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 Post subject: To Smart people (and to future Chemists & Mathematicians)
PostPosted: Thu Apr 03, 2008 6:59 pm 
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Edit: April 12, 2008 with new question:

A) Solve for x: e^(2x) - e^x + 5 = 11

B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground.

C) What is the molarity (M) of a solution of 50% NaOH?

D) What is the concentration of a strong acid with a pH of 1.2?





Old Questions - Solved
-----------------------------------------------------


1) A bicycle traveled 1 mile. During that period, the wheel on the bicycle made 1260 rotations. What is the diameter of the wheel?

2) A car is traveling at a velocity (in ft/sec) represented by the equation f(t)= 3t + 8 where t is in seconds. How far did the car travel after 24 seconds?

3) Write the equation of the function if:
f(1)=11
f(2)=20
f(3)=29
f(4)=38
f(5)=47
...

Thanks.

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Last edited by ThiefzV2 on Mon Apr 14, 2008 1:11 am, edited 6 times in total.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:09 pm 
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((1 mile * 63360in) / 1260 rotations)/3.14 pie = 16.014in diameter

80? for 2


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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:09 pm 
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playing too much sro made ur math brain dead?

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:12 pm 
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InsertName wrote:
playing too much sro made ur math brain dead?



+1 dude :P

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:13 pm 
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1) 2 * PI * r = The circumference of the circle. Here, r is the radius.

1 mile = 63360 inches.

2 * PI * r = 63360/1260, the latter being there because I divided the number of inches in a mile to the number of rotations performed.

2 * PI * r = 50,285 ; PI is roughly 3.1415, and 2*r = 16. Consequently, the Wheel has the Diameter of 16 centimeters. Hope the calculations are all right.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:20 pm 
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3) f(x) = 9x+2

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:20 pm 
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Ryoko wrote:
((1 mile * 63360in) / 1260 rotations)/3.14 pie = 16.014in diameter

80


That's what I got too. My answer was 63360/(1260pi) since my teacher wants an exact answer without rounding. But I wasn't sure if I set the problem up right though.

What is the 80 btw?

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:21 pm 
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Stress wrote:
1) 2 * PI * r = The circumference of the circle. Here, r is the radius.

1 mile = 63360 inches.

2 * PI * r = 63360/1260, the latter being there because I divided the number of inches in a mile to the number of rotations performed.

2 * PI * r = 50,285 ; PI is roughly 3.1415, and 2*r = 16. Consequently, the Wheel has the Diameter of 16 centimeters. Hope the calculations are all right.



Thanks all of you. Did anyone solve #2?

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:23 pm 
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ThiefzV2 wrote:
Ryoko wrote:
((1 mile * 63360in) / 1260 rotations)/3.14 pie = 16.014in diameter

80


That's what I got too. My answer was 63360/(1260pi) since my teacher wants an exact answer without rounding. But I wasn't sure if I set the problem up right though.

What is the 80 btw?


80 was a guess at number 2; not sure why stress switched from inches to centimeteres


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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:24 pm 
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Ilove2quack wrote:
3) f(x) = 9x+2


That's right. Just figured it out too. Here's the proof to that.
Suppose f(x) = a*x + b

Let's replace this into the first two equations:

a*1 + b = 11
a*2 + b = 20.

Multiply the first equation with (-1) and then add the two equations. The result is:

- a + 2*a - b + b = -11 + 20 => a=9. Replacing a=9 into the first equation, gives b=2.

:)

EDIT: here in Europe the centimeter is the basic unit of measure. Didn't switch to centimeters while I calculated it though. I got exactly the same result as Ryoko.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:26 pm 
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2) f(t)= 3t + 8
f(24)= 80 ft/s

x=v.t (=) x = 80*24 (=) x = 1920 ft


not sure if it's right, i work with metters here >_< (btw 1920 ft = 585,216m)

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:32 pm 
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Ilove2quack wrote:
2) f(t)= 3t + 8
f(24)= 80 ft/s

x=v.t (=) x = 80*24 (=) x = 1920 ft


not sure if it's right, i work with metters here >_< (btw 1920 ft = 585,216m)


Meters or feet don't make any difference here. The basic formula of the velocity (v = x / t) is, of course, correct, whatever the unit of measure is.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:35 pm 
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rawr :wink:

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:37 pm 
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Ryoko wrote:
ThiefzV2 wrote:
Ryoko wrote:
((1 mile * 63360in) / 1260 rotations)/3.14 pie = 16.014in diameter

80


That's what I got too. My answer was 63360/(1260pi) since my teacher wants an exact answer without rounding. But I wasn't sure if I set the problem up right though.

