VCE Specialist 3/4 Question Thread!

[Shadowxo]

For these 2 calc questions http://m.imgur.com/a/cwa1L

Q8f) why are they treating e^x as if it is just x? What's the logic behind what the answer did? I was abit lost doing this question.

Q12a) when you flip 6/x -> x/6, why do both of the inequalities flip? I thought you'd only flip the inequality sign of -1 because you'll be dividing by a negative number? Whereas for 1, you won't be so you don't flip the sign? Do you flip both both regardless if you're flipping 'the angle' (6/x)?

8. f) The derivative of tan^-1(f(x)) is \(\frac{1}{1+f(x)^2}\times f'(x)\)
In this case, f(x) = e^x. This results in the answer provided
12. Intuitive way: -1≤x≤1. \(\frac{6}{x}\) = 1 when x = 6, and it equals -1 when x = -6. As x becomes greater than 6, \(\frac{6}{x}\) becomes less than 1 and closer to zero. As x becomes less than -6, \(\frac{6}{x}\) becomes greater than -1 and closer to zero. Therefore x ≥ 6 or x ≤ -6.

[de]

For these 2 calc questions http://m.imgur.com/a/cwa1L

Q8f) why are they treating e^x as if it is just x? What's the logic behind what the answer did? I was abit lost doing this question.

Q12a) when you flip 6/x -> x/6, why do both of the inequalities flip? I thought you'd only flip the inequality sign of -1 because you'll be dividing by a negative number? Whereas for 1, you won't be so you don't flip the sign? Do you flip both both regardless if you're flipping 'the angle' (6/x)?

For the first one-they are treating e^x as just x (or "u" to avoid confusion) so as to use the chain rule.
The second one-try each of the inequalities separately. 6/x>=-1 is definitely true if x is positive, so worry about when x is negative then -6/x<=1 and when you multiply through by the negative "x" you get x<=-6. 6/x<=1 is definitely true if x is negative so worry about it being positive and we get x>=6

[geminii]

Hey guys,

I've attached a part of my textbook as a picture. It says that a = rcos theta, and that b = rsin theta, but I have no idea why that is. Could anyone please explain this?

Thanks so much!

[Shadowxo]

Hey guys,

I've attached a part of my textbook as a picture. It says that a = rcos theta, and that b = rsin theta, but I have no idea why that is. Could anyone please explain this?

Thanks so much!

Hi,
So cos(x) = adjacent/ hypotenuse. In this case, the hypotenuse is r, adjacent is a, opposite is b. cos(x) = a/r so a = rcos(x)
And sin(x) = opposite / hypotenuse. sin(x) = b/r so b = rsin(x)
Hope this helps
Also, this is useful to know for physics, and speeds etc (finding the horizontal and vertical speeds), and resolution of forces

[deStudent]

http://m.imgur.com/a/XBYv1

For Q9) how do I determine that it's C, more specifically how for we know cos is negative? Given the domain, it can also be positive right?

SA part b) just to confirm that the vertex is (6,-1)? Not sure if it's question marked because I got it wrong or my messy handwriting.

Thanks

[lzxnl]

http://m.imgur.com/a/XBYv1

For Q9) how do I determine that it's C, more specifically how for we know cos is negative? Given the domain, it can also be positive right?

SA part b) just to confirm that the vertex is (6,-1)? Not sure if it's question marked because I got it wrong or my messy handwriting.

Thanks

9. To get a full curve, you need a full period of sine and cosine. The period of the sine and cosines functions in C is pi.

b. Here's a neat trick. To find vertices of a hyperbola, you can solve either dy/dx = 0 or dx/dy = 0, depending on which one is allowed to be zero. In practice, what you do is you set the perfect square brackets to zero and see if that's allowed. So here, try x = 3. You run into complex numbers so then try y = -1. You'll find x = 6 or x = 0. The vertex is certainly right.

[Jimmonash1991]

"Quantum"

His avatar says I didnt need knowledge of it. I am aware of what quantum means being a Physics Major. I challenge ye to a duel to reclaim my honor! En guard! Oh c**p my biggoron sword broke 

[samuelbeattie76]

Please see the question attached, I am looking to prove this identity in a straight-forward way, not involving any mathematical induction.

Any help is much appreciated.
Thanks.

[de]

Please see the question attached, I am looking to prove this identity in a straight-forward way, not involving any mathematical induction.

Any help is much appreciated.
Thanks.

So using your hint-clearly the right hand side is equivalent to the number of ways of choosing n from 2n.
Now look at the left hand side. If we pick 0 blue pens and n red pens the number of ways of doing this is and then choosing 1 blue pen and n-1 red pens we get and so on... each time considering the number of blue and red pens chosen separately and using the fact that (which is intuitively true since the number of ways of choosing k objects from n should be the same as the number of ways of choosing the objects not to be chosen or n-k from n.

