Is this a differential equation?

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Mod note: HTML size = ... tags are not needed. You can make things look just fine using [ tex ] tags instead of [ itex ] tags.
So, I was doing some stuff, messing around when I thought of something. What if I took a random physics formula and integrated it into the original function? Then I was like, whoa! Is this a differential equation? I worked it like this... [tex]\Delta y = V_i + \frac{1}{2}gt^{2}[/tex]

So that is my formula, now I shall make it be equal to the derivative of velocity with respect to time using the form of dy/dx[tex]\frac{dv}{dt}= V_i + \frac{1}{2}gt^{2}[/tex]

Then I multiply dt to the other side.[tex]dv = (V_i + \frac{1}{2}gt^{2}) dt[/tex]

Then I integrated both sides.[tex]\int dv = \int (V_i + \frac{1}{2}gt^{2}) dt[/tex]

And I obtained...[tex] v = \frac{1}{2}V_i^{2} + \frac{1}{2}(\frac{1}{3}gt^{3}) + C [/tex]

Once I differentiated back, I got the original vertical motion formula! Is this a differential equation? If so, what kind?
 
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  • #2
LCKurtz
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So, I was doing some stuff, messing around when I thought of something. What if I took a random physics formula and integrated it into the original function? Then I was like, whoa! Is this a differential equation? I worked it like this...

[itex]\Delta y = Vi + \frac{1}{2}gt^{2}[/itex]

So that is my formula, now I shall make it be equal to the derivative of velocity with respect to time using the form of dy/dx

[itex]\frac{dv}{dt}= Vi + \frac{1}{2}gt^{2}[/itex]

Then I multiply dt to the other side.
[itex]dv = (Vi + \frac{1}{2}gt^{2}) dt[/itex]

Then I integrated both sides.
[itex]\int dv = \int (Vi + \frac{1}{2}gt^{2}) dt[/itex]

And I obtained...

[itex] v = \frac{1}{2}Vi^{2} + \frac{1}{2}(\frac{1}{3}gt^{3}) + C [/itex]

Once I differentiated back, I got the original vertical motion formula! Is this a differential equation? If so, what kind?

Please do not use super size fonts in your posts. Your second equation is a differential equation -- an equation involving derivatives. I don't know what your variables are, but I suspect that your last equation is incorrect. Why would ##\int i\, dt## be ##\frac 1 2 i^2##?
 
  • #3
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Oh no that i isn't seperate. That's Velocity initial. Hence the Vi I just couldnt figure out the subscript thing. Oh and sorry about the big font thing, it's just kind of hard to read for me when it's smaller than like 5 or so so I wanted to set an example because I was having trouble reading other people's stuff because it was small.
 
  • #4
LCKurtz
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Oh no that i isn't seperate. That's Velocity initial. Hence the Vi I just couldnt figure out the subscript thing. Oh and sorry about the big font thing, it's just kind of hard to read for me when it's smaller than like 5 or so so I wanted to set an example because I was having trouble reading other people's stuff because it was small.

So then how does ##\int V_i\, dt## become ##\frac 1 2 V_i^2##? And you can probably increase the font size in your browser. Ctrl-plus works in Firefox.
 
  • #5
HallsofIvy
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The initial velocity doesn't change with time, does it? Then
[tex]\int V_i dt= V_i\int dt= V_it+ C[/tex].
 
  • #6
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Oh, that's the power rule. So, Vi has an invisible exponent of 1. By the power rule of integration, you add 1 to the exponent, hence the two. Also, you divide by the exponent plus 1. However, it is easier for me to, instead of writing Vi^2/2 to instead write 1/2Vi^2 because remember, there is an invisible one before Vi since it is Vi and Vi is also 1Vi so 1Vi^2/2 or 1/2Vi^2.
 
  • #7
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Ivy, what do you mean? I thought Velocity changed with time, you know? I mean, acceleration is [itex] a = \frac{dv}{dt}[/itex] and velocity is [itex] v = \frac{ds}{dt}[/itex]. I would argue that velocity changes with time, though I may be confused ha ha.
 
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Ivy, what do you mean? I thought Velocity changed with time, you know? I mean, acceleration is [itex] a = \frac{dv}{dt}[/itex] and velocity is [itex] v = \frac{ds}{dt}[/itex]. I would argue that velocity changes with time, though I may be confused ha ha.

Velocity does change with time, initial velocity does not. It's just like position, if I drop a ball from 10 meters, the position will change, but the initial height will always be 10 meters.
 
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  • #9
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Guys, it works though because it differentiates back. Watch the magic!

[tex]\frac{dv}{dt}(\frac{1}{2}V_i^{2} + \frac{1}{2}(\frac{1}{3}gt^{3}) + C)[/tex]
[tex]\frac{dv}{dt}= \frac{1}{2}\frac{d(V_i^{2})}{dt} + \frac{1}{2}\frac{d(\frac{1}{3}gt^{3})}{dt} + \frac{1}{3}gt^{3}\frac{d(\frac{1}{2})}{dt} + \frac{dv}{dt}(C)[/tex]

Since 2 * 1/2 is 1 we end up with Vi again, and since 1/2 times the derivative of gt, which is gt^2 is just 1/2gt^2 we get that back! Then, 1/2 is a constant, as is C so the rest is zero! So, what do we have? The first thing! WOO!

[tex]\frac{dv}{dt}= V_i^{2} + \frac{1}{2}gt^{2}[/tex]

See? See? I think it should be right, I followed the rules of differentiation and integration.
 
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Velocity does change with time, initial velocity does not. It's just like position, if I drop a ball from 10 meters, the position will change, but the initial height will always be 10 meters.
Ah okay, well, what should it be with respect to? I'm pretty sure of the math, kind of new to the physics lol.
 
