Author Topic: VCE Specialist 3/4 Question Thread! (Read 2578777 times) Tweet Share

yoloswag69 · « **Reply #1845 on:** June 18, 2013, 10:00:31 pm »

One more question this has me so depressed
let f(x)=ax^2 where a and b are constants and give g(x)=x^2+8
(a) Find in terms of a and b the gradient of the curve y=f(x) at x=2
b Find the gradient of the curve y=g(x) at x =2
c If the tangent line to the curve y=f(x) and y=g(x) are parallel at x=2
i find an euation relating to a and b
ii hence find the vlaues of a and b, given that the curves y=f(x) and y=g(x) in fact touch at x=2

Aelru · « **Reply #1846 on:** June 18, 2013, 10:21:18 pm »

Question~:

Q1: For equation $\frac {1-x}{x+1}$ , how do i show algebraically, that y=x does not intersect?

lzxnl · « **Reply #1847 on:** June 18, 2013, 10:22:30 pm »

Try and solve y=x and y=(1-x)/(1+x)
You should not get any real solutions.

Aelru · « **Reply #1848 on:** June 19, 2013, 06:51:42 pm »

Alright just posting my steps on the previous question I asked, to see if it checks out with you guys. I think there's a easier method, but I'm not sure what it is.

I actually gave the wrong equation. It asked me to prove algebraically, that $y=x$ , does not intersect with $y= \frac {1}{\sqrt{1-x^2}}$

1: Looking at original equation of $y= \frac {1}{\sqrt{1-x^2}}$ , It can be seen that $-1<x<1$ .
2 : Subbed $y=x$ into the equation $y= \frac {1}{\sqrt{1-x^2}}$ , and it became $x= \frac {1}{\sqrt{1-x^2}}$

3: Moved denominator on RHS over to LHS. Thus, $x\sqrt{1-x^2}} = 1$ . But, for the LHS to be equal to RHS, $x$ and $\sqrt{1-x^2}}$ must be equal to 1. From step one, we've determined that equation $y= \frac {1}{\sqrt{1-x^2}}$ has to have values of x within implied domain $-1<x<1$ . Therefore, no solutions exist.

Does that.. check out okay? $:-\$

Ancora_Imparo · « **Reply #1849 on:** June 19, 2013, 07:07:16 pm »

Not quite... Your logic in step 3 is the problem.
You said that $x$ and $\sqrt{1-x^2}$ have to equal 1. This is not true. What if $x=1/2$ and $\sqrt{1-x^2}=2$ ? By inspection, you'll see that this can't be true as well, but you still need to prove it. You could try and prove this by testing all types of cases, but this may take some time.

What I would do is:
$x\sqrt{1-x^2}=1$
$x^2(1-x^2)=1$
$x^2-x^4=1$
$x^4-x^2+1=0$

Let $A=x^2$ :
$A^2-A+1=0$

Discriminant $=(-1)^2-4*1*1=1-4=-3$
Since discriminant < 0, there are no real solutions.

jono88 · « **Reply #1850 on:** June 19, 2013, 07:09:46 pm »

Find unit vector perpendicular to both a=2i-3j+k and b=-2i+3y+k.

Alwin · « **Reply #1851 on:** June 19, 2013, 07:10:18 pm »

Quote from: Aelru on June 19, 2013, 06:51:42 pm

Alright just posting my steps on the previous question I asked, to see if it checks out with you guys. I think there's a easier method, but I'm not sure what it is.
....
3: Moved denominator on RHS over to LHS. Thus, $x\sqrt{1-x^2}} = 1$ . But, for the LHS to be equal to RHS, $x$ and $\sqrt{1-x^2}}$ must be equal to 1. From step one, we've determined that equation $y= \frac {1}{\sqrt{1-x^2}}$ has to have values of x within implied domain $-1<x<1$ . Therefore, no solutions exist.