[Jimmonash1991]

Hello fellow humans,

Again I am pestering you about the Cambridge text, just to make sure I am not going crazy (though the quest for the meaning of pi probably does that to everyone mathematically inclined).

If someone doesnt mind, could this special person please tell me how to write in mathematical text on here? I see a "pi" symbol that highlights as "LaTeX", but I am not very savvy when it comes to computer coding.....

I have attached the question concerned, so I will run through my effort thus far:

Part (a)

(i) Just asks for vector OD. If AD = 1/2AB, then OD = 1/2AB + OA = 1/2(b+a)

(ii) Asks for DE, so that will be OE - OD and since OE = kOB, then

DE = kb - 1/2(b+a)
      = (k-1/2)b -1/2a

Part (b)

Showing the value of lamda (k) given if DE is perpendicular to OB just requires the dot product of the two vectors being zero

DE.OB = [(k-1/2)b -1/2a].b

            = (k-1/2)b.b -1/2a.b

            = kb.b -1/2b.b -1/2a.b = 0

kb.b = 1/2(a.b + b.b)

k = (a.b + b.b)/(2b.b)

Part (c)

(i) Finding the value of the cosine just requires fiddling with the expression for lamda in (b)

5/6 = (a.b + b.b)/(2b.b)

5/3 = a.b/b.b +1

2/3 = a.b/b.b

If a.b = (a)(b)cosx, with x the angle AOB, then

(a)(b)cosx/(b)^2 = 2/3

(a)/(b)cosx = 2/3

As OA = OB, then (a) = (b), thus

cosx = 2/3 Q.E.D.

(ii) This final part is the most excruciating, though it shouldnt be if I had added the vectors correctly. Please tell me I am not going crazy?

Firstly OD = 1/2(b+a)

           DE = (k-1/2)b -1/2a, and since k = 5/6

                 = 1/3b - 1/2a

Then DF = 1/2DE = 1/6b - 1/4a

DF = OF - OD, thus

OF = DF + OD
      = 1/6b - 1/4a + 1/2a + 1/2b
      = 2/3b + 1/4a

AE = OE - OA = 5/6b - a

If AE and OF are perpendicular, then their dot product SHOULD be zero

AE.OF = [2/3b + 1/4a].[5/6b - a] = 10/18(b)^2 - 2/3a.b + 5/24a.b - 1/4(a)^2

Taking the fact (b) = (a) and (b)^2 out as a common factor as a result

AE.OF = (b)^2[10/18 - 2/3 + 5/24 - 1/4]

Making 72 the common denominator

AE.OF = (b)^2[40/72 - 48/72 + 15/72 - 18/72] = -11/72(b)^2

The above is obviously not zero, and I have been through the calculation a number of times to make sure I did not miss anything.

Give it a go. I hope I am wrong.

James

[Shadowxo]

Hello fellow humans,

Again I am pestering you about the Cambridge text, just to make sure I am not going crazy (though the quest for the meaning of pi probably does that to everyone mathematically inclined).

If someone doesnt mind, could this special person please tell me how to write in mathematical text on here? I see a "pi" symbol that highlights as "LaTeX", but I am not very savvy when it comes to computer coding.....

I have attached the question concerned, so I will run through my effort thus far:

Part (a)

(i) Just asks for vector OD. If AD = 1/2AB, then OD = 1/2AB + OA = 1/2(b+a)

(ii) Asks for DE, so that will be OE - OD and since OE = kOB, then

DE = kb - 1/2(b+a)
      = (k-1/2)b -1/2a

Part (b)

Showing the value of lamda (k) given if DE is perpendicular to OB just requires the dot product of the two vectors being zero

DE.OB = [(k-1/2)b -1/2a].b

            = (k-1/2)b.b -1/2a.b

            = kb.b -1/2b.b -1/2a.b = 0

kb.b = 1/2(a.b + b.b)

k = (a.b + b.b)/(2b.b)

Part (c)

(i) Finding the value of the cosine just requires fiddling with the expression for lamda in (b)

5/6 = (a.b + b.b)/(2b.b)

5/3 = a.b/b.b +1

2/3 = a.b/b.b

If a.b = (a)(b)cosx, with x the angle AOB, then

(a)(b)cosx/(b)^2 = 2/3

(a)/(b)cosx = 2/3

As OA = OB, then (a) = (b), thus

cosx = 2/3 Q.E.D.

(ii) This final part is the most excruciating, though it shouldnt be if I had added the vectors correctly. Please tell me I am not going crazy?

