[Calum Campbell]

So, it seems like there are a lot of threads on the forums about members with math questions (some of which are made by me, I must confess), so let's see how a thread dedicated to them goes.

Let's start out with one I've been having trouble with in my calculus class.

A leaky bucket that weighs 5 lb and a rope of neglible weight are used to draw water from a well that is 493 ft deep. The bucket is filled with 38 lb of water and is pulled up at a rate of 5 ft/sec, but water leaks out of the bucket at a rate of 0.4 lb/sec. Find the work done in pulling the bucket to the top of the well (in ft-lb).

All I know on this one is that it helps to find the work done pulling the water in the bucket up the well and the work done pulling the bucket itself up individually, and then to add them together.

[NEGRO]

So, it seems like there are a lot of threads on the forums about members with math questions (some of which are made by me, I must confess), so let's see how a thread dedicated to them goes.

Let's start out with one I've been having trouble with in my calculus class.

A leaky bucket that weighs 5 lb and a rope of neglible weight are used to draw water from a well that is 493 ft deep. The bucket is filled with 38 lb of water and is pulled up at a rate of 5 ft/sec, but water leaks out of the bucket at a rate of 0.4 lb/sec. Find the work done in pulling the bucket to the top of the well (in ft-lb).

All I know on this one is that it helps to find the work done pulling the water in the bucket up the well and the work done pulling the bucket itself up individually, and then to add them together.

EDIT: Da Nang found a mistake in my calculation, I corrected it and striked the wrong parts through. I wrote the edited parts italic

Since this is linear we can do it as you suggest and superpose the work to bring the bucket up and the work to bring the water up

First some basics:
W = F * s ( --> Work = Force * path)
This works if the Force is constant.
If it changes, we need to make an integral:
W = INTEGRAL( F(t) * s)

The force to lift something within a gravitational field is F = m*g
If the mass changes, we have to use F(t) = m(t)*g
g is the gravity of earth.

So let's start:

The work for pulling the bucket up the well:
W_bucket = F*h = m*g*h = 5 lb * 32.174 ft/s^2 * 493 ft
W_bucket = 79311.375 ft^2*lb/s^2
Note: Since F is constant, there is no need for an integral here


The water:
Due to the water leaking out, the mass and therefor the force changes over time, so we need an integral:
W_water = INTEGRAL_OVER_TIME( F(t) * h) = INTEGRAL( m(t) * g * h * dt)
W_water = INTEGRAL( F(t) * dh) = INTEGRAL( m(t) * g * v * dt)
with
dh = v*dt
and
m(t) = 38lb - 0.4lb/s * t
As you can see, we have to make the mass depend on the time.
Here you can just take a linear approximation: m(t) is a line with t as parameter that starts with 38lb at t=0s and decreases by 0,4lb per second.

Let's have a look at the integration interval:
Lower border: 0s: that's when we start to pull the bucket up
Upper border: 493 ft / (5 ft/sec) = 98.6 secs. That's the time it takes to pull the bucket out of the well.

Now we have to be careful: it takes 98.6s to pull the bucket up, but after 95s all the water has leaked out: 0.4lb/s*95s = 38lb!
We've chosen a linear approximation for m(t), that means after 95 seconds, m(t) becomes negative (if you imagine it graphically, the line m(t) is under the x-axis then).
http://www.wolframalpha.com/input/?i=38-0.4t+from+0+to+100 is a graph of m(t), on the y-axis is the mass of water, on the x-axis the time.
So for 3.6s we have a negative mass. That cannot happen, so our upper integration border is 95 seconds! After that, the water is gone and there is no work done for the water anymore!
So the integration interval is [0; 95]s.

The water is gone after 95s, that means at a speed of 5 ft/s the water is gone after 475 ft.
So the height we have to consider when calculating the work done by pulling up the water is 475ft.

So we get:
W_water = 32.174 ft/s^2 * 475 ft * INTEGRAL( 38 lb-0.4 lb/s * t * dt) in the Intervall [0;95]s
W_water = 32.174 ft/s^2 * 475 ft * (38 t - 0,4*t^2/2)*lb in the intervall t=[0;95]
W_water =27585200 ft^2*lb/s^2
W_water = g*v*INTEGRAL( (38lb - 0.4lb/s * t) dt) in the intervall [0;95]s
W_water = 32.174 ft/s^2 * 5 ft/s * (38 t - 0,4*t^2/2)*lb in the intervall t=[0;95]s


And then you have to add W_bucket and W_water.
I always calculate in SI units, never in imperial ones, so I don't know if I have the correct values for g or so..

To sum this up:
For the first 475ft you have water in the bucket. So the work done is the sum of the bucket and the water.
After the 475ft there is no water in the bucket anymore and you just have the bucket to pull out.


