Andiio's Questions Thread :)

[Andiio]

End points are differentiable.

E.g. r = (t+1)^2 for t >= 0
Find velocity v(0).

Ah, but why are they denoted with an open circle? (In the above case, anyways)

And also, for the graph of say, log e (x), how would you determine the gradient function (without differentiating the function, purely through graphical interpretation)

[evaever]

End points are differentiable.

E.g. r = (t+1)^2 for t >= 0
Find velocity v(0).

Ah, but why are they denoted with an open circle? (In the above case, anyways)

What case did you refer to?

[Andiio]

End points are differentiable.

E.g. r = (t+1)^2 for t >= 0
Find velocity v(0).

Ah, but why are they denoted with an open circle? (In the above case, anyways)

What case did you refer to?

The image I posted up (above Pixon's post)

[evaever]

(0,2) is not an endpoint. It is not in the function (open circle).
The other two are endpoints(solid dots).

[Andiio]

Yeah I meant the endpoints (filled in circles).. i'm just not sure why there are 2 open circles at (-3, 2/3) and (3, -5/3) in the gradient function.
I've attached the answer.

[vgardiy]

I thought endpoints weren't differentiable?

[brightsky]

They aren't. Thinking about it graphically, you can't really get the tangent of an end point. Thinking about it through limits, you can only find the limit in one direction, but not the other. Hence, no you can't differentiate endpoints.

[Pixon]

Yeh, just always go back to first principles if you are stuck and think about what the gradient function means with limits and you'll see it clearly. Strong first principles will basically get you everywhere

[evaever]

End points are differentiable.
The problem is: It is the way that differentiability is defined in a simple way in the Study Design, rendering end points not differentiable. A first year calculus textbook gives you a proper definition.
In the past no questions were asked in the vcaa exams about differentiability of end points. You may find them in practice exams or some text books. Some said yes, some no.

[Pixon]

End points are differentiable.
The problem is: It is the way that differentiability is defined in a simple way in the Study Design, rendering end points not differentiable. A first year calculus textbook gives you a proper definition.
In the past no questions were asked in the vcaa exams about differentiability of end points. You may find them in practice exams or some text books. Some said yes, some no.

Yeh, I know what you are talking about. But it's also not worth confusing students with it.

Now...I hope I don't confuse you in typing this but this is important

There is a certain type of VCAA question which sort of addresses this differentiation issue but basically no student will pick up on it and if they do they won't understand it but will simply know what to do.

I'll make up an example.
    For what interval is f(x) decreasing?

Now usually you would simply think of when
and so your answer would be

But the answer actually should include 0 is a point. Now I won't go into it because I deem myself insufficient in knowledge to explain it, but look out for these questions that refer to DECREASING/INCREASING and remember to include those endpoints or stationary points (if i didn't restrict the domain, the answer still remains the same including when as the start of when the function decreases

For more info: http://en.wikipedia.org/wiki/Monotonic_function

Just be aware of this because it is a vcaa question...I think I saw it in 2009 and it really annoyed me but then it was explained to me clearly with limits and such.

[Andiio]

Zzz LOL so basically, wrt (with respect to) VCE Maths, (stupid VCAA!), we cannot differentiate endpoints?

Really gotta revise all the limit stuff, what do you think are the main things/points I should focus on reinforcing?

[Andiio]

What's the general procedure for a question like this?

Taking note of the domain and subbing in values from there?

Like, e.g., with the first question,(part a) would you 'test' x=0?



Thanks!

[brightsky]

For a) there's obviously a discontinuity at x = 0. but , hence there is a step discontinuity at x = 0.

Approach the other ones in the same way.

EDIT: woops xD

[Andiio]

Could anyone please provide the proof for this?

I understand the a = tan(theta 2) <-- compound angle derivation, but not sure about why m = tan (theta) and why a = theta₂ - theta₁.

Thanks!


EDIT: MY BAD LOL DIDN'T NOTICE

[brightsky]

You sure this is the right image, lol?