Firstly OD = 1/2(b+a)

           DE = (k-1/2)b -1/2a, and since k = 5/6

                 = 1/3b - 1/2a

Then DF = 1/2DE = 1/6b - 1/4a

DF = OF - OD, thus

OF = DF + OD
      = 1/6b - 1/4a + 1/2a + 1/2b
      = 2/3b + 1/4a

AE = OE - OA = 5/6b - a

If AE and OF are perpendicular, then their dot product SHOULD be zero

AE.OF = [2/3b + 1/4a].[5/6b - a] = 10/18(b)^2 - 2/3a.b + 5/24a.b - 1/4(a)^2

Taking the fact (b) = (a) and (b)^2 out as a common factor as a result

AE.OF = (b)^2[10/18 - 2/3 + 5/24 - 1/4]

Making 72 the common denominator

AE.OF = (b)^2[40/72 - 48/72 + 15/72 - 18/72] = -11/72(b)^2

The above is obviously not zero, and I have been through the calculation a number of times to make sure I did not miss anything.

Give it a go. I hope I am wrong.

James

Hi
First of all: LaTeX can be used on this board, there are a few threads around explaining it LaTeX Guide This is by RuiAce but there are a few others Type [*tex] to start and [/*tex] to finish (without the asterisk) with the code in between. Alternatively, on a mac you can use option-(key) eg option - p for π, option - v for √ etc. Also up the top there are a few things like a box that says "sup", highlight what you want to be up the top then press it to get powers eg ²⁴, same for sub.

With your question:
I started from scratch to make sure everything was right (which it was).
At the end, you took out b² as a common factor.
a*a = |a|², b*b = |b|²,  |b| = |a| therefore b*b = |a|²
But a does not equal b, so a*b does not equal a*a, so you can't make it b*b and take it out as a factor
OF*AE =


Hope this helps

[Jimmonash1991]

Yeah definitely made sense once I realised the fact the dot product is not equal to the product of the of their magnitudes! Unless they are collinear of course.....

Speaking of which, there is another question I would like help with. The question is attached, and I am looking for help finding the dot product that is required in the second part of (a)

Looking at the diagram, if I draw a line from point A to E, then it forms a non right angled triangle with sides AB, and BE.

In terms of vectors u and v, AE = mu + nv

Since it is a non right angled triangle, I thought to use the cosine rule, because we are given an angle between AB and BE, which is 60 degrees (from it being an equilateral triangle), and two sides, being AB and BE



Taking the square root on both sides



The answer was



But this would suggest the dot product between u and v was 1, which is surely not the case, because the angle between them is 60 degrees in an equilateral triangle isnt it? Since they are not parallel?

If yourself or someone could kindly alleviate the issue, I am sure the book is superior to me here.

Thanks again,

James


       

[Shadowxo]

Yeah definitely made sense once I realised the fact the dot product is not equal to the product of the of their magnitudes! Unless they are collinear of course.....

Speaking of which, there is another question I would like help with. The question is attached, and I am looking for help finding the dot product that is required in the second part of (a)

Looking at the diagram, if I draw a line from point A to E, then it forms a non right angled triangle with sides AB, and BE.

In terms of vectors u and v, AE = mu + nv

Since it is a non right angled triangle, I thought to use the cosine rule, because we are given an angle between AB and BE, which is 60 degrees (from it being an equilateral triangle), and two sides, being AB and BE



Taking the square root on both sides



The answer was



But this would suggest the dot product between u and v was 1, which is surely not the case, because the angle between them is 60 degrees in an equilateral triangle isnt it? Since they are not parallel?

If yourself or someone could kindly alleviate the issue, I am sure the book is superior to me here.

Thanks again,

James


       

Hi,
Your working out is mostly right, your error was taking the vectors and putting them in the formula instead of just the lengths. With a²=b²+c²-2bc(cosx), a b and c are just the lengths of the sides. So you end up with
|AE|²=m²+n²-2mn(cosx)
x = 60 so cos(x) = 1/2
|AE|²=m²+n²-mn
|AE| = √(m²+n²-mn)

Also just a little side note: when attempting this problem I did |AE|²=√(mu+nv)² and ended up with the same answer as the book (u.v = cosx = cos(120) = -1/2)

Hope this helps

[Jimmonash1991]

Thanks mate,

I call myself mathematically inclined? Jesus. You know what it might be? I have done so little vector work so far, and back in High School my math was so s**t that seeing the basics in this form, and making mistakes, is highlighting that, but more of a blessing than a curse.

That said though, how pathetic is it not to just take the length of a vector and pump it in there? F**k me.

Can I ask how you got the angle as being 120? Since it is a non right angle triangle you could use true north as a guide, but the bearing would have been a pain in the arse to find.

Cheers,

James

[Shadowxo]

3/4 Specialist is hard, it's easy to make little mistakes that ruin your answer, so don't worry
And since the angle in the triangle is 60, as we multiply u and v we take the angle between them, as long as their start points are aligned (so we move v down to join u) and that angle is 120, as it's 180-60. Not sure if I've explained that well enough but hope it makes a bit of sense