Horizontal tangents to an implicitly defined function

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Oct 1st 2008, 10:37 AM
Hweengee

Horizontal tangents to an implicitly defined function

Consider the implicit function: y-y^5=x^4-2x^3+x^2

dy/dx = 2x(2x^2-3x+1)/1-5y^4

Solving for dy/dx=0, there are 3 solutions, namely x=0,1/2 and 1.

This implies that there are 3 horizontal tangents to y-y^5=x^4-2x^3+x^2.

But i suspect that there are more than 3 horizontal tangents to the curve, since a circle given by the equation x^2+y^2=1 has two horizontal tangents at
(0,1) and (0,-1), even though the derivative -x/y has only one solution. Is there any way that this can be shown? Or I am wrong?
Oct 1st 2008, 11:16 AM
Jhevon

Quote:

Originally Posted by Hweengee

Consider the implicit function: y-y^5=x^4-2x^3+x^2

dy/dx = 2x(2x^2-3x+1)/1-5y^4

Solving for dy/dx=0, there are 3 solutions, namely x=0,1/2 and 1.

This implies that there are 3 horizontal tangents to y-y^5=x^4-2x^3+x^2.

But i suspect that there are more than 3 horizontal tangents to the curve, since a circle given by the equation x^2+y^2=1 has two horizontal tangents at
(0,1) and (0,-1), even though the derivative -x/y has only one solution. Is there any way that this can be shown? Or I am wrong?

lets use $x^2 + y^2 = 1$ . indeed, the derivative is given by $- \frac xy$ which has one solution for $x$ in order to be zero. the thing is, we have an implicit function here. it is not enough to solve only for x, we care about solutions for $x$ AND $y$ to our curve. so how do i know that there will be only two horizontal tangents to this curve? simple, plug in my one solution for $x$ into the curve. i get $y^2 = 1 \implies y = \pm 1$ . so though i had one solution for $x$ , i have TWO solutions for $x$ and $y$ , and hence i have TWO horizontal tangents. namely at the points $(0,1)$ and $(0,-1)$

apply this principle o your curve
Oct 1st 2008, 11:24 AM
Hweengee

I've tried this, and end up with the following result.

x=0,1 y=0,1,-1 ,implying 3 tangents

when x=1/2, y-y^5=1/16, and I have no idea how to solve quintic equations.

Would appreciate any more advice you guys have.

Thanks.
Oct 1st 2008, 11:43 AM
Moo

Hello,

Quote:

Originally Posted by Hweengee

I've tried this, and end up with the following result.

x=0,1 y=0,1,-1 ,implying 3 tangents

when x=1/2, y-y^5=1/16, and I have no idea how to solve quintic equations.

Would appreciate any more advice you guys have.

Thanks.

It's better writing this way :
horizontal tangents at points :
- (0,0)
- (0,1)
- (0,-1)
- (1,0)
- (1,1)
- (1,-1)

that makes 6!

Now, $y-y^5=\frac{1}{16}$
It is not possible (or very difficult) to solve for it.. But using Descartes' rule of signs, you may be able to find how many solutions there are. (my calculator gives 3)
Oct 1st 2008, 11:50 AM
Jhevon

Quote:

Originally Posted by Hweengee

I've tried this, and end up with the following result.

x=0,1 y=0,1,-1 ,implying 3 tangents

when x=1/2, y-y^5=1/16, and I have no idea how to solve quintic equations.

Would appreciate any more advice you guys have.

Thanks.

as Moo said, there are 6 so far. why did you pick only 3? you have 3 for x = 0 and 3 for x = 1

as for the quintic, it is indeed a naughty quintic. the solutions are not integers, and are probably not rational. there are 3 solutions though. i guess you can try to estimate them using the Newton-Raphson method, or something. i will check if there are rational solutions

EDIT: yup, no rational solutions

EDIT 2: if you are just required to find how many horizontal tangents there are, then we are ok, but if you need to find the points, you have some hard work ahead of you
Oct 1st 2008, 12:05 PM
Hweengee

i didn't realise that there were 6 possible combinations of x and y such that dy/dx=0. thanks for the help so far, so if there are 3 solutions to the quintic would it be right to say that implies 3 horizontal tangents (0.5,y1)(0.5,y2) and (0.5,y3)? making for a total of 9? or would it be best to use a CAS to solve this problem, as this is a question that's supposed to be solved using MAPLE. maybe i'll try using MAPLE to check the number of solutions to the quintic the next time i get access to the computer lab.
Oct 1st 2008, 12:12 PM
Jhevon

Quote:

Originally Posted by Hweengee

i didn't realise that there were 6 possible combinations of x and y such that dy/dx=0. thanks for the help so far, so if there are 3 solutions to the quintic would it be right to say that implies 3 horizontal tangents? making for a total of 9?

yes, it would be right to say that

you would have 3 nw solutions corresponding to a third x-value. making 9 pairs of solutions in all

Quote:

or would it be best to use a CAS to solve this problem as this is a question that's supposed to be solved using MAPLE.

you can use Maple to find the solutions to the quintic and thus find the specific points. if you are required to do so, then do it

if my memory of Maple serves me correctly,

Code:

f:=y-> y^5 - y + (1/16); fsolve(f(y)= 0);

should work