) which would be 1/(1-(3/4)) which is 4 multiplied by (3/2)h * (3/4) which is 4*(9/8)h which is 9/2h which we would add to the initial drop of (3/2)h which would give us (12/2)h or 6 h.
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Given Solution: `a** If the ball starts from height h: • It falls to the floor then bounces up to 3/4 of its original height then back down, covering distance 3/2 h on its first complete bounce. • Then it bounces to 3/4 of that height, covering in the round trip a distance of 3/4 ( 3/2 h). • Then it bounces to 3/4 of that height, covering in the round trip a distance of 3/4 [ 3/4 ( 3/2 h) ] = (3/4)^2 * (3/2 h). • On the next bounce the round trip will be 3/4 of this, or (3/4)^3 * (3/2 h). • All the bounces give us a distance of (3/4) * (3/2 h) + (3/4)^2 * (3/2 h) + (3/4)^3 * (3/2 h) + ... = [ 3/4 ( 3/2 h) ] * ( 1 + 3/4 + (3/4)^2 + ... ) = [9/8 h] * ( 1 + 3/4 + (3/4)^2 + ... ) = (9/8 h) ( 1 / ( 1 - 3/4) ) = 9/2 h. • There is also the initial drop h, so the total distance is 11/2 h. But the total distance isn't the question. The question is how long it takes the ball to stop. • The time to fall distance y is ( 2 * y / g ) ^ .5, where g is the acceleration of gravity. • This is also the time required to bounce up to height h. • The distances of fall for the complete round-trip bounces are 3/4 h, (3/4)^2 h, (3/4)^3 h, etc.. • So the times of fall are ( 2 * 3/4 h / g ) ^ .5, ( 2 * (3/4)^2 h / g ) ^ .5, ( 2 * (3/4)^3 h / g ) ^ .5, ( 2 * (3/4)^4 h / g ) ^ .5, etc.. • The times for the ‘complete’ round-trip bounces will be double these times; the total time of fall will also include the time sqrt( 2 h / g) for the first drop. We can factor the ( 3/4 ) ^ .5 out of each expression, getting times of fall (3/4)^.5 * ( 2 * h / g ) ^ .5 = `sqrt(3)/2 * ( 2 * h / g ) ^ .5, ( sqrt(3)/2 ) ^ 2 * ( 2 * h / g ) ^ .5, ( sqrt(3)/2 ) ^ 3 * ( 2 * h / g ) ^ .5, etc.. Adding these terms up, multiplying by 2 since the ball has to go up and down on each bounce, and factoring out the (2 * h / g ) ^ .5 we get total time = 2 * (2 * h / g ) ^ .5 * [ ( sqrt(3)/2 ) + ( sqrt(3)/2 ) ^ 2 + ( sqrt(3)/2 ) ^ 3 + ( sqrt(3)/2 ) ^ 4 + ... ] . The expression in brackets is a geometric series with r = ( sqrt(3)/2 ), so the sum of the series is sum of series = 1 / ( 1 - ( sqrt(3)/2 ) ) h and g are given by the problem so 2 * (2 * h / g ) ^ .5 is just a number, easily calculated for any given h and g. We also add on the time to fall to the floor after the drop, obtaining total time sqrt(2 h / g) + 2 * (2 h / g)^.5 (1 / (1 - sqrt(3)/2) ). Rationalizing the last fraction and factoring out sqrt(2 h / g) we have • total time = sqrt(2 h / g) * ( 1 + 2 * (4 + 2 sqrt(3) ) = sqrt(2 h / g ) ( 9 + 4 sqrt( 3) ).** 22:45:05
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how do you verify that the ball stops bouncing after 1/4 `sqrt(10) + 1/2 `sqrt(10) `sqrt(3/2) (1 / (1-`sqrt(3/4)) sec? You verify that the ball stops bouncing after this time by solving but when I solve I don’t get zero I get .7906 + 1.9365(1/.1339) **&&&Which is 15.2529 I am not sure how to answer this question????***** 22:45:05 22:48:05
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What geometric series gives the time and how does this geometric series yield the above result? **&&The geometric series that gives the time is sqroot (2h/g)+ 2*squareroot of ((2*h)/g)*(1/(1-sqroot (3/2))) Which yields the total time to stop by summing all of the roundtrip drops which requires multiplying by 2 and also adding the initial bounce to the geometric series???****
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22:48:05 22:49:09
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How far does the ball travel on the nth bounce? **&& for the nth bounce we use a(1-x^n) / (1/x) which would be 1(1-squareroot (3/2)^n)/(1-squareroot (3/2)) So we have 1(1-squareroot (3/2)^n)/(1-squareroot (3/2)) as how far the ball would travel on the nth bounce????****
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22:49:09 22:49:32
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How long does it takes a ball to complete the nth bounce? &&**It takes the ball sqroot (2h/g)+ 2*squareroot of ((2*h)/g)* 1(1-squareroot (3/2)^n)/(1-squareroot (3/2)) to complete the nth bounce????**** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### for the total time to stop I get squareroot of (2h)/g * [1+2 * (1/1-squareoot of (3/2))] Rather than sqrt(2 h / g ) ( 9 + 4 sqrt( 3) ). Am I close to the correct answer??? I am also not sure If I answered all of the supplemental questions correctly I’m not sure if I answered correctly for how long it takes the ball to complete the nth bounce and how far the ball travels on the nth bounce????
