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Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164685 times)  Share 

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deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8580 on: February 22, 2017, 12:09:41 am »
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7.a)c), there are 3 points of inflection but (π,0) is the minimum, as your working shows, meaning at (π,0) the gradient is a minimum (what the question asks for)
7.b)b)Looking at the graph you can see there is a turning point at x=-1, a minimum, which is also found when f'(x)=0. As f''(x) > 0 at this point, this means the gradient is increasing meaning it is a minimum.
8b)So if it's a local minimum at x=a, it means before x=a (ie when x=a-h) the gradient of f(x) is negative and after x=a (ie when x=a+h) the gradient is positive. We're looking at the gradient of y=f'(x), ie what f''(x) is. Since h is a small value, I believe they intend it such that it's close to zero meaning there's no point of inflection (as I believe a non-SP point of inflection could affect the "positive" statement). So as the gradient is always increasing, the gradient of f'(x) must be positive. This can be drawn as a graph of f'(x) and the y value continuously going up, similar to a linear graph.
Their linear graph shows y=f'(x) and since it's going up, the gradient of that graph is positive.
8.c) The gradient of f(x) is (presumably) always increasing, meaning f''(x) will always be positive. f''(x) is the same as the gradient of f'(x).

Hope this helps, a bit tired so apologies if I'm not too concise with my answer
Thanks a lot!

For 7a)c) I don't think I fully understand what the gradient minimum is? At x=0,pi and 2pi, wouldn't they all be the gradient minimum? They're the same value right?

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8581 on: February 22, 2017, 12:17:18 am »
+3
Thanks a lot!

For 7a)c) I don't think I fully understand what the gradient minimum is? At x=0,pi and 2pi, wouldn't they all be the gradient minimum? They're the same value right?

Drawing a graph, or using a graphing app such as desmos. com might help.
For all 3 of those values, they're a turning point on the gradient graph. But x=π is a minimum (on the gradient graph), meaning that's when the gradient is the lowest/minimum. x=0,2π are both maximums (on the gradient graph) so those are the largest gradients
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deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8582 on: February 22, 2017, 05:05:41 pm »
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Drawing a graph, or using a graphing app such as desmos. com might help.
For all 3 of those values, they're a turning point on the gradient graph. But x=π is a minimum (on the gradient graph), meaning that's when the gradient is the lowest/minimum. x=0,2π are both maximums (on the gradient graph) so those are the largest gradients
Thanks. Last thing: Q8c) why can the second derivative also equal zero (greater and equal to zero), this was what the book had with no explanation.

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8583 on: February 22, 2017, 08:09:12 pm »
+3

I have another teaser for you, which I have been brooding over. It involves a pentagon inscribed in a semi circle, and I have to show that the diametr of the circle from A to E in the diagram attached, is p = sin(4x)/sin(x).

Now I can see that the length AB is 1, and that the vector AO is the radius of the circle. By drawing a vector bisecting the angle from O to a point M, which is halfway between A and B, we get two right angled triangles. I have attached a sketch of what I mean.

You can do this to the four triangles in the semi circle, to get eight right angled triangles, but how you get the ratio I have little clue. ..

If you don't mind explaining the geometry a little bit, because my geometry was NEVER good AT ALL  ::)

James



I've attached my working / the solution :) let me know if there's anything you want me to explain further!

« Last Edit: February 22, 2017, 08:23:38 pm by Shadowxo »
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Jimmonash1991

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Re: Specialist 3/4 Question Thread!
« Reply #8584 on: February 22, 2017, 10:47:39 pm »
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Thanks Shadow,

The rest of the question was fine, but they could have made the diagram a bit more specific couldn't they? They had a similar question involving two tangents to a circle that intersected at a point such that the shaded area between the two points the tangents joined at the circle, A and B, was equal to the rest of the area of the circle. That is where I have seen 2*pi - (remaining angle) before.

What I am trying to illustrate is though Shadow, is the f***ing diagram "LOOKED" like it was the radius of the circle, but whatever.

Sorry to bother you again mate, but since my geometry is so bad, could you help me with another one?

