Bazza's 3/4 Question Thread

[paulsterio]

but isn't there a force of mg (tension) acting on both sides of the rope?
Should you add these together of is tension a force / unit area?

Yes, there are, but look at it this way.

Let's do this situation instead:



You see that of course, in both situations, the force meter will show the same value of mg?

[WhoTookMyUsername]

ah, yeah i think i see it now , when it's nailed there's also a "pulling force" at the top perse, which N already takes into account, thanks paul

[nina_rox]

Sorry, I have a quick question.
I'm getting confused with signed and unsigned graphs. So for a velocity time graph is the distance the whole area under the graph (disregarding the negative areas and calculating them as positive - so unsigned) while the displacement regards the negative area (so signed)?

Thank you very much

That's correct

Yeah I thought my logic was correct and then I went to do some checkpoint questions and there was a question that asked for the distance and the area below the x axis has been subtracted from the area above the x axis in the answers? Is this incorrect?

[paulsterio]

Yes that is incorrect.

[nina_rox]

Yes that is incorrect.

Thanks very much Paul

[WhoTookMyUsername]

If the second derivative = 0 at x=a
And when x <a f''(x) >0 and when x>a
F''(x) is < 0
Is the point x=a a point of inflection?

Is there a point of inflection for the graph x^(1/3)? ( gradient is undefined at 0,0, as is second derivative)

Thanks

[brightsky]

1) yes cos there's a change in concavity
2) no x^(1/3) has no point of inflection. you should be able to visualise the graph as the inverse of x^3. the tangent at x =0 is x = 0.

[WhoTookMyUsername]

thanks for help
but if the answer for 1) is yes answer for 2) has to be yes double deriving x^(1/3) on CAS gives -2/(9x^(5/3)) and subbing in x = 1 and x = -1 gives -2/9 and 2/9 respectively

[Phy124]

I would have to disagree and say that does have a point of inflection when

Remember; does not necessarily need to equal for the coordinate to be a point of inflection, it can also be non existent.

In this case, the point in does not exist and at on a change of concavity occurs, so it is a point of inflection.

[brightsky]

thanks for help
but if the answer for 1) is yes answer for 2) has to be yes double deriving x^(1/3) on CAS gives -2/(9x^(5/3)) and subbing in x = 1 and x = -1 gives -2/9 and 2/9 respectively

note that f''(a) = 0 is a NECESSARY condition for x = a to be a point of inflection.

so basically when verifying whether or not x = a is a point of inflection:
1) verify that f''(a) = 0 (NECESSARY condition, but not SUFFICIENT, example:y = x^4)
2) ensure that f''(a - some small number) and f''(a+some small number) have opposite signs

only when both conditions are satisfied can x = a have the privilege of being called a point of inflection.

[WhoTookMyUsername]

I think what ranga said is correct, i don't think the first point needs to be satisfied as concavity would still change regardless?

[brightsky]

I think what ranga said is correct, i don't think the first point needs to be satisfied as concavity would still change regardless?

http://www.encyclopediaofmath.org/index.php/Point_of_inflection

scroll down to where they talk about necessary and sufficient existence conditions.

[WhoTookMyUsername]

http://math.uchicago.edu/~vipul/teaching-0910/152/infinitycuspsasymptotes.pdf

Control f points of vertical tangent

[brightsky]

http://math.uchicago.edu/~vipul/teaching-0910/152/infinitycuspsasymptotes.pdf

Control f points of vertical tangent

ahh i stand corrected. never knew that!

[Phy124]

ahh i stand corrected. never knew that!

Not surprising, I know of many people who have already studied this level of mathematics and either don't know or have forgotten this rule. It's great that you've learnt it before you need it and will definitely never forget it