IARTV

[wombifat]

I was given the 2009 IARTV methods practice exams by my teacher today and have a couple of questions about the technology free one

firstly, are these exams credible? Because one the questions I'm not sure if it's really correct. Also, the exam seems much more difficult than the VCAA technology free exams.

Also with regards to this question

Find the point on the graph of y=2Inx which is closest to the line with equation y=2x+1

how are we supposed to know that it's the point with the same gradient? I don't remember ever learning anything like that. Is this likely to come up on the exam?

Also, it asks you to sketch the graph of (x-4)^3\5
are we required to know how to do this without technology? do we just have to memorise the shapes of all these graphs?

[GerrySly]

Also, it asks you to sketch the graph of (x-4)^3\5
are we required to know how to do this without technology? do we just have to memorise the shapes of all these graphs?

If they give you a plane to draw on, just put in all the values they give you (for a graph with domain [-4, 4] put in -4, -3, -2, -1, 0, 1, 2, 3, 4) and then just put dots for all those points. You will start to see them linking up and you can draw it. Though ideally you would want to memorise a bunch of graphs basic forms then be able to apply translations to these.

[khalil]

Or...you could even recall that it is a basic cubic graph, translated 4 to the right and dilated 1/5 from the y axis.

[GerrySly]

Or...you could even recall that it is a basic cubic graph, translated 4 to the right and dilated 1/5 from the y axis.

I thought it was , but if I missed the mark, then yeah that'll be the way to do it

[naved_s9994]

Find the point on the graph of y=2Inx which is closest to the line with equation y=2x+1

I believe this is examinable, as I did a question similar in Essential textbook,
quite some time ago.

[TrueTears]

Find the point on the graph of y=2Inx which is closest to the line with equation y=2x+1

how are we supposed to know that it's the point with the same gradient?

I wouldn't do it that way....

It's just the minimum of the modulus difference, ie

let y1 = 2lnx and y2 = 2x+1

Sketch the graph just to check

the point closet is just the minimum value of |y2-y1|

[khalil]

Or...you could even recall that it is a basic cubic graph, translated 4 to the right and dilated 1/5 from the y axis.

I thought it was , but if I missed the mark, then yeah that'll be the way to do it

lol. Yeh, I think you're right.

[TrueTears]

Also, it asks you to sketch the graph of (x-4)^3\5
are we required to know how to do this without technology? do we just have to memorise the shapes of all these graphs?

Memorise the basic graph of each of those functions, then apply transformations.

[kamil9876]

Find the point on the graph of y=2Inx which is closest to the line with equation y=2x+1

how are we supposed to know that it's the point with the same gradient?

I wouldn't do it that way....

It's just the minimum of the modulus difference, ie

let y1 = 2lnx and y2 = 2x+1

Sketch the graph just to check

the point closet is just the minimum value of |y2-y1|

no it's not. It's actually the minimum value of . But gradient method is easier computationally.

proof for gradient method in attatchement:

refer to the figure, suppose we want to find the minimum distance between the bold line and the bold curve. Treat the Bold Line as the as a new x-axis. The new y axis will be perpendicular to this (as drawn). The minimum distance between the Bold curve and the new x axis happens at the point where it's a minimum turning point (looking at it as a graph in the new x-y axis). When this happens the gradient is 0 (in the new axis) and hence parralel to the x axis. Hence it happens when the tangent is parralel to the line, as required.

[TrueTears]

Find the point on the graph of y=2Inx which is closest to the line with equation y=2x+1

how are we supposed to know that it's the point with the same gradient?

I wouldn't do it that way....

It's just the minimum of the modulus difference, ie

let y1 = 2lnx and y2 = 2x+1

Sketch the graph just to check

the point closet is just the minimum value of |y2-y1|

no it's not. It's actually the minimum value of . But gradient method is easier computationally.

wdf?? If you wanna find the distance should be ... Why you mod the whole thing...

[kamil9876]

just realised, modulus sign was unnecesary since the sum of two squares is always positive. Btw minimum value of occurs when the minimum value of occurs so it woudl still give you the correct thing.

[TrueTears]

just realised, modulus sign was unnecesary since the sum of two squares is always positive. Btw minimum value of occurs when the minimum value of occurs so it woudl still give you the correct thing.

It's not meant to be a mod sign, It's meant to be a square root.

[kamil9876]

I meant for the purpose of answering the question the square root can be ommited since we're after when the value is at a minimum, and it's at a minimum when square root is at a minumum.

[wombifat]

Find the point on the graph of y=2Inx which is closest to the line with equation y=2x+1

how are we supposed to know that it's the point with the same gradient?

I wouldn't do it that way....

It's just the minimum of the modulus difference, ie

let y1 = 2lnx and y2 = 2x+1

Sketch the graph just to check

the point closet is just the minimum value of |y2-y1|

no it's not. It's actually the minimum value of . But gradient method is easier computationally.

proof for gradient method in attatchement:

refer to the figure, suppose we want to find the minimum distance between the bold line and the bold curve. Treat the Bold Line as the as a new x-axis. The new y axis will be perpendicular to this (as drawn). The minimum distance between the Bold curve and the new x axis happens at the point where it's a minimum turning point (looking at it as a graph in the new x-y axis). When this happens the gradient is 0 (in the new axis) and hence parralel to the x axis. Hence it happens when the tangent is parralel to the line, as required.

and we are just supposed to know this?

[/0]

very nice... rotating the axes n all