What is the 80 btw?


80 was a guess at number 2; not sure why stress switched from inches to centimeteres


I see. I'm sure that's not correct since the formula only gives the velocity that car is traveling at 24 seconds. It's asking how far the car has traveled alltogether after 24 seconds.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:39 pm 
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Ilove2quack wrote:
2) f(t)= 3t + 8
f(24)= 80 ft/s

x=v.t (=) x = 80*24 (=) x = 1920 ft


not sure if it's right, i work with metters here >_< (btw 1920 ft = 585,216m)


That can't be right. The car velocity is increasing every second, so you simply can't take the velocity of the car at t=24 (80 ft/sec) and multiply it by 24. LOL. But however, what you did what find the upper limit. But that is NOT the answer.


Stress wrote:
Meters or feet don't make any difference here. The basic formula of the velocity (v = x / t) is, of course, correct, whatever the unit of measure is.


Yeah, but the problem stated it in ft/sec though. :)

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 7:52 pm 
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ThiefzV2 wrote:
Ilove2quack wrote:
2) f(t)= 3t + 8
f(24)= 80 ft/s

x=v.t (=) x = 80*24 (=) x = 1920 ft


not sure if it's right, i work with metters here >_< (btw 1920 ft = 585,216m)


That can't be right. The car velocity is increasing every second, so you simply can't take the velocity of the car at t=24 (80 ft/sec) and multiply it by 24. LOL. But however, what you did what find the upper limit. But that is NOT the answer.



LOL, true my bad. Well if you do the primitive of the velocity equation you get the position equation:

x(t)=(3/2)t²+8t
x(24)=1056ft

T_T.

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Last edited by Ilove2quack on Thu Apr 03, 2008 8:25 pm, edited 1 time in total.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:00 pm 
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Write this down on your homework, for epic lulz:
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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:04 pm 
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f(1) = 11
f(2) = 14
f(3) = 17

etc.

f(24) = 80

v1 = x1 / t1
v2 = x2 / t2

etc.

v24 = x24 / t24

Here, the v-s are the 24 different velocities of the car, in the 24 different seconds of movement. The t-s are the time durations. Here, they are all equal to 1, because we're taking the speed of the car into consideration, in every separate section of 1 second each.

That is, we are dividing the total 24 seconds into sections of 1 second each, and we calculate the distance covered each second, and then sum them up.

t1 = t2 = ... = t24 = 1 second.

Conclusively, v1 + v2 + v3 + ... + v24 = x1 + x2 + x3 + ... x+24.
f(1) + f(2) + ... + f(24) = 3(1 + 2 + 3 ... + 24) + 24*8 = 1092 feet.

That's my best guess at it. It's basic math. Calculus can also be used, I think. Don't know if it's correct though.

EDIT: why did Cruor's integral give a different result?

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:19 pm 
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I wish you guys would help me with my school work.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:21 pm 
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Stress wrote:
EDIT: why did Cruor's integral give a different result?


no idea. i did what you did and add it up and got 1092.


dom wrote:
I wish you guys would help me with my school work.


what kind? lol

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:28 pm 
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Oh.. duh at 3...

Just add 9 everytime you go up..

11 + 9 = 20..
20 + 9 = 29...

So, after 47 comes 56.. then comes 65, 74, 83, 92, 101...

Just continue with counting one 9 up everytime...

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:30 pm 
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Stress wrote:
EDIT: why did Cruor's integral give a different result?

Approximation:
Image

Exact:
Image

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:53 pm 
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Icealya wrote:
Oh.. duh at 3...

Just add 9 everytime you go up..

11 + 9 = 20..
20 + 9 = 29...

So, after 47 comes 56.. then comes 65, 74, 83, 92, 101...

