Calculus - finding a maximum

Ana NgAna Ng Registered User regular
I'm working on a rather nasty calculus problem involving finding a maximum velocity. I understand the basic idea of it, that I need to set my equation to 0 and solve for the variable, trouble is my equation looks like this...

-0.0157x^5 - 0.39x^4 - 2.29x^3 + 55.5t^2 - 407t + 1400 = 0

I tried plugging into wolframalpha and got a rather unhelpful answer. Can anyone offer any guidance?

Posts

  • SavantSavant Registered User regular
    edited October 2014
    Is that your original equation, or some derivative? Since you are dealing with a multivariable equation, unless x is a function of t then you are probably trying to find critical points, which can local maximums or minimums or saddle points. You can do that by setting all the partial derivatives of your original equation with respect to the each variable to zero, then solve the resultant equations. So take what your original equation is and find the partial derivative with respect to x and set that to zero, and the partial derivative with respect to t and set that to zero, and find which points solve those equations.

    Just eyeballing that equation there isn't a global maximum on it since it will grow unbounded as t grows without limit, so maybe there is a missing constraint or local maximum you are looking for?

    Savant on
  • Ana NgAna Ng Registered User regular
    Savant wrote: »
    Is that your original equation, or some derivative? Since you are dealing with a multivariable equation, unless x is a function of t then you are probably trying to find critical points, which can local maximums or minimums or saddle points. You can do that by setting all the partial derivatives of your original equation with respect to the each variable to zero, then solve the resultant equations. So take what your original equation is and find the partial derivative with respect to x and set that to zero, and the partial derivative with respect to t and set that to zero, and find which points solve those equations.

    Just eyeballing that equation there isn't a global maximum on it since it will grow unbounded as t grows without limit, so maybe there is a missing constraint or local maximum you are looking for?

    I just realized I mixed x's and t's.. it should be all t's in there. Ugh. Sorry, tired brain making dumb typing mistakes.

    That aside, yes, that is the derivative of the original. So can I just take each piece of the derivation and set to zero, is that what you mean? So like:

    -0.0157t^5 = 0 and solve for t and then do that for each? I had suspected that each piece represented a local min or max, and that I'd need to just look at them all and figure which of the local max's was the largest.

  • enlightenedbumenlightenedbum Registered User regular
    edited October 2014
    You're setting the whole thing = 0 and then in a perfect world factoring it to get (t - a)(t - b)(t - c)(t - d)(t - e) = 0 so your critical points would be a, b, c, d, and e. Then you check to see for which of those values is the first derivative changing from positive to negative (a local max). Plug any of those values into the original function along with any boundary points to get the global maximum.
    -
    But in this case, that would be insane because fuuuuuck factoring that mess. Should be able to get some (probably ugly) solutions to that equation mechanically. Or graphing it to find any zeroes.

    EDIT: I did the latter using Wolfram and it works.

    enlightenedbum on
    Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
    Warren 2020
    KetBra
  • Ana NgAna Ng Registered User regular
    You're setting the whole thing = 0 and then in a perfect world factoring it to get (t - a)(t - b)(t - c)(t - d)(t - e) = 0 so your critical points would be a, b, c, d, and e. Then you check to see for which of those values is the first derivative changing from positive to negative (a local max). Plug any of those values into the original function along with any boundary points to get the global maximum.
    -
    But in this case, that would be insane because fuuuuuck factoring that mess. Should be able to get some (probably ugly) solutions to that equation mechanically. Or graphing it to find any zeroes.

    EDIT: I did the latter using Wolfram and it works.

    Damn. Alright. I'll have to look into that... this sucks, it's part of a large group calculus assignment. I'm really surprised at this problem, it really seems out of hand :\

  • KetBraKetBra Dressed Ridiculously Registered User regular
    edited October 2014
    Yeah if you plot in wolfram you get a max

    Wolfram doesn't seem to think there's a solution.

    So uh... Newton's method is probably what your prof is angling at you to use, assuming she just doesn't want a computer to do all your work for you.

    KetBra on
    KGMvDLc.jpg?1
  • enlightenedbumenlightenedbum Registered User regular
    Maybe? My experience with computerized math problems is that they throw in a bunch of obnoxious numbers for no reason other than to make things incredibly hard to do by hand, because computerized education is kind of dumb at this point in its existence.

