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

Sorry I was assuming that the constant was zero b/c that makes fewer things to deal with. So yes you have it except there is no requirement to be continuous. I am just asking about a concept. So if I do a definite integral then I get a value. It seems to me that a definite integral of a part of any piecewise function should give the same result as an indefinite integral of the equation defining that segment and then plugging in values, they are after all the same process, right?

It has been some time since I took calc, but I was helping someone and the answer they were given had something where they integrated from zero to the variable, not the limit, then plugged in a value later and somehow the answer said it was double what I thought it would be.

So here if I integrate
dx/dt=b*sin(a*t) integrate from 0-->t
x=-b/a*cos(a*t)+b/a then if I plug in 5 for t I get x=-b/a*cos(a*5)+b/a

If I just directly integrate I get
-b/a*cos(a*5)+b/a*cos(0) which matches

I just wanted to double-check there was no weird thing with piecewise functions where you needed to do something different. If you are between the limits it is totally irrelevant as far as I remember.
 
Sxotty said:
Sorry I was assuming that the constant was zero b/c that makes fewer things to deal with. So yes you have it except there is no requirement to be continuous. I am just asking about a concept. So if I do a definite integral then I get a value. It seems to me that a definite integral of a part of any piecewise function should give the same result as an indefinite integral of the equation defining that segment and then plugging in values, they are after all the same process, right?
Click to expand...

Yes, but for a piecewise function, each interval has their own constants (as they might not be continuous), so when applying limits, it should be applied separately in order to eliminate the constants.
 
orangpelupa said:
https://arstechnica.com/science/2019/11/limits-schlimits-its-time-to-rethink-how-we-teach-calculus/
Click to expand...

I remembered that when I was young my sister bought some calculus books written by a Japanese author. He put cute pictures with humorous examples to explain the basics of calculus.
The book was written for people with practical use of calculus, so it does not focus on rigorous proof and the basic theories of calculus, but on the real world examples of how calculus might be needed to calculate the correct answers.
I think because I read these books, it's easier for me to grasp the idea of calculus and thus easier to understand the rigorous proof and basic theories of calculus. Some might say that this is not "pure" but I guess it worked for me and at least I'm able to handle some simple calculus problems when I encountered them (I have to say I don't really encounter much of them in my job, I'm a computer programmer :p )

Of course, problems requiring calculus are really not that rare. A recent example someone asked me is, suppose that you have an airplane, burning 0.1 ton of fuel for each ton (of the plane's weight) per hour cruising, how much fuel do you need for an 80 ton plane if you need 10 hours of cruising time? Hint: it's not 80 ton ;)
 
TheAlSpark said:
hm... what about the weight of the fuel adding to the load and decreasing over the flight time? :eek:
Click to expand...

Yes :) On real airplanes the fuel actually do not account for that much weight (but still significant: it's common to have 30%~50% of the total takeoff weight of a large airplane in fuel).

orangpelupa said:
Now I can't stop wondering that fuel question was asked in what context / situation / environment...
Click to expand...

Imagine that you put 80 ton of fuel on the plane, but now the plane has to carry 160 ton weight... :p
 
Yeah, but when was that question raised?

I keep imagining it was over a big family dinner, and one of them brought a fiancee.

A: hey this chicken is tots delicious
B: INDEED! That reminds me, [the fuel question]

Then family member C D E and F also got into the fuel question, leaving the fiancee.

Fiancee, thinking: I think I'm gonna marry into a family that are an awesome cook, close together, and also super smart.
 
orangpelupa said:
Yeah, but when was that question raised?

I keep imagining it was over a big family dinner, and one of them brought a fiancee.

A: hey this chicken is tots delicious
B: INDEED! That reminds me, [the fuel question]

Then family member C D E and F also got into the fuel question, leaving the fiancee.

Fiancee, thinking: I think I'm gonna marry into a family that are an awesome cook, close together, and also super smart.
Click to expand...

LOL I think it's from an intern :p
 
pcchen said:
Of course, problems requiring calculus are really not that rare. A recent example someone asked me is, suppose that you have an airplane, burning 0.1 ton of fuel for each ton (of the plane's weight) per hour cruising, how much fuel do you need for an 80 ton plane if you need 10 hours of cruising time? Hint: it's not 80 ton ;)
Click to expand...

150 tons?
 
DuckThor Evil said:
150 tons?
Click to expand...

It's close.
Let w(t) be the total weight of the plane at hour t, so
w(t) = w0 + f(t)
where w0 is the weight without fuel and f(t) is the weight of fuel over time.

w0 does not change over time, so the rate of change of w(t) is the same as f(t), so
w'(t) = -r*w(t)
(r is the fuel consumption rate, 0.1 in this case)
So we know that
w(t) = C*e^(-r*t)
(C is a constant)

To determine C, we know that w(10) = w0 because the plane needs 10 hours of fuel so f(10) = 0.
So w(10) = C*e^(-0.1*10) = C*e^-1 = w0
=> C = w0 * e
Therefore
w(t) = w0 * e^(1-r*t)

Now we want to know how much fuel we need at t = 0 so
f(0) = w(0) - w0 = w0 * (e - 1) = w0 * 1.71828...