Just continue with counting one 9 up everytime...


Well duh! It's formulating an equation, not recognizing what number to add to get the next one.



Cruor wrote:
Stress wrote:
EDIT: why did Cruor's integral give a different result?

Approximation:
Image

Exact:
Image


Can you explain like Stress does to all of his solutions?

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 8:56 pm 
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ThiefzV2 wrote:
Well duh! It's formulating an equation, not recognizing what number to add to get the next one.


Could you say that in english please??

I can't speak french....

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 9:33 pm 
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Icealya wrote:
ThiefzV2 wrote:
Well duh! It's formulating an equation, not recognizing what number to add to get the next one.


Could you say that in english please??

I can't speak french....


Good then GTFO.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 9:39 pm 
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Icealya wrote:
ThiefzV2 wrote:
Well duh! It's formulating an equation, not recognizing what number to add to get the next one.


Could you say that in english please??

I can't speak french....

Hes saying that, the question isn't asking what number is added to get to the next one. Hes asking the equation, so you can find the nth value. Using your method, it would take a while to find the nth value.

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 10:03 pm 
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ThiefzV2 wrote:
Cruor wrote:
Stress wrote:
EDIT: why did Cruor's integral give a different result?

Approximation:
Image

Exact:
Image


Can you explain like Stress does to all of his solutions?

Ilove2quack took the primitive of your velocity function to find the position function, then evaluated it at 24 seconds. This works so long as the function is linear and always positive, and assuming you start at 0 seconds.

I took the definite integral of your velocity function. This is essentially the same as what Ilove2quack did. Technically I should have used the absolute value of the velocity function, but it is always positive on that interval so it may be omitted. Note that the definite integral can also be expressed as the limit of a Riemann Sum:
Image is equivalent to Image

Stress approximated the displacement using a right-endpoint Riemann Sum with 24 subdivisions.

The total displacement of a velocity function can be found by finding the area beneath it on a graph. With Stress' solution, you are actually finding the area beneath a stair-step function approximating the velocity function, and so you end up with an answer too large. The integral on the other hand uses an infinite number of subdivisions, and so it is exact.

Of course, your teacher probably expects a geometric calculation using the graph of the velocity function; after all, it is essentially just a triangle.

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 Post subject: Re: Math HW Help
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Cruor wrote:
I took the definite integral of your velocity function. This is essentially the same as what Ilove2quack did. Technically I should have used the absolute value of the velocity function, but it is always positive on that interval so it may be omitted. Note that the definite integral can also be expressed as the limit of a Riemann Sum:
Image is equivalent to Image


How did you convert the integral to the Riemann Sum using the infinite series? What are the steps?

Quote:
Of course, your teacher probably expects a geometric calculation using the graph of the velocity function; after all, it is essentially just a triangle.


What do you mean, and how is a quadratic equation "essentially just a triangle"?

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 Post subject: Re: Math HW Help
PostPosted: Thu Apr 03, 2008 11:33 pm 
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ThiefzV2 wrote:
How did you convert the integral to the Riemann Sum using the infinite series? What are the steps?

First you find the length of the subdivisions. This is given by:
Image
Where b is the upper bound of the interval, a the lower bound, and n the number of subdivisions.

Then you find an equation for x at any given subdivision:
Image

Finally, you plug everything in:
Image
The limit is of course only necessary if you plan on having an infinite number of subdivisions.

In this case, here are the steps you would go through:
  1. Image
  2. Image
  3. Image
  4. Image

ThiefzV2 wrote:
What do you mean, and how is a quadratic equation "essentially just a triangle"?

That is not a quadratic function. It is linear. Plotting it on a graph, you will find that it is a straight line with a slope of 3 that's shifted 8 units above the x-axis. Therefore, you can subdivide the area beneath it into a rectangle with base 24 and height 8, and a triangle with base 24 and height 72.

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 Post subject: Re: Math HW Help
PostPosted: Fri Apr 04, 2008 6:39 pm 
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Cruor wrote:
Therefore, you can subdivide the area beneath it into a rectangle with base 24 and height 8, and a triangle with base 24 and height 72.


Can you demonstrate what you mean?

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