I hope I haven't made a mistake!

[Nikki Morse]

So we get:
W_water = 32.174 ft/s^2 * 475 ft * INTEGRAL( 38 lb-0.4 lb/s * t * dt) in the Intervall [0;95]s
W_water = 32.174 ft/s^2 * 475 ft * (38 t - 0,4*t^2/2)*lb in the intervall t=[0;95]
W_water =27585200 ft^2*lb/s^2

The general idea is good but the integral's units don't add up. Instead of lb*(ft/s)², you're getting lb*ft²/s, which is not a unit for energy.

Now I won't plug in the numbers (imperial units! Aghh!) but here's how I would've done it symbolically.

Start from Newton's second law of motion. Since the bucket is moving at constant speed, then the sum of all external forces equal zero. Due to the nature of the problem we'll separate these forces for the different objects (up is positive direction):

T_bucket - G_bucket = 0 and T_water - G_water = 0 (1.1). The total work done bringing the bucket and water up is the sum of the work done by the tension forces. Clearly, T_bucket is constant thus the work it does is trivial, W_bucket = T_bucket*h, h = height/depth of the well.

T_water however is not, and thus requires a path integral W = ∫T_water·ds = ∫T_water*ds (1.2) ∵ T_water and ds have the same direction.

From (1.1): T_water = G_water = M(t)g = (M - c*t)g, c is the change of the water's mass over time, which is constant, M is the initial mass of the water.

(1.2): ∫T_water*ds = ∫(M - c*t)g*ds = g*∫(M - c*t)ds = g*∫(M - c*s/v)ds (1.3) ∵ t = s/v when v is constant

T_water will begin from an initial value of Mg and linearly decrease to 0, where it will no longer act on the system. ∴ The time interval is from t=0 to M-ct=0 ? t = M/c and the distance interval [0,Mv/c].

Thus for (1.3): W_water = g*₀∫^Mv/c(M - c*s/v)ds = g*(M(Mv/c) - ?c*(Mv/c)²/v) = g(M²v/c) - ?M²v/c) = ?*M²*g*v/c

∴ W_total = W_bucket + W_water = mgh + ?*M²*g*v/c, m is the mass of the bucket.

(Alternatively, since T_water decreases linearly you could've simply done it geometrically by multiplying the mean value Mg/2 by the distance that the force acts on, Mv/c, and skipped the calculus part altogether. But it's always good to practice one's calculus-fu from time to time.)

[Dalley hussain]

The general idea is good but the integral's units don't add up. Instead of lb*(ft/s)², you're getting lb*ft²/s, which is not a unit for energy.

Yep, there is a mistake, I found it and correct it in the original post, thanks!
If I substitute my equation now (with t=s/v) I get the same as you in (1.3).

[cheryl wright]

if bill has 6 tacos, and I have three, how much time does it take for me to decide to kill bill for his tacos?

ON TOPIC: Linear displacement. A movie takes 2 hours to finish. Find the displacement of the minute hand and the hour hand. I need to check my work...

[chinadoll]

I read this one on yaho a while back, you would be surprised how many people will get it wrong.

A bat and a ball cost $1.10 in total. The bat costs $1 more than the ball. How much does the ball cost?

[Victoria Bartel]

I read this one on yaho a while back, you would be surprised how many people will get it wrong.

A bat and a ball cost $1.10 in total. The bat costs $1 more than the ball. How much does the ball cost?

$0.05, if you're not including taxes.

[Zach Hunter]

;O Bout damn time ;D for a math thread

[Sudah mati ini Keparat]

if bill has 6 tacos, and I have three, how much time does it take for me to decide to kill bill for his tacos?

ON TOPIC: Linear displacement. A movie takes 2 hours to finish. Find the displacement of the minute hand and the hour hand. I need to check my work...

bumpy bump bump...

[REVLUTIN]

bumpy bump bump...

0. Displacement, if I recall, is the change in position of the object in question. Which, in this case, if the movie takes exactly two hours, would be zero since both hands would end up exactly where they started.

Edit: To answer the taco question, you wouldn't kill Bill. Exactly 3.14159 seconds after you decided you wanted his tacos, you would think of pie so you'd go and get some delicious apple pie instead. By the time you'd be done, Bill will have eaten his tacos so you'd have no reason to kill him.

[Tanika O'Connell]

ON TOPIC: Linear displacement. A movie takes 2 hours to finish. Find the displacement of the minute hand and the hour hand. I need to check my work...

If we're talking about clocks, then it only makes sense to use angular displacement Δθ = θ_f - θ_i unless you're only interested in the translational displacement of a certain point on each hand, in which case you're going to get a symbolic answer since you do need to know the distance from the rotational axis to that point to get a numerical answer.

