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?
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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.

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.
Warren 2020
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 :\
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.
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.
Warren 2020
Unless what you're trying to teach is how to use a graphing calculator/Wolfram Alpha
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.
Though actually, I just reread 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.
Warren 2020
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 antiderivative 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 antiderivative 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.
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
Warren 2020
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.
Warren 2020