Calculus help

Monroe Claxton

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My kid will be taking a calculus 2 final soon and her teacher gave her a bunch of practice problems. Only one made any sense and it don't make no sense.

A ball is dropped from a height of 2 meters. The height after each bounce is 3/4 of the previous height. Find the total distance traveled by the ball.

How can you find the distance when the ball never stops bouncing?
 
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Ron Mehico

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I would assume they mean find the total distance traveled after the ball has stopped bouncing there Larry.
 

LineSkiCat14

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I'm just here to ask, how can school's give multiple classes of calculus, which maybe 3% of the class will use.. while ZERO classes of Personal Finance are offered, which EVERYONE could use?

Tell your kid to just cheat his way through this useless class (unless he aspires to go into Science/engineering) and focus his efforts on something else.
 

UKGrad93

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Its a geometric series.

1 bounce = 2 m

2 bounce = 2(3/4) + 2(3/4) = 3 m

3 bounce = 2(3/4)(3/4) + 2(3/4)(3/4) 2.25 m

N bounce = 2h (3/4)^n

Total = 2 + Sum of [20 *(3/4)^n] from zero to infinity

(copy & paste from equation writer doesn't work)

As n gets bigger, the term that you are adding gets smaller and smaller, eventually it approaches zero.

2 + [(2*2) / (1-(3/4)] = 18 m
 

UKGrad93

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I'm just here to ask, how can school's give multiple classes of calculus, which maybe 3% of the class will use.. while ZERO classes of Personal Finance are offered, which EVERYONE could use?

Tell your kid to just cheat his way through this useless class (unless he aspires to go into Science/engineering) and focus his efforts on something else.
I guess I'm in the 3%, because much of my work relies on calculus principles. I never took a finance class, but compounding interest is best solved by using calculus.
 
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UKserialkiller

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Its a geometric series.

1 bounce = 2 m

2 bounce = 2(3/4) + 2(3/4) = 3 m

3 bounce = 2(3/4)(3/4) + 2(3/4)(3/4) 2.25 m

N bounce = 2h (3/4)^n

Total = 2 + Sum of [20 *(3/4)^n] from zero to infinity

(copy & paste from equation writer doesn't work)

As n gets bigger, the term that you are adding gets smaller and smaller, eventually it approaches zero.

2 + [(2*2) / (1-(3/4)] = 18 m


Damn 93.
 

LineSkiCat14

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I guess I'm in the 3%, because much of my work relies on calculus principles. I never took a finance class, but compounding interest is best solved by using calculus.

Someone who understands Calc probably doesn't need Credit and Debt 101.
 

LineSkiCat14

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I'm still advocating the cheating idea, btw. And if she gets caught just say "Well our future presidents can lie and cheat? Why can't I?"
 
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allabouttheUK

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Its a geometric series.

1 bounce = 2 m

2 bounce = 2(3/4) + 2(3/4) = 3 m

3 bounce = 2(3/4)(3/4) + 2(3/4)(3/4) 2.25 m

N bounce = 2h (3/4)^n

Total = 2 + Sum of [20 *(3/4)^n] from zero to infinity

(copy & paste from equation writer doesn't work)

As n gets bigger, the term that you are adding gets smaller and smaller, eventually it approaches zero.

2 + [(2*2) / (1-(3/4)] = 18 m

Dude?! You can't talk like that in here! He just wanted some calculus help, dayum!
 
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LineSkiCat14

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I'm still advocating the cheating idea, btw. And if she gets caught just say "Well our future presidents can lie and cheat? Why can't I?"

Note: When I say "Presidents".. we all know which one we are talking about, right?

And before the "Polidicks thread is that way" comment... It's a damn Calculus thread.. it should be ruined by politics.
 

funKYcat75

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My kid will be taking a calculus 2 final soon and her teacher gave her a bunch of practice problems. Only one made any sense and it don't make no sense.

A ball is dropped from a height of 2 meters. The height after each bounce is 3/4 of the previous height. Find the total distance traveled by the ball.