Spoiler

It's only 0 for one of the hands

[Kelsey Hall]

If we're talking about clocks, then it only makes sense to use angular displacement Δθ = θ_f - θ_i unless you're only interested in the translational displacement of a certain point on each hand, in which case you're going to get a symbolic answer since you do need to know the distance from the rotational axis to that point to get a numerical answer.

Spoiler

It's only 0 for one of the hands

Yeah...

Spoiler

Don't know why I thought the hour hand would go all the way around. Oops. >_>

[Del Arte]

Hey math, why do you insist that you are more important than sociology and psychology?

^thinks we should start teaching both of those before math, albeit in a very, very simplified manner, like most subjects.

[Rachel Hall]

Does no one own a phone with a calculator on it? ^^

[Lisa Robb]

Hey math, why do you insist that you are more important than sociology and psychology?

^thinks we should start teaching both of those before math, albeit in a very, very simplified manner, like most subjects.

http://xkcd.com/435/

[Project]

http://xkcd.com/435/

I actually got a good laugh out of this.

[Meghan Terry]

I always calculate in SI units, never in imperial ones, so I don't know if I have the correct values for g or so..

g is not involved in this. lbs is force, not mass, so you could say it's "already" multipled by acceleration due to gravity (in imperial units.)

IE if I told you the bucket weighed 6 newtons, you wouldn't multiply by g. It's the same case here, the imperial unit for mass is the slug. If the question said the bucket had a mass of 5 slugs, only then would you need to multiply by imperial g to get a lb weight. And I assure you, you'll probably never be in that situation.

[Wayland Neace]

/threadrevive

Here's a tricky one! I got a lot of the way through it and I can't seem to find my mistake...

Find the length of the curve

x = 3y^(4/3) - (3/32)y^(2/3) from [-125 , 27]

From here you can take the derivative.

x' = 4y^(1/3) - (1/16)y^(-1/3)

Square both sides and add 1

(x')^2 + 1 = (4y^(1/3) - (1/16)y^(-1/3))^2 + 1

Now if we expand and complete the square it turns out that the expression simplifies to...

(x')^2 + 1 = (4y^(1/3) + (1/16)y^(-1/3))^2

Now take the square root, and that should be the function that we integrate from -125 to 27.

4y^(1/3) + (1/16)y^(-1/3)

I would think that this turns out to be

12 + 1/(16*3) + 20 + 1/(16*5)

However, this does not appear to be the right answer in the back of the book.... Where did I make my mistake?! This is driving me crazy!

[Mrs Pooh]

x = 3y^(4/3) - (3/32)y^(2/3)

Integrate
sqrt(1 + [f'(x)]^2)

where f'(x) is your derivative, over your interval.


Since you correctly found the derivative to be 4y^(1/3) - (1/16)y^(-1/3), square it for: (or not, you could enter it in your calculator without actually expanding, which is probably best since I did this quickly)

1/256y^(-2/3) - .5 + 16y^(2/3)

Now add one to that

1/256y^(-2/3) + .5 + 16y^(2/3)

Square root

sqrt(1/256y^(-2/3) + .5 + 16y^(2/3))

and that is the function you integrate from -125 to 27.

[aisha jamil]

Hey guys, I don't really need help on it, but I just need to know exactly what it's trying to ask:

Sand is falling from a rectangular box container whose base measures 40 inches by 20 inches at a constant rate of 300 cubic inches per minute.

How is the depth of the sand in the box changing?

Note: I'm pretty sure "falling from a rectangular box container" was meant to be "falling into a rectangular box container".

I might need help with the next part, I haven't started yet, but it looks like basic related rates stuff.

EDIT: Nevermind, I overlooked something at the beginning of the problem.

[Robert DeLarosa]

if bill has 6 tacos, and I have three, how much time does it take for me to decide to kill bill for his tacos?

ON TOPIC: Linear displacement. A movie takes 2 hours to finish. Find the displacement of the minute hand and the hour hand. I need to check my work...

2 seconds. That's how much time it would take to do the 5 Point Palm Exploding Heart Technique.

[Hannah Barnard]

Hey guys, I don't really need help on it, but I just need to know exactly what it's trying to ask:

Sand is falling from a rectangular box container whose base measures 40 inches by 20 inches at a constant rate of 300 cubic inches per minute.

How is the depth of the sand in the box changing?

Note: I'm pretty sure "falling from a rectangular box container" was meant to be "falling into a rectangular box container".

I might need help with the next part, I haven't started yet, but it looks like basic related rates stuff.

EDIT: Nevermind, I overlooked something at the beginning of the problem.

Yea, I don't know if you caught it already, but it is falling from the box, not into it. Imagine a tupperware container with a small hole in the bottom through which the sand is leaving the tupperware.