# Calcus homework help because I'm too dumb to help my kid, please.

Discussion in 'General Discussion' started by digitalwanderer, Oct 8, 2019.

1. ### pcchen Moderator ModeratorVeteranSubscriber

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Congratulations! #41
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hey guys!!!

i wanted to say thank you all for the encouragement and support last semester, i'm still in disbelief that i passed and that i'm in calc ii right now. i have a peer mentor now and um. declared my major a little over 2 weeks ago (mechanical engineering major with applied math minor for now, math part might evolve as time goes on... we'll see)...

but i'm back with another question. i have it actually figured out, but i still don't quite understand WHY it had to be done this way. it might just be because in my lecture today we went over it at the very end, and my professor might've forgotten this step? dunno. either way, since i'm on a school computer (too lazy to get my laptop out), i'm gonna type out the problem and whatnot. it's a work problem.

"An aquarium 2m long, 1m wide, and 1m deep is full of water. Find the work needed to pump half of the water out of the aquarium. Use the fact that the density of the water is 1000 kg/m^3."

the way we were taught in class was with this OTHER problem:
"A cylindrical tank with a total height of 10 ft and a radius of 4 ft is full of water up to 8 ft. Find the work required to empty the tank by pumping all water to the top of the tank. The density of the water is 62.5 lb/ft^3."

we were instructed to split the tank up into horizontal layers and use the formula W = (force to lift a layer)*(distance by which the layer is lifted). the force to lift a layer breaks down into (volume of layer)*(density of layer), so the complete formula is (volume of layer)*(density of layer)*(distance by which the layer is lifted).

plugging the given numbers in, you get pi*(4^2)*(delta y)*(62.5)*(10-yi). you write a reimann sum, rearrange it a bit, and then make a definite integral. yay. i don't know how to type out an integral, but if needed, i can definitely take pictures of what the hell i mean later.

this is where my problem comes up. with the original problem, with the aquarium (that i assume is rectangular?), after struggling with it and, um, looking up how to do it, everything pointed to me needing to use gravity (9.8).
ex) W = 2*9.8*delta y*1000*(1-yi) CORRECT
W= 2*delta y*1000*(1-yi) NO

i saw this, thought it was bullshit, checked the back of the book for the answer, and it was the number that got multiplied by 9.8. after just thinking about it, i do understand that... gravity is a thing. it's important. especially in this scenario. however, i don't quite understand how i was supposed to know that i needed gravity and what the value of it means. from googling shit, i can see that in the end, we need the mass of the layer the force, since work is equal to the... ok. i know that gravity is acceleration but i'm not sure about acceleration. i know i'm in calculus, but i've never taken a physics class in my life, so all the physics i'm encountering now is really foreign to me (i'm in two straight-up engineering classes now... we're building a bridge in one of them).

gotta run, being driven home and don't wanna hold the driver up. thanks for any help lol

#42
3. ### pcchen Moderator ModeratorVeteranSubscriber

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Disclaimer: I am not good at physics It looks to me that your question is why gravity (g) is involved in this question. The answer is that in the formula W you mentioned, there's an element "force to lift a layer". In an environment with gravity, you need the force to counter the gravity in order to lift the mass. That's why g is involved.

#43
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4. ### 3dilettante LegendAlpha

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Work is described as being force applied over a displacement. In terms of units, it is measured in Joules, which are Newton*meters, which are kg*m/s^2 * m = kg*m^2/s^2.
Just from the equations, there is a lack of agreement in units if g is not included.
W needs to be have quantities of meters, kilograms, and seconds.
W (kg*m2/s2) = 2m^2 * delta y m * 9.8(m/s2) * 1000 kg/m^3 * (1-yi)m
Without g, it is immediately obvious there is is no s^-2, and so no Joules.
Next, getting the mass of water from kg/m^3 cancels out too many units of length to give the necessary m^2 in kg*m^2/s^2.

As an attempt to give a reason outside of the equation itself, one thing to look at is what is being calculated, or how it is treated in experiments. Usually, scenarios in physics where objects are being lifted against gravity are going for a minimum amount of work needed to lift the mass a given distance. In that case, there is an attempt to apply an upward force sufficient to get the mass moving up, but at a constant velocity from t=0 to the end of the time interval.
No acceleration means no net force, meaning any force applied is counterbalanced by gravity. The work the lifter performs can be calculated in terms of the known mass and acceleration due to gravity that has been perfectly matched.
There was an initial acceleration or excess force, although that can be excluded if measurement starts after, and in a theoretical setup can be an amount of excess force that can be arbitrarily reduced towards zero. (Newton's 2nd Law, an object's motion will not change in the absence of a net force.)

In the absence of gravity, that trivial initial bit of force that can be reduced arbitrarily close to zero would be able to give the same displacement as a pump laboring against the acceleration due to gravity.
Since work is force over distance, a lack of gravity means force goes to 0, and so essentially no work is necessary to lift the water. In the problem scenario, there is force applied to continually keep gravity from accelerating the water downwards, and it is applied over the vertical displacement.

Why this is useful comes when asking how much gravitational potential energy is in this system due to the water being raised, which would also be the amount kinetic energy measured if the water were allowed to fall back to the initial level. The quantities of energy are not coincidentally measured in Joules. A change in the amount of energy in a specific system, or a transformation of energy from one form or another is known as work.

#44

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pcchen- yeah, that's what i was asking. my brain was kinda fried and i couldnt figure out the best way to put it. it also turns out that when you're given the density of the water, you're not being given a force. since you need a force for the work problem, you have to multiply the density by rhe gravity to make a force. to get the problem to, y'know... work.

also hey. to anyone. i had my first calc ii exam and this integral is stuck on my mind. everything i saw via calculators showed it ending up with like... stuff on the numerator with 5 as the denominator. im too lazy to pull out any of the chicken-scratch work i tried doing afterwards, but... what the FUCK is this integral? ^the bastard in question^

#45

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this was posted while i was fiddling with my last post but yeah. this. im gonna have to read this again when my brain isn't as fried because i really need to understand how physics work better.

#46
7. ### 3dilettante LegendAlpha

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Since I was passing through:

The most straightforward solution that I could think of is to rewrite it as the integral of x^-1/2 (x + 1)^2 dx.
Before integrating, expanding everything out and distributing allows for term by term integration.
integral of x^-1/2 * (x^2 + 2x + 1) dx
integral of x^3/2 + integral of 2x^1/2 + integral of x^-1/2

2/5 * x^5/2 + 4/3 * x^3/2 + 2 x^1/2 + C
There's some factoring and manipulation to get a common denominator, depending on how far you need to take it.

There's a substitution (u=x^1/2) that I initially tried that after a bit of fiddling works, but in the end I'm not sure it's worth the hassle in this case.

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8. ### Davros Legend

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If someone tries to talk to you about quantum mechanics - RUN..............

#48
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