VCE Specialist 3/4 Question Thread!

[Eiffel]

Yeah i know how to use all applications of differentiation, however the problem is that it takes up the whole page (not complaining here), so what i was hoping for was someone doing it in very little steps (least steps - may be doing more than required).

I know you let u = cos(2x), v = x, use your quotient rule for the first differentiate. Then you have this nasty thing on the numerator, in which you let that equal u = 3 terms of something, and then within "u", you need to differentiate some of the terms through product/chain, which gets annoying (is this what you have to do?) and confusing since you have a derivative, then you derive that again but while you derive that you need to also derive whats inside the first derivative. Make sense?

[kinslayer]

Personally, this is how I would do it:

First, use the product rule to get:

d/dx(cos(2x)/x) = d/dx(1/x * cos(2x)) = -2sin(2x)/x - cos(2x)/x^2

Then you can use the product rule again:

d/dx(-2sin(2x)/x - cos(2x)/x^2) = -4cos(2x)/x + 2sin(2x)/x^2 + 2sin(2x)/x^2 + 2cos(2x)/x^3

Now you're done, but you can make it neater by writing everything over x^3, like in the answer key.

I never use the quotient rule -- I split it up like I've done there, and use the chain rule on the 1/( ... ) part if necessary.

[Eiffel]

How can one interpret this?

i know there is ALWAYS (?) a stationary point of inflection for f(x) when f''(x) = 0, but what does the + 0 - values mean? (i know + means local min etc) but in this context how does this verify a point of inflection ?

e.g.
For example, if f (x) = x4:
f′(x) = 4x3
= 0 when x = 0. Hence a stationary point occurs at x = 0.
For f′′(x) = 12x2 = 0, a solution occurs when x = 0.
Thus, it appears, there is a stationary point of inflection at x = 0.
But the sign diagram of the second derivative (see figure at right) shows
that f ′′(x) does not change sign at x = 0.
Therefore, there is not a stationary point of inflection at x = 0. --- but when i graph it, we do see a stationary point of inflection at x = 0

[Eiffel]

Additionally. I am aware that f''(x) > 0 = local min..... < 0 local max..... = 0 what is this (poi or s poi) ?

As well going about finding the minimum gradient of the curve with equation

[Cogglesnatch Cuttlefish]

I'm having some trouble wrapping my head around linear dependence.  Can someone explain how any 3 non-parallel vectors in a plane are linearly dependent while 3 non-parallel vectors in 3D are not?

[keltingmeith]

How can one interpret this?

i know there is ALWAYS (?) a stationary point of inflection for f(x) when f''(x) = 0, but what does the + 0 - values mean? (i know + means local min etc) but in this context how does this verify a point of inflection ?

e.g.
For example, if f (x) = x4:
f′(x) = 4x3
= 0 when x = 0. Hence a stationary point occurs at x = 0.
For f′′(x) = 12x2 = 0, a solution occurs when x = 0.
Thus, it appears, there is a stationary point of inflection at x = 0.
But the sign diagram of the second derivative (see figure at right) shows
that f ′′(x) does not change sign at x = 0.
Therefore, there is not a stationary point of inflection at x = 0. --- but when i graph it, we do see a stationary point of inflection at x = 0

I'm... Honestly confused by all of this. y=x^4 has no points of inflection, stationary or otherwise. Also, when f''(x)=0, there isn't ALWAYS a stationary point of inflection. In most cases, this refers to a point of inflection, yes, but it's not ALWAYS stationary.

Additionally. I am aware that f''(x) > 0 = local min..... < 0 local max..... = 0 what is this (poi or s poi) ?

As well going about finding the minimum gradient of the curve with equation

First answered above, for the second:

Classic max/min problem. So, you want to find the minimum of the gradient, let's use our trusty 3-part method for max/min problems:

1. Derive the equation for what you want to maximise/minimise.
So, we want the minimum of the gradient, which means we need the equation for the gradient - which is dy/dx:



2. Differentiate this equation and find its zeroes.
So, we differentiate again:



And find its zeroes:



3. Substitute this into the equation from step 1, and compare to the end-points.

When x=2, dy/dx=3(2)^2-12(2)-8=12-24-8=-20

The end-points both go off to +infinity, so we can see that the minimum gradient is -20.

I'm having some trouble wrapping my head around linear dependence.  Can someone explain how any 3 non-parallel vectors in a plane are linearly dependent while 3 non-parallel vectors in 3D are not?

Well, why are they linearly dependent in the plane? What's different about 2D and 3D that might mean the third vector is no longer dependent on the other two?

[IndefatigableLover]

So a question asked me to find the points of intersection of two ellipses and then go on to show that the points of intersection are the vertices of a square.

If my "vertices" were points such that they were: (-1,-1), (-1,1), (1,1), (1,-1) with the centre being the origin, is it enough to say that all four points are equal distance apart from each other thus the "vertices" form a square or would I need to prove it using a method such as vectors (E.g Equal Length followed by Dot Product)?

[keltingmeith]

So a question asked me to find the points of intersection of two ellipses and then go on to show that the points of intersection are the vertices of a square.

If my "vertices" were points such that they were: (-1,-1), (-1,1), (1,1), (1,-1) with the centre being the origin, is it enough to say that all four points are equal distance apart from each other thus the "vertices" form a square or would I need to prove it using a method such as vectors (E.g Equal Length followed by Dot Product)?

Doesn't have to be a vector method, but you do also need to show that each length is perpendicular to the other. Same magnitude could be a rhombus, after all. (And technically, a square is just a special case of a rhombus. )

[IndefatigableLover]

Doesn't have to be a vector method, but you do also need to show that each length is perpendicular to the other. Same magnitude could be a rhombus, after all. (And technically, a square is just a special case of a rhombus. )

Haha well my school is doing vectors first so I wanted to relate it to vectors rather than just use "m" and "-1/m" I suppose but I definitely agree (since the solutions just assume it's a square which is fair enough but it was a "show that" question which left me a bit confused on why they left it like that I suppose)
Thanks though EulerFan101!

[Chang Feng]

For question 6.c) how is AC=BC=1  and how is abc an isosceles triangle??
Thanks

[abcdqd]

(Image removed from quote.)
For question 6.c) how is AC=BC=1  and how is abc an isosceles triangle??
Thanks

Since ABE is an isosceles triangle, .

Now we are given that ACE is also an isosceles triangle with AC=AE, so

That means . Therefore

We see that , so ABC is an isosceles triangle, and hence AB=AC=1

[TheAspiringDoc]

Hi ,
could someone explain to me how to integrate x^3/(x-1) with boundary points 5 and 2? its as though its completely different from integrating x^2/(x-1)
Thanks!!

[keltingmeith]

Hi ,
could someone explain to me how to integrate x^3/(x-1) with boundary points 5 and 2? its as though its completely different from integrating x^2/(x-1)
Thanks!!

... Where did you get this question from? I don't think it belongs in the Further Question Thread...

[TheAspiringDoc]

... Where did you get this question from? I don't think it belongs in the Further Question Thread...

Thought it up, I've been learning integration recently.

[keltingmeith]

Thought it up, I've been learning integration recently.

Welp, I'd do it by parts, but I think in future it'd be best to leave non-relevant maths questions out of the further/methods/specialist question threads.