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when i say 30 i mean 40
what does "reducing the length of the curve" mean? I mean you can draw the shortest possible path from (10,10) to (30,0) that also happens to be a straight line.
yea but it needs to be smooth

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so polynomial
what does that mean? a linear function is a polynomial too. what exactly do you mean by smooth?
mena no rough edges and its curve
like continuous function
ahhh, better.
is there function that looks like the black line i drew
so you need a polynomial function that makes a smooth transition and is also the shortest
so it has curve to smothly attach to another track but at the smae shorter length
yes
but at the same time *
what's the context of this question? it's really specific
i have done the question let me send the link
but what i am trying to figure out is the shorter length than this
sorry, my OS stopped working.
so on xy plane i want a function that give me length <27.37
|dw:1443516744812:dw|
ok
You're permitted to use calculators right?\[g(x) = ax^4 + bx^3 + cx^2 + dx + e\]\[10^4 a + 10^3 b + 10^2 c + 10 d + e = 10\]\[30^4 a + 30^3 b + 30^2 c + 30 d + e=0\]\[\int_{10}^{30} g(x)dx=200\]\[4a(10)^3 + 3b(10)^2 + 2c(10) + d(10) =0\]Now the last one is tricky.
yess
You know how to calculate arc length using integrals right?
yea but i use wolfram alpha normaly u mind using it sending the link plz
Because you really are looking to minimise arc-length given the above constraints.
ok wait do i do the same steps as i did for my previous function find a,b,c,d
and then put them in an expression and use arc length to get length
I think Wolfram is gonna have some good fun with this one.
heheh
Here, you have a, b, c, d and e.
yea kinda forgot about e
thank you so much for ur help
I'm curious to see how you'll input so much into Wolfram. It's scary.
what if i find an expression then too
sorry, the last equation is\[4a(10^3) + 3b(10^2 ) + 2c(10) + d = \color{red}1\]
ok
wow the expression is way too long... you'll have to be subscribe to pro
i have subscribe pro
wow, cool.
@ParthKohli for arguments sake, (specifics not necessary) i would write the function for arc length squared lets call it L=f(a b c d e) now what?to minimize, we need to substitute variables in terms of each other so that they represent their inter-dependencies right? how will you do that? you have four other equations ? or am i way off, and talking crap?
yes - we have four equations. in the arc length function, we'll eliminate four variables and will be left with a function that is in terms of only one variable.
is that a general thing, as in if there are n unknowns and (n-1) equations we can bring it down to 1 variable?
I suppose. Let's take a smaller example.\[f(x,y,z) = x+y-z\]\[x=y+1\]\[z = -x\]\[f(x,y,z) =x+(x-1) - (-x) = 3x - 1\]
Three variables, two equations. Now we can minimise the function with one-variable very easily.
ok. thanks :)
@ParthKohli one other thing, i get why you have picked an equation with 5 unknowns since we have 5 pieces of information, but the equation could just as easily been \[ax^7 + bx^6 +cx^5 +dx +e\] or something else equally arbitrary so why specifically have you chosen the one that you have?
that is a very good observation, one that I did give some thought to but skipped. unfortunately, it may be impossible to find *the* polynomial that satisfies the above constraints because the degree may go as high as... anything we want it to reach. the degree-seven polynomial you wrote may not be the best degree-seven polynomial to satisfy the conditions because it's missing some terms and unless we're sure while solving for the seventh-degree that those terms turn out to be zero, we'll never be sure. so choose a degree-four polynomial adds certainty to what we get - we can be sure that we'll get the best fourth-degree polynomial. of course there are polynomials much better than that one, but we can only go as far as we can look.
ultimately, we're looking for a polynomial that resembles a straight-line the most in the region x = 10 to x = 30, so we can add as many terms as we like, but we can't go that far - can we?
ah..makes sense. thanks a lot!!
by the way, we still can approach this problem through multivariable calculus for polynomials that have more terms than five. that way, we can get an even closer polynomial.
hey its not making sense
thanks for the question - I really had to think about it to come up with an answer.
5 variables how do i get values for 5 unknown variable
plug it into WolframAlpha
i only have 4 equation
i found a ,b, c ,d but not e
were you able to find a, b, c, d in terms of e?
\[0.0001852x ^{4}-0.01246x ^{3}+0.2076x ^{2}-0.01555x+e\]
http://math.bd.psu.edu/~jpp4/finitemath/4x4solver.html
wait what? you were able to uniquely determine a, b, c, d?!
from here
it's not a 4x4 equation so
yea i thought so help plz
plug the stuff into Wolfram and it should give you all other variables in terms of one variable
@ayeshaafzal221 where did you find this question?
it didnt
its my practise question for exams
what did it return?
you don't have a Wolfram pro subscription do you?
on my ipad yes
even on my ipad not working
there's this data entry or something feature which you can use
ok
or u know the values i have found cant i just sub it into any equation and get the e value
no, you can't do that. it's not a 4x4 equation so you most likely input the wrong equations.
yea true i am gonna go old fashion , and do em manually
@ParthKohli , when you say multi-variable calculus approach: are you talking about Lagrange multipliers?(i've just reached this topic)
Obviously. Yes.
alright..thanks
@ParthKohli srry i am disturbing u again when u say we have five variables and one of them was minimum arc length what do u mean?

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