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Given Solution: `a **The integral was: int(1/x, x, 1, 4n-3). The integral does not converge. ** `q Explain in terms of a graph how you set up rectangles to represent the series, and how you oriented these rectangles with respect to the graph of your function in order to prove your result. `a ** STUDENT ANSWER: I didn't really set up a graph, but if I had to it would be a series of rectangles which get smaller with each successive one. The area of each rectangle would be equal to the respective terms in the series. So, the area of the first would be 1, the second would be 1/5, the third would be 1/9, and so on. The series of rectangles would go on forever and thus the series would not converge.** INSTRUCTOR SOLUTION *&*& The denominators are 1, 5, 9, 13, ... . These numbers increase by 4 each time, which means that we must include 4 n in the expression for the general term. For n = 1, 2, 3, ..., we would have 4 n = 4, 8, 12, ... . To get 1, 5, 9, ... we just subtract 3 from these numbers, so the general term is 4 n - 3. For n = 1, 2, 3, ... this gives us 1, 5, 9, ... . So the sum is sum( 1 / ( 4n - 3), n, 1, infinity). Knowing that the integral of 1 / (4 x) diverges we set up the rectangles so they all lie above the y = 1 / (4x) curve. Since 1 / (4n - 3) > 1 / (4 n) we see that positioning the n = 1 rectangle between x = 1 and x = 2, then the n = 2 rectangle between x = 2 and x = 3, etc., gives us a series of rectangles that lie above the y = 1 / (4x) curve for x > 1. Since an antiderivative of 1 / (4x) is 1/4 ln | x |, which as x -> infinity approaches infinity, the region beneath the curve has divergent area. Since the rectangles lie above the curve their area also diverges. Since the area of the rectangles represents the sum of the series, the series diverges. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### I took a different approach on the graph, would my approach work, or do I need to graph 1/(4x) instead of 1/(4x-3)????
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Given Solution: `a **The integral was: int(1/x, x, 1, 4n-3). The integral does not converge. ** `q Explain in terms of a graph how you set up rectangles to represent the series, and how you oriented these rectangles with respect to the graph of your function in order to prove your result. `a ** STUDENT ANSWER: I didn't really set up a graph, but if I had to it would be a series of rectangles which get smaller with each successive one. The area of each rectangle would be equal to the respective terms in the series. So, the area of the first would be 1, the second would be 1/5, the third would be 1/9, and so on. The series of rectangles would go on forever and thus the series would not converge.** INSTRUCTOR SOLUTION *&*& The denominators are 1, 5, 9, 13, ... . These numbers increase by 4 each time, which means that we must include 4 n in the expression for the general term. For n = 1, 2, 3, ..., we would have 4 n = 4, 8, 12, ... . To get 1, 5, 9, ... we just subtract 3 from these numbers, so the general term is 4 n - 3. For n = 1, 2, 3, ... this gives us 1, 5, 9, ... . So the sum is sum( 1 / ( 4n - 3), n, 1, infinity). Knowing that the integral of 1 / (4 x) diverges we set up the rectangles so they all lie above the y = 1 / (4x) curve. Since 1 / (4n - 3) > 1 / (4 n) we see that positioning the n = 1 rectangle between x = 1 and x = 2, then the n = 2 rectangle between x = 2 and x = 3, etc., gives us a series of rectangles that lie above the y = 1 / (4x) curve for x > 1. Since an antiderivative of 1 / (4x) is 1/4 ln | x |, which as x -> infinity approaches infinity, the region beneath the curve has divergent area. Since the rectangles lie above the curve their area also diverges. Since the area of the rectangles represents the sum of the series, the series diverges. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ### I took a different approach on the graph, would my approach work, or do I need to graph 1/(4x) instead of 1/(4x-3)????