limits and hyperbolas (1 Viewer)

[kurt.physics]

I am using MIF right know. In the "the hyperbola" section, example 1 says "sketch xy = -1. I am slightly perplexed by the second method,

look at limits
Denominator cannot equal 0
y cannot equal 0

As x --> 0 on LHS, y --> ∞
As x --> 0 on RHS, y --> –∞

As x--> +∞, y--> 0
As x--> –∞, y--> 0


My question is, how does one know?, Substitution doesnt work, that gives -1÷0 or -1÷∞

 

[undalay]

simply test a number close to zero (or infinity)

so if x approaches 0 from the negative side (that is from the left)

test like -0.01.

-1 / -0.01 = 100
-1 / -0.0001 = 10,000
-1 / -0.0000001 = 10,000,000

 

[Slidey]

xy=-1 is actually a rotation by 45 degrees of x^2-y^2=-1

Anyway, you don't really need to test numbers. Just try these values in your head:
x -> infinity, y -> 0 (x=-1/y so as x gets bigger, y gets smaller)
y -> infinity, x -> 0 (y=-1/x so as y gets bigger, x gets smaller)

And it's an easy matter of noticing that sign(x) = -sign(y) (i.e. when x is negative y is positive and vice versa).

So for example when x is near infinity, y is an infinitesimally small negative amount below zero.

Try this one: y(x^3-x)=1

 

[tacogym27101990]

thats a pretty funky looking graph

 

[kurt.physics]

Try this one: y(x^3-x)=1

See, here is my problem, how do i know if it is plus or minus infinity?

I got

x--> 0, y--> ∞
x--> 1, y--> ∞
x--> -1, y--> ∞

Having used my math graph program, i know that it should be

x--> 0, y--> ∞
x--> 1, y--> plus and minus ∞
x--> -1, y--> plus and minus ∞

 

[undalay]

using xy = -1. for example.

you know it'll reach infinity at x = 0 (or y = 0)
But the thing about limits is, you need to approach it from both sides.

So if you approach x =0 from the negative side.

i.e. x = -negative infinity -> x = 0
then x will be negative for all values in this region.

so the two negatives will cancel out, thus making the limit from this side positive infinity.

However if u come from the other side.
I.e. x = positive infinity -> x = 0
then x will always be positive (but there's still the negative on the 1) so thus the limit on this side will be negative infinity.

However since the righthand limit does not equal the left hand limit, a general limit does not exist.

 

[Slidey]

See, here is my problem, how do i know if it is plus or minus infinity?

I got

x--> 0, y--> ∞
x--> 1, y--> ∞
x--> -1, y--> ∞

Having used my math graph program, i know that it should be

x--> 0, y--> ∞
x--> 1, y--> plus and minus ∞
x--> -1, y--> plus and minus ∞

You use a bit of intuition and case testing.

You know that in general cubics |x^3| approach infinity as |x| approaches infinity. So we can say that in general 1/(x^3-x)
approaches 0 as |x| -> infinity (because of the limit of 1/x as x approaches infinity)
So that's the horizontal asymptotes out of the way (we know there aren't any others because you can isolate y, making this a function - meaning at most 2 asymptotes). Now for the vertical asymptotes:

As |y| -> infinity, x^3-x -> 0, right?
Find the roots: x(x^2-1)=0, thus funny business at x= -1, 0, 1.

As you can probably imagine since x^3-x is a polynomial (always continuous), inverting it will only produce discontinuities at its roots, or x^3-x=0, thus it's safe to say that for x < -1 and x > 1 it begins to behave essentially like y=1/x. That is: if x = 1.001, y = a large positive number, while if x = -1.001, y = a large negative number.

So what about -1 < x < 0 and 0 < x < 1 ?

Well you could graph y=x^3-x and figure out the x value of the turning points and sign of the y value from that. Probably easiest to just derive y=1/(x^3-x) though:
y=(x^3-x)^-1
y'=-(3x^2-1)(x^3-x)^-2=0, stat points at x = 1/sqrt(3), -1/sqrt(3)
(and if y=x^3-x, y'=3x^2-1, so stat pts at x = 1/sqrt(3), -1/sqrt(3)... neat hey?)

Now you can graph either side of the stat points. This will tell you not only if it is a minimum or maximum, but also the sign of the asymptotes it creates (because x=-1, 0 and 1 are vertical asymptotes remember? Confirm this if you want by testing x=0.001, x=0.999, etc)

Don't worry if you don't understand it fully. It's a fairly easy topic, but one which doesn't translate so well over digital text; just keep trying and suddenly a lightbulb will go off and it will feel all nice and fuzzy inside. Besides, 'advanced' graphing techniques like function inversion are generally taught in 4unit.

Blue is y=x^3-x
Yellow is y=1/(x^3-x) or y(x^3-x)=1

And it is possible to derive relations without isolating "y=". It is a 4unit technique called 'implicit differentiation' of which normal differentiation is actually a trivially case. Deriving the function implicitly instead of explicitly:
y(x^3-x)=1
y'(x^3-x)+y(3x^2-1)=0 (product rule + function of a function rule, because y is a function of x)
y'=-y(3x^2-1)/(x^3-x)
y'=(3x^2-1)/(x^3-x)^2 (yay!)

As you can see it is generally pointless to use implicit differentiation if you have the form "y=". However, exceptions do exist:

y=e^(x^2)*(3x-1)^5/(2x-5)^14
ln(y) = x^2 + 5ln(3x-1) - 14ln(2x-5)
y'/y = 2x + 15/(3x-1) - 28/(2x-5)
y' = y(2x + 15/(3x-1) - 28/(2x-5)), which we could simplify if we wanted:
y' =e^(x^2)*(3x-1)^5*(2x + 15/(3x-1) - 28/(2x-5))/(2x-5)^14

Which again could be simplified by finding a common denominator, but that'd be pretty fiddly. Either way it's still shorter than explicit differentiation which would involve 3 applications of the chain rule and 3 applications of the product rule before you even started to collect and simplify.