How can you find the distance when the ball never stops bouncing?
You sure there's not a pic showing how many bounces the teacher wants? The ball will stop bouncing, but you probably gotta deal with gravity and all that jazz.
 
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KentuckyStout

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The answer is 14 from the answer key. How did you even get 18? How come it is not like 500?

The answer is actually 13.890243 but I guess 14 works as well.

The initial drop is from 6.56 feet and the first bounce reaches a height of 4.92 feet.

To put it in layman's terms and convert it to imperial numbers - the overall distance traveled is just over 45' - 6".
 
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UKGrad93

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The answer is 14 from the answer key. How did you even get 18? How come it is not like 500?
Its a geometric series.

1 bounce = 2 m

2 bounce = 2(3/4) + 2(3/4) = 3 m

3 bounce = 2(3/4)(3/4) + 2(3/4)(3/4) 2.25 m

N bounce = 2h (3/4)^n

Total = 2 + Sum of [20 *(3/4)^n] from zero to infinity

(copy & paste from equation writer doesn't work)

As n gets bigger, the term that you are adding gets smaller and smaller, eventually it approaches zero.

2 + [(2*2) / (1-(3/4)] = 18 m

Actually I had a mistake, should have been:

-2 + [(2*2) / (1-(3/4)] = 14 m

The ball will bounce a shorter and shorter distance. In the real world, things like friction are at work and the ball will lose momentum and stop. As far as the problem goes, it keeps bouncing, but those bounces become really tiny, at infinity bounces, it approaches zero height. Calculus works by taking limits.

A fraction raised to some power gets smaller. 3/4 = 0.75. 0.75^2 = 0.5625, 0.75^3 = 0.422, 0.75^infinity ~=0.

You can check it out with a scientific calculator. Enter a number less than 1. Then push the y^x button and enter larger and larger numbers. The answer gets smaller and smaller. Everntually zero.
 

LineSkiCat14

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There's actually a better way to solve this. "55378008" and flip that sucker upside down. Then hit the teacher with one of these:

 
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Tinker Dan

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Not Calculus, but my daughter was taking an ethics class in college and complaining how much time the assignments took. I suggested she just get the Cliff's Notes for it.

She did NOT see the hilarity in my suggestion. She got that **** from her mother.....
 

Ohiocatfan826

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Its for calculus, this a limit problem ie there is an equation where the upper limit is the initial height and the lower limit is zero, use the equation the person above laid out above and plugs in the upper and lower limit. Have your kid go back to that chapter and review ...this is actually not one of the more difficult concepts of Calc 1
 
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UKGrad93

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Its for calculus, this a limit problem ie there is an equation where the upper limit is the initial height and the lower limit is zero, use the equation the person above laid out above and plugs in the upper and lower limit. Have your kid go back to that chapter and review ...this is actually not one of the more difficult concepts of Calc 1
+1

I'm starting to think that very few paddock posters took calculus.
 

gamecockcat

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UKgrad93 had the general gist of the solution, just didn't do the math correctly as he pointed out. It is a pretty basic calculus concept, that of limits. As numbers are plugged into a function, the sum of all the answers approaches some number - that number is the limit.

For this problem, when dropped the ball falls 2 m. On the first bounce it returns 3/4 of the original height or 1.5 m. Then descends the 1.5 m to the second bounce. It would then reach a height of 1.125 m, etc. Each bounce gets lower and lower until, on the 1,000th bounce or 1,000,0000th bounce, the height it reaches is a small fraction of a meter. Hence, the sum APPROACHES 14 total meters but would never, in theory, actually reach it. In theory, the ball would never stop bouncing even a miniscule fraction of a meter. Hence the concept of limits. The sum will never go above 14 in this example and will only approach 14 no matter how many bounces you want to calculate.

Having taken 4 semesters of calculus and spent almost 10 years as an electrical engineer in a plant, I can tell you I have NEVER used it since college. If I had ever been in design, I most likely would have used it at some point. Now, that I'm in financial planning, I use basic finance and personal finance concepts every day. So, why are engineers made to take so much damn calculus?
 
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