    I feel like if you show you knew to take a derivative, solve for 0, and then you knew why something is a local max/min, you know what you're doing, as long as you know how to get zeroes of a polynomial.

    Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
    Warren 2020
  • KetBraKetBra Dressed Ridiculously Registered User regular
    Eh, I dunno. It seems like a decent Newton's method problem, and a dumb problem if you are trying to teach anything else

    Unless what you're trying to teach is how to use a graphing calculator/Wolfram Alpha

    KGMvDLc.jpg?1
  • Ana NgAna Ng Registered User regular
    Maybe? My experience with computerized math problems is that they throw in a bunch of obnoxious numbers for no reason other than to make things incredibly hard to do by hand, because computerized education is kind of dumb at this point in its existence.

    I feel like if you show you knew to take a derivative, solve for 0, and then you knew why something is a local max/min, you know what you're doing, as long as you know how to get zeroes of a polynomial.

    This is why I'm very frustrated. I understand what I'm supposed to do, this just seems near impossible to do.

    For a contrast, the example in class was y= 2x^3 - 15x^2 + 24x + 7
    Which made sense to take y' and y'' for and do the whole thing.
    So this particular problem just seems prohibitively intense and a little pointless.

  • enlightenedbumenlightenedbum Registered User regular
    edited October 2014
    Yeah, if you're doing this for a group assignment on real paper that you're turning in, and you're not actually in a Newton's Method unit, would be to take the derivative as you did, write that down, set it equal to 0, graph the derivative in a graphing calculator or online, draw that graph, and then write down the solution(s) you get, noting that at that point the graph of the first derivative goes from positive to negative, so hooray that's a local max.

    Though actually, I just re-read the OP again. You say you're looking for a maximum velocity, so you need to double check whether you're given x(t) [position] or v(t) [velocity] as your initial function. If you start with x(t), you actually want to start by getting x'(t), which is velocity, then find the max there.

    Also notably real world physics kind of questions frequently have a built in constraint of t = 0, because we usually don't care about the past. If this was a physics class I'd definitely include that, in a calculus class, you don't always get points for common sense.

    EDIT: For clarity.

    enlightenedbum on
    Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
    Warren 2020
  • Ana NgAna Ng Registered User regular
    edited October 2014
    Yeah, if you're doing this for a group assignment on real paper that you're turning in, and you're not actually in a Newton's Method unit, would be to take the derivative as you did, write that down, set it equal to 0, graph the derivative in a graphing calculator or online, draw that graph, and then write down the solution(s) you get, noting that at that point the graph of the first derivative goes from positive to negative, so hooray that's a local max.

    Though actually, I just re-read the OP again. You say you're looking for a maximum velocity, so you need to double check whether you're given x(t) [position] or v(t) [velocity] as your initial function. If you start with x(t), you actually want to start by getting x'(t), which is velocity, then find the max there.

    Also notably these kinds of functions usually have a built in constraint of t = 0, because we usually don't care about the past.

    it's dealing with angular acceleration σ (or whatever, just random symbol I don't remember) of a pitchers arm during time interval of 20 milliseconds before a pitch to 10 milliseconds after the ball is released.

    So I'm given σ(t) = -0.0785t^4 - 1.56t^3 + 6.88t^2 + 111t - 407

    at t = 0 when the ball is released, angular velocity is 1400 degrees per second.

    So I have to find the equation of angular velocity as a function of time... so my understanding is that the anti-derivative of that big mess will be the equation for velocity for the time period. Then I set it to 0 and I can find the critical points.

    Edit: To elaborate more on what I'm doing. So I ended up with all the junk of the anti-derivative then + C, so I solved for C and got 1400 which makes sense.

    So I've found myself with

    v(t) = -0.0157t^5 - 0.39t^4 - 2.29t^3 + 55.5t^2 - 407t + 1400

    So if I set this to 0 and solve for t that'll give me the critical points..

    ugh I'm so tired, I need to shelf this for the night and come back when I'm fresh.

    Ana Ng on
  • KetBraKetBra Dressed Ridiculously Registered User regular
    So what exactly are you solving for here? The maximum angular velocity?