I have attached the question and my little sketch. It basically involves two isosceles triangles, one inscribed in another, and I really would like to see how the length BD = 1 + sin(18)

I got all the angles right, but I cannot seem to get the magnitudes of the sides right required. I know BE = 1, then CD = sin(18) and DE = sin(18) from my diagram.

So the length BD should be 1 - 2sin18

If I used the triangle BAD, letting the angle BAD be 54 degrees, then BD = sin(54) = sin(18 + 36) = sin(18)cos(36) + cos(18)sin(36)

Finally, letting angle ABD be 36 degrees, and using BA as the hypotenuse, then BD = cos36 = 1 - 2sin^2(18)

So you can see why I am at a loss? :)

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8585 on: February 22, 2017, 11:16:48 pm »
+2
Hint:
For BC:
We know AC=AE=1
We know ABC=BAC=36º

For CD:
We know CAD=18º
CD = sin18º as you worked out

BD=BC+CD
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8586 on: February 23, 2017, 03:28:00 pm »
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Thanks Shadow,

The rest of the question was fine, but they could have made the diagram a bit more specific couldn't they? They had a similar question involving two tangents to a circle that intersected at a point such that the shaded area between the two points the tangents joined at the circle, A and B, was equal to the rest of the area of the circle. That is where I have seen 2*pi - (remaining angle) before.

What I am trying to illustrate is though Shadow, is the f***ing diagram "LOOKED" like it was the radius of the circle, but whatever.

Sorry to bother you again mate, but since my geometry is so bad, could you help me with another one?

I have attached the question and my little sketch. It basically involves two isosceles triangles, one inscribed in another, and I really would like to see how the length BD = 1 + sin(18)

I got all the angles right, but I cannot seem to get the magnitudes of the sides right required. I know BE = 1, then CD = sin(18) and DE = sin(18) from my diagram.

So the length BD should be 1 - 2sin18

If I used the triangle BAD, letting the angle BAD be 54 degrees, then BD = sin(54) = sin(18 + 36) = sin(18)cos(36) + cos(18)sin(36)

Finally, letting angle ABD be 36 degrees, and using BA as the hypotenuse, then BD = cos36 = 1 - 2sin^2(18)

So you can see why I am at a loss? :)

First of all, yes the diagram for that question was pretty misleading, it did look like AE was the diameter.

As to why the ways you did it didn't work:

First Way:
BD = BE - DE
We don't know the length of BE (I think you got confused by this, AE and AC are the ones that have a length of 1)
And DE = sin(18º).
So BD = BE - sin(18º). This has an unknown so it won't work (could be found via a long way)

Second way:
Using BAD = 54º
BD = AB*sin(54º) (not just sin54º)
sin(x) =O/H -> O=H*sin(x)
sin(x) = BD/AB -> BD = AB*sin(x)
Again, we don't know BA so this won't work (could be found via a long way)

Third way:
Using ABD = 36º
BD=AB*cos(36º)
Same answer as before, and we don't know the length of AB.

The solution:
BD = BC + CD
As BAC = ABC = 36º, ABC is an isosceles triangle, with the length BC = AC. As we know AC = 1, we therefore know BC = 1.
We know CAD = 18º from earlier, and as AC = 1
CD = AC*sin(18º) = 1*sin(18º) = sin(18º)

BD = BC + CD = 1 + sin(18º) as required
:)
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Mattjbr2

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Re: Specialist 3/4 Question Thread!
« Reply #8587 on: February 24, 2017, 03:44:36 pm »
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Hi guys,
For the life of me I can't seem to figure out why my answer differs from the book's answer.
I tried simplifying my method as much as possible.
Any help would be appreciated!
Link to image: http://i.imgur.com/eDm9yb3.jpg

Edit: The book subtracted the area of the left side from the area of the right side in step 6 rather than adding them together like i did. But why? The area of the left side is already positive! And we're asked to find the total area of both sides, so why subtract rather than add?
Edit: Omg even my CAS is agreeing with my book. This question seems as basic as 1+1=2, yet I can't wrap my head around what's wrong with my logic.
« Last Edit: February 24, 2017, 03:59:53 pm by Mattjbr2 »
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8588 on: February 24, 2017, 05:11:48 pm »
+3
Hi guys,
For the life of me I can't seem to figure out why my answer differs from the book's answer.
I tried simplifying my method as much as possible.
Any help would be appreciated!
Link to image: http://i.imgur.com/eDm9yb3.jpg

