Post by: /0 on May 02, 2008, 07:32:36 pm
Post by: Mao on May 02, 2008, 07:40:15 pm
that reasoning doesnt make sense, because that practically goes against stationary points in general: for a turning point such as
if you are applying the first principles, you must realise that regardless how small
its really hard to explain...
Post by: /0 on May 02, 2008, 08:20:30 pm
Post by: Mao on May 02, 2008, 08:40:31 pm
but you should keep in mind that:
gradient = derivative = tangent =
they are all infinite.
Post by: bigtick on May 07, 2008, 10:48:43 pm
Post by: Fitness on May 21, 2008, 12:34:04 am
so it can't be a function?
It's a 'one-to-one' graph. Of course it's a function.
Consider the graph of... it has an infinite gradient at x = 0.
NO! It's not infinite! It's undefinable!
Post by: Neobeo on May 21, 2008, 06:59:46 am
so it can't be a function?
It's a 'one-to-one' graph. Of course it's a function.Consider the graph of
NO! It's not infinite! It's undefinable!
Actually the gradient is well-defined at x = 0. So it is in fact infinite.
Post by: /0 on May 21, 2008, 07:14:09 pm
Post by: Glockmeister on May 21, 2008, 07:22:30 pm
Post by: bigtick on May 21, 2008, 10:05:30 pm
Post by: Fitness on May 22, 2008, 02:38:41 am
Actually the gradient is well-defined at x = 0. So it is in fact infinite.
see
Since f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.
=> f'(0) = 1/0
Therefore it is undefined. kthx.
Post by: Mao on May 22, 2008, 08:45:15 pm
but infinitity is a well defined concept!Actually the gradient is well-defined at x = 0. So it is in fact infinite.
seeSince f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.
=> f'(0) = 1/0
Therefore it is undefined. kthx.
Post by: Fitness on May 22, 2008, 09:35:46 pm
but infinitity is a well defined concept!Actually the gradient is well-defined at x = 0. So it is in fact infinite.
seeSince f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.
=> f'(0) = 1/0
Therefore it is undefined. kthx.
Yes, infinity is a well defined concept.
BUT IT ISN'T INFINITE IN THIS CASE!
That would be just like saying 1/0 is infinity when, in fact, it isn't.
Take a hyperbola, for example: y=1/x
Now, at x=0, the y value is not 'infinity', it is undefined and visa-versa.
Post by: Mao on May 22, 2008, 09:40:50 pm
why cant it have a tangent with an infinite gradient? [its equation is x=0]but infinitity is a well defined concept!Actually the gradient is well-defined at x = 0. So it is in fact infinite.
seeSince f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.
=> f'(0) = 1/0
Therefore it is undefined. kthx.
Yes, infinity is a well defined concept.
BUT IT ISN'T INFINITE IN THIS CASE!
That would be just like saying 1/0 is infinity when, in fact, it isn't.
Take a hyperbola, for example: y=1/x
Now, at x=0, the y value is not 'infinity', it is undefined and visa-versa.
Post by: Fitness on May 22, 2008, 09:43:24 pm
why cant it have a tangent with an infinite gradient? [its equation is x=0]but infinitity is a well defined concept!Actually the gradient is well-defined at x = 0. So it is in fact infinite.
seeSince f '(x) = 1/x^(2/3), so f '(0) is undefined. But as x -> 0, f '(x) -> oo, this means that the tangent lines become steeper and steeper as x -> 0.
=> f'(0) = 1/0
Therefore it is undefined. kthx.
Yes, infinity is a well defined concept.
BUT IT ISN'T INFINITE IN THIS CASE!
That would be just like saying 1/0 is infinity when, in fact, it isn't.
Take a hyperbola, for example: y=1/x
Now, at x=0, the y value is not 'infinity', it is undefined and visa-versa.
Because 1/0 is not infinity. The end lol :P
Post by: Mao on May 22, 2008, 09:52:37 pm
have you checked that this function complies with the rules of differentiability at x=0? [i'm pretty sure vertical tangents are exceptions to differentiation, and this is marked by its derivative's domain. this, however, does not imply it doesn't have a gradient]
if you differentiate this with vector calculus, i'm sure you'll end up with a unit tangent of
doing vector calculus now [Captain: dont laugh, haha]
the relationship definining this function is continuous, and can be described by the relationship:
this relationship can then be defined as a space-curve relative to O as:
hence its unit tangent can be described as:
yay vertical tangent =]
Post by: Fitness on May 22, 2008, 10:10:21 pm
the relationship definining this function is continuous, and can be described by the relationship:
Incorrect.
if y3 = x
then y = x1/3 and y = -x1/3
you can't tell me that x1/3 = -x1/3
Post by: Mao on May 22, 2008, 10:11:24 pm
it is absurd to say that a unit circle does not have a gradient at x=1, yet this certainly cannot be found by differentiation.
differetiation is based on the change of one variable against another. one the independant variable does not change, then it fails. i.e. vertical tangents.
Post by: Fitness on May 22, 2008, 10:13:46 pm
what I meant to say was, derivatives are not the be-all and end-all when it comes to gradients.
it is absurd to say that a unit circle does not have a gradient at x=1, yet this certainly cannot be found by differentiation.
differetiation is based on the change of one variable against another. one the independant variable does not change, then it fails. i.e. vertical tangents.
An end point doesn't have a vertical tangent... It has no tangent at all. That's what I am trying to say.
Post by: bigtick on May 22, 2008, 11:22:06 pm
Infinity is a well defined concept, but 1/0 is undefined.
1/0 not= oo, but lim(x->0) of 1/x = oo.
The gradient of a vertical tangent is not= 1/0, it is = oo.
Post by: Mao on May 23, 2008, 08:42:47 am
the relationship definining this function is continuous, and can be described by the relationship:
Incorrect.
if y3 = x
then y = x1/3 and y = -x1/3
you can't tell me that x1/3 = -x1/3
absolutely absurd
but it doesnt have an end-point. the cubic root function is a continuous function defined forwhat I meant to say was, derivatives are not the be-all and end-all when it comes to gradients.
it is absurd to say that a unit circle does not have a gradient at x=1, yet this certainly cannot be found by differentiation.
differetiation is based on the change of one variable against another. one the independant variable does not change, then it fails. i.e. vertical tangents.
An end point doesn't have a vertical tangent... It has no tangent at all. That's what I am trying to say.
I think you are confusing the cubic root function with the square root function.
Post by: Fitness on May 23, 2008, 01:04:41 pm
I realised just before I went to bed last night what I did :/
Post by: Captain on May 24, 2008, 10:59:34 pm
doing vector calculus now [Captain: dont laugh, haha]
I laughed hard.