Since w0 = 80 so f(0) ~ 137.5 ton.
 
pcchen said:
It's close.
Let w(t) be the total weight of the plane at hour t, so
w(t) = w0 + f(t)
where w0 is the weight without fuel and f(t) is the weight of fuel over time.

w0 does not change over time, so the rate of change of w(t) is the same as f(t), so
w'(t) = -r*w(t)
(r is the fuel consumption rate, 0.1 in this case)
So we know that
w(t) = C*e^(-r*t)
(C is a constant)

To determine C, we know that w(10) = w0 because the plane needs 10 hours of fuel so f(10) = 0.
So w(10) = C*e^(-0.1*10) = C*e^-1 = w0
=> C = w0 * e
Therefore
w(t) = w0 * e^(1-r*t)

Now we want to know how much fuel we need at t = 0 so
f(0) = w(0) - w0 = w0 * (e - 1) = w0 * 1.71828...

Since w0 = 80 so f(0) ~ 137.5 ton.
Click to expand...
:oops:
 
i like whatever happened here. i understand some of it.

it also leads me to ask, though: does anyone know any good calculus books? the one my college is currently using kinda blows (yes, i am just using an amazon link. no, i'm definitely not sponsoring it and don't recommend that anyone buys it because it SUCKS). the department used to use this one, which seems a bit more... understandable. but our current one is definitely more of a pure math kind of book, and my brain doesn't click with it. it could also just be that i'm not used to reading college textbooks, ESPECIALLY college math textbooks, but i feel like there's gotta be something out there that i can get more out of.

and yeah, this is just... me wanting to understand math better. i still can't believe i actually LIKE math and that i'm apparently good at it now? i mean, my grade in that class definitely isn't the best, but i'm passing ok i just checked and i have either a C+ or a B- so i'm just absolutely BLOWN AWAY right now. i've learned that there really are a lot of students who don't pass calc 1 their first try... but i probably really will?

yeah. ok. either way. i like the problem that was discussed... it's cool. does anyone know of calc books that aren't as full of only proofs and theorems and actually help walk you through them/apply them/have more of a visual aspect to it? i'm doing good but i'm always just wanting to continue to understand things better and whatnot.

i'm gonna go pay attention to my philosophy class instead of talking about and thinking about math.
 
Davros said:
Hang on, as the plane lightens the fuel consumption rate decreases

ps: how does w'(t) differ from w(t) ?
Click to expand...

The rate is the fuel consumption per ton per hour, so it's constant. The rate per hour depends on the weight of the plane at the time, so it's -r * w(t).
w'(t) is a shorthand for dw/dt (i.e. the derivative of w(t) ).
 
As an engineer with a PhD, for me the best part in this thread is the reminder of the fact that I don't have to deal with hand-written calculus anymore.

I still remember a 1st year class (Matter State Physics I think) where we learned how to calculate the probability of an electron jumping between atoms/ions.
Fuck that shit.
 
how the hell do you solve this??
20191125_100527.jpg
after trying to look

edit: i'm about to throw up because of how mad i am at trying to use the mobile site (and just struggling with math). oh my god.

i put down what i found after attempting to look up how you solve the problem. i don't know where to go from here. this is so frustrating. i've never taken physics before and i'm bad at related rates problems. please help.
 
maddy said:
how the hell do you solve this??
View attachment 3449
after trying to look

edit: i'm about to throw up because of how mad i am at trying to use the mobile site (and just struggling with math). oh my god.

i put down what i found after attempting to look up how you solve the problem. i don't know where to go from here. this is so frustrating. i've never taken physics before and i'm bad at related rates problems. please help.
Click to expand...

I think the idea is because f is constant (focal length depends on the shape of the lens), you know the derivative of u (the speed of the object), and you want to find out the derivative of v (the speed of the lens).
So

1/u+1/v = 1/1
-> 1/u = 1-1/v
-> u = 1/(1-1/v)
-> u' = -1/[(1-1/v)^2*v^2] * v' = -1/(v-1)^2 * v'
(from du/dt = du/dv * dv/dt)

Now we know u' = 20, and we want to solve v', but we don't know v yet. Fortunately we know u = 50, so by 1/u = 1-1/v we know v = 1/(1-1/50) = 50/49.
So v' = -u' * (v-1)^2 = -20 * (50/49 - 1)^2 = -20/49^2

That means in order for the image to stay in focus, the lens has to move toward the screen at 20/49^2 feet per second.
It's quite possible I made some mistakes in the calculation but you got the general idea :)
 
Last edited:
the results are in...

i took my final this monday after getting 4 1/2 hours of sleep (i was just anxious about taking my first final). it was 4 hours long (double time :oops:) and probably around 20 questions? definitely wasn't too confident in my performance... i basically had a B- before taking the final, so either way i knew i wasn't going to fail the class.

my professor put the final exam grades in on wednesday. i got 132/200... a 66%...
...but i still ended up with a 78 overall. i passed calc i. :cool:

i currently have it on my schedule, but as of right now, i'm still not sure if i'm going to take calc ii yet. from what i've seen and heard, it's usually viewed as the hardest calc course, since students need a strong understanding of the material in calc i ? i think i'm just being hard on myself for getting a 66% on the final (still trying to get myself to not panic over 'bad' grades), which is a little ridiculous. i forget that a lot of people usually don't pass the class on their first attempt. :oops: to anyone that's had to take calc i more than once: you're stronger than me, because i don't know how well i'd handle doing it again...

but yeah! um. i wanted to say a quick thank you to anyone who's helped, whether it be with specific questions or finding resources. i've been a bit too busy to actually say it got sick over thanksgiving break, lasted through most of the last week of the semester, and this week is finals. i finished the two finals i had to go in to do, and now i just have to actually get myself to write my english paper on planned obsolescence and tech waste... :| i just have trouble starting papers.
 
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