  • #11
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Also, differential equations are normally written in terms of one function. So if you wanted to set these equations (involving y, v and a) up as something more like "textbook" differential equations, it would be common to write:

[itex] u(t) = y = position [/itex]
[itex] u'(t) = \frac{dy}{dt} = velocity [/itex]
[itex] u''(t) = \frac{d^{2}y}{dt^{2}} = acceleration [/itex]

It's also common to use y, y', and y''.

When you take Diff EQ, you will probably see this form when talking about modeling spring vibrations.
 
  • #12
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Ah okay, well, what should it be with respect to? I'm pretty sure of the math, kind of new to the physics lol.

Initial velocity is just with respect to whichever way you define the positive and negative axis. For 1 dimensional problems like throwing a ball up in the air, etc. it's really up to you whether you make down direction positive or negative. It will always work out the same as long as you are consistent throughout the problem.
 
  • #13
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Also, keep in mind the fact that something "differentiates back" doesn't mean it's corrected when in integrated form.

If I have a function [itex] y'(t) = k = 5 [/itex]

Then if I say [itex] \int y'(t) dt = y = \frac{1}{2} k^{2} [/itex]

Notice that this is explicitly wrong since we're supposed to be integrating with respect to t.

We know I'm wrong intuitively, since we know we should have [itex] y(t) = kt + C = 5t + C[/itex]

It is wrong even though it still "differentiates back" if I continue to treat it as a function of this incorrect "variable" k: [itex] \frac{d}{dk}(\frac{1}{2} k^{2}) = k [/itex]
 
  • #14
LCKurtz
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Oh, that's the power rule. So, Vi has an invisible exponent of 1. By the power rule of integration, you add 1 to the exponent, hence the two. Also, you divide by the exponent plus 1.
I don't think you need to teach anyone here how to integrate, although you have something to learn.
However, it is easier for me to, instead of writing Vi^2/2 to instead write 1/2Vi^2 because remember, there is an invisible one before Vi since it is Vi and Vi is also 1Vi so 1Vi^2/2 or 1/2Vi^2.

Are you being serious or just trolling? Even if ##V_i## wasn't a constant and you meant ##V(t)##, the power rule does not give ##\int V(t)dt = \frac {V^2}{2}##.
 
  • #15
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I wasn't trying to teach anybody how to integrate, I assume you know how. No, I am not trolling, I hate being accused of "trolling" because somebody disagrees with me and/or I'm incorrect. And I am not treating it as a constant, everything is a variable in this equation. So please don't get upset at my ignorance, the only mistake I see is that I integrated it as a function of the wrong variable.
 
  • #16
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I wasn't trying to teach anybody how to integrate, I assume you know how. No, I am not trolling, I hate being accused of "trolling" because somebody disagrees with me and/or I'm incorrect. And I am not treating it as a constant, everything is a variable in this equation. So please don't get upset at my ignorance, the only mistake I see is that I integrated it as a function of the wrong variable.

We've covered why initial velocity is constant, not a function. Unless you want to get technical and call it a constant function of t, which is just a constant. It is certainly not a variable, the value does not change. It's the same as if I had ∫5(dt) ... this would be (5t+C), except here you don't know the constant's value so you are representing it with (Vi). Remember, you are taking the integral of with respect to t (hence the "dt").
 
  • #17
LCKurtz
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And I am not treating it as a constant, everything is a variable in this equation. So please don't get upset at my ignorance, the only mistake I see is that I integrated it as a function of the wrong variable.

So at this point, after reading various responses, do you understand that your last equation does not follow from the previous equation?
 
  • #18
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I feel pretty damn stupid to be honest, wow, I was just trying to figure out how a differential equation works and I messed up on the most basic thing... why didn't I just use some theoretical function that wasn't a part of physics so that I wouldn't have to worry about the physics and just concentrate on the math. I'm in high school physics and algebra, I've studied ahead in intermediate algebra and figured out the things at the back of the book and stuff. I am studying Calculus on my own in order to be prepared for college, and so naturally I'm going to make errors. You know, I think I messed up mainly because I haven't completed everything there is to know about differentiation or integration ha ha, so, yeah, sorry for sounding ignorant; I'm just trying to learn.

And yes, I understand now. Wow, what a foolish error on my part. What I think I did wrong was not treating Vi as a constant, and g is also a constant because it is acceleration due to gravity. But I know why I cannot integrate or differintiate velocity with respect to time. Okay, I was just reading my Technical Calculus book(it has physics problems) and I read about how "Velocity is ds/dt s=distance and t=time" It would make no sense to divide velocity by time to get velocity, that is acceleration. Also, I understand that Vi is a constant because it is the starting velocity of the problem, as such it does not change. So yeah, I made a mistake.
 
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  • #19
LCKurtz
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One of the problems we run into on this forum is we often don't know the background of the students we are helping. Being in high school and working ahead studying calculus on your own is to be commended. So feeling "pretty damn stupid" on your part is not called for. Keep studying. Work lots of problems in your calculus book. You will find that once you understand calculus, lots of the physics theory will be easier to deal with.
 
  • #20
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Is it bad to get frustrated with certain areas of calculus then go back to them later? Like I skip around but always go back eventually to make sure I got it.
 
  • #21
LCKurtz
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Is it bad to get frustrated with certain areas of calculus then go back to them later? Like I skip around but always go back eventually to make sure I got it.

Calculus texts are arranged in the order they are for a reason. You really need to get familiar with differentiation before you tackle integration. Some topics, such as limits, may seem a bit obscure to you studying on your own, but I imagine your Technical Calculus book does them once over lightly anyway. Full understanding of that may have to wait until you have a teacher in a university calculus course. Or perhaps there is an AP Calculus teacher in your school that you could talk to about the rough spots.
 

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