    KGMvDLc.jpg?1
  • enlightenedbumenlightenedbum Registered User regular
    edited October 2014
    OK, so there is actually a mistake in how you're understanding things. They want the maximum in velocity, right?

    So first you need the velocity function, which you've done.

    But we're interested in finding the maximum velocity from t = -.02 to t = .01. Since we're finding the maximum velocity, to get critical points we want the derivative of the velocity function, which is just the acceleration function. So you solve that for 0, between the two relevant times. Then put any critical points and the two end points into the velocity function to see what the maximum is.

    EDIT: I should add that you should double check what units they're giving the acceleration in. If it's in seconds what I say above is correct, if it's in milliseconds you have to replace the -.02 and .01 with -20 and 10, and do some crazy stuff to the 1400 initial value in your velocity function. I mention this because of where the critical points are (there are some between -20 and 10, but not between -.02 and .01

    enlightenedbum on
    Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
    Warren 2020
  • Ana NgAna Ng Registered User regular
    OK, so there is actually a mistake in how you're understanding things. They want the maximum in velocity, right?

    So first you need the velocity function, which you've done.

    But we're interested in finding the maximum velocity from t = -.02 to t = .01. Since we're finding the maximum velocity, to get critical points we want the derivative of the velocity function, which is just the acceleration function. So you solve that for 0, between the two relevant times. Then put any critical points and the two end points into the velocity function to see what the maximum is.

    EDIT: I should add that you should double check what units they're giving the acceleration in. If it's in seconds what I say above is correct, if it's in milliseconds you have to replace the -.02 and .01 with -20 and 10, and do some crazy stuff to the 1400 initial value in your velocity function. I mention this because of where the critical points are (there are some between -20 and 10, but not between -.02 and .01

    The units of the angular acceleration are degrees per second per millisecond.

    Part A is to find an equation for angular velocity as a function of time, so that part I've done.

    And then for Part B I have to find the max angular velocity over the interval of -20 millisec to 10 millisec. So you're saying to find that I'm using the acceleration function set to 0, not velocity?

  • DaenrisDaenris Registered User regular
    Ana Ng wrote: »
    OK, so there is actually a mistake in how you're understanding things. They want the maximum in velocity, right?

    So first you need the velocity function, which you've done.

    But we're interested in finding the maximum velocity from t = -.02 to t = .01. Since we're finding the maximum velocity, to get critical points we want the derivative of the velocity function, which is just the acceleration function. So you solve that for 0, between the two relevant times. Then put any critical points and the two end points into the velocity function to see what the maximum is.

    EDIT: I should add that you should double check what units they're giving the acceleration in. If it's in seconds what I say above is correct, if it's in milliseconds you have to replace the -.02 and .01 with -20 and 10, and do some crazy stuff to the 1400 initial value in your velocity function. I mention this because of where the critical points are (there are some between -20 and 10, but not between -.02 and .01

    The units of the angular acceleration are degrees per second per millisecond.

    Part A is to find an equation for angular velocity as a function of time, so that part I've done.

    And then for Part B I have to find the max angular velocity over the interval of -20 millisec to 10 millisec. So you're saying to find that I'm using the acceleration function set to 0, not velocity?

    Yes. Because if you set the velocity equation to 0 that's not going to find you a maximum velocity, it's going to find where the velocity is 0. To find the maxima/minima of the velocity, you need to find the points where the derivative of velocity is 0 (the peaks and troughs on the velocity graph). Then take those local maxima/minima points and plug them into your velocity equation to find the maximum value in that interval -- also don't forget to test the endpoints (-20 and 10) as well, because even though they might not be a maxima/minima based on the derivative, since you're on a closed interval they could still be the maximum in that range.

    enlightenedbumAna NgKetBratynic
  • Ana NgAna Ng Registered User regular
    Thank you so much everyone that helped! Hopefully I did well on the assignment, I think it turned out correctly so we'll see :)

  • enlightenedbumenlightenedbum Registered User regular
    Did you remember to adjust the 1400 degrees/second to match the units from the acceleration part?

    Herbert Hoover got 40% of the vote in 1932. Friendly reminder.
    Warren 2020
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