Edit: The book subtracted the area of the left side from the area of the right side in step 6 rather than adding them together like i did. But why? The area of the left side is already positive! And we're asked to find the total area of both sides, so why subtract rather than add?
Edit: Omg even my CAS is agreeing with my book. This question seems as basic as 1+1=2, yet I can't wrap my head around what's wrong with my logic.

It's just a little thing, all your working is right :)
they want you to find the integral from -1 to 2, but the integral from -1 to 0 will be negative. You did all the right steps in finding the area, but as the integral from -1 to 0 is negative, you have make the area negative - so subtract the area π/6+√3-2 (as this area is positive). The mistake you made is finding the area instead of the integral.
Example for clarification:
Integral from 0 to 2pi of sin(x) is 0. The first half is positive and the second half is negative, so overall it amounts to 0.

Does this make sense? You can pm me if it doesn't :D
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Mattjbr2

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Re: Specialist 3/4 Question Thread!
« Reply #8589 on: February 24, 2017, 07:18:25 pm »
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Huh. I assumed they wanted the literal area, not the integral. So an integral is all the positive areas minus all the negative areas? Huh. That makes sense in terms of finding the displacement of a V-T graph in Physics, but its weird because i assumed i must pretend the areas beneath the graph are included in the overall area as positive areas. I guess its better this way because when i get to kinematics, I'd have to subtract the area from the bottom anyway. It makes sense now. Thanks! :)
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8590 on: February 24, 2017, 07:32:46 pm »
+1
No problem :)
But just remember, when you find the integral underneath the x axis, it'll be negative already. It was only positive this time because you calculated using the positive values and areas.
The integral from -1 to 2 in this case would be the integral from -1 to 0 (which will be negative if you calculate it) plus the integral from 0 to 2 (which will be positive)
So the integral will be all the sections added together, but the ones underneath the x axis will be negative already. If you're only taking the positive areas, then you have to change the regions under the x axis to negative because it's below the x axis
Hope this clears things up  :D
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excelsiorxlcr

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Re: Specialist 3/4 Question Thread!
« Reply #8591 on: February 25, 2017, 01:01:59 pm »
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Hi, how would you solve something like cos(sin-1(4/5)) by hand? I think it's something to do with drawing a triangle, but I don't know how exactly to go about it. Questions like Page 139 question 6) I mean.

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Re: Specialist 3/4 Question Thread!
« Reply #8592 on: February 25, 2017, 01:18:29 pm »
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Hi, how would you solve something like cos(sin-1(4/5)) by hand? I think it's something to do with drawing a triangle, but I don't know how exactly to go about it. Questions like Page 139 question 6) I mean.

excelsiorxlcr

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Re: Specialist 3/4 Question Thread!
« Reply #8593 on: February 25, 2017, 01:33:49 pm »
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Thanks, also how do you solve sin2x cos3x = cos x? I took cos x to the other side then took it out as a common factor and ended up with cosx (cos2x sin2 x - 1) = 0. I know how to solve cos x = 0 but not cos2x sin2 x - 1 = 0.

Sine

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Re: Specialist 3/4 Question Thread!
« Reply #8594 on: February 25, 2017, 01:48:04 pm »
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Thanks, also how do you solve sin2x cos3x = cos x? I took cos x to the other side then took it out as a common factor and ended up with cosx (cos2x sin2 x - 1) = 0. I know how to solve cos x = 0 but not cos2x sin2 x - 1 = 0.
you probs can't solve it because there aren't any solutions  :P

Think about at what point must cos2x sin2 x  = 1. Only when cos2x and sin2x are equal to 1. cos2x and sin2 x will equal 1 only when cos(x)=1,-1 same goes with sin(x)=1,-1. However this never happens at the same time hence no solutions.
« Last Edit: February 25, 2017, 01:57:46 pm by Sine »