Does that.. check out okay? $:-\$

No, your step three is incorrect sorry. It only works for Null Factor Law,
$ab=0 \Rightarrow a=0 \text{ or } b=0$

$ab=1 \ne > a=1 \text{ and } b=1. \text{ for example, }a=10\text{ and } b=0.1 \Rightarrow ab=1$

You have to square both sides and solve. Use the discriminate to show there are no solutions

EDIT: Beaten, look at Ancora_Imparo's post for solution

b^3 · « **Reply #1852 on:** June 19, 2013, 07:19:14 pm »

Quote from: jono88 on June 19, 2013, 07:09:46 pm

Find unit vector perpendicular to both a=2i-3j+k and b=-2i+3y+k.

Let $\begin{alignedat}{1}\underset{\sim}{c} & =a\underset{\sim}{i}+b\underset{\sim}{j}+c\underset{\sim}{k}\end{alignedat}$
Now if you take the dot product of that vector and let it equal to zero, you will get two equations. To get the third you need to find the magnitude of this vector, which as it is a unit vector will be equal to 1.

Spoiler

$\begin{alignedat}{1}\left(2\underset{\sim}{i}-3\underset{\sim}{j}+\underset{\sim}{k}\right).\left(a\underset{\sim}{i}+b\underset{\sim}{j}+c\underset{\sim}{k}\right) & =0 \\ 2a-3b+c & =0 \\ c & =-2a+3b...[1] \\ \left(-2\underset{\sim}{i}+3\underset{\sim}{j}+\underset{\sim}{k}\right).\left(a\underset{\sim}{i}+b\underset{\sim}{j}+c\underset{\sim}{k}\right) & =0 \\ -2a+3b+c & =0...[2] \\ \text{Substitute [1] into [2]} \\ -2a+3b-2a+3b & =0 \\ 6b & =4a \\ b & =\frac{2}{3}a....[3] \\ c & =-2a+3\left(\frac{2}{3}a\right) \\ c & =0 \\ \sqrt{a^{2}+b^{2}+c^{2}} & =1....[4] \\ a^{2}+b^{2}+c^{2} & =1 \\ a^{2}+\frac{4}{9}a^{2} & =1 \\ \frac{13}{9}a^{2} & =1 \\ a & =\pm\frac{3}{\sqrt{13}} \\ & =\pm\frac{3\sqrt{13}}{13} \\ b & =\pm\frac{2}{3}\frac{3\sqrt{13}}{13} \\ & =\pm\frac{2\sqrt{13}}{13} \\ a\underset{\sim}{i}+b\underset{\sim}{j} & =\frac{3\sqrt{13}}{13}\underset{\sim}{i}+\frac{2\sqrt{13}}{13}\underset{\sim}{j} \\ a\underset{\sim}{i}+b\underset{\sim}{j} & =-\frac{3\sqrt{13}}{13}\underset{\sim}{i}-\frac{2\sqrt{13}}{13}\underset{\sim}{j} \end{alignedat}$

Aelru · « **Reply #1853 on:** June 19, 2013, 07:24:53 pm »

Quote from: Ancora_Imparo on June 19, 2013, 07:07:16 pm

Not quite... Your logic in step 3 is the problem.
You said that $x$ and $\sqrt{1-x^2}$ have to equal 1. This is not true. What if $x=1/2$ and $\sqrt{1-x^2}=2$ ? By inspection, you'll see that this can't be true as well, but you still need to prove it. You could try and prove this by testing all types of cases, but this may take some time.

What I would do is:
$x\sqrt{1-x^2}=1$
$x^2(1-x^2)=1$
$x^2-x^4=1$
$x^4-x^2+1=0$

Let $A=x^2$ :
$A^2-A+1=0$

Discriminant $=(-1)^2-4*1*1=1-4=-3$
Since discriminant < 0, there are no real solutions.

Ah, I see your point! Excellent method actually. The discriminant bit really opened my eyes haha. Thanks!

Alwin, thanks for the help as well

By the way ; you mentioned about testing different types of cases. Would you mind explaining what tests are appropriate for this type of question as well? (apart from discriminants)

Alwin · « **Reply #1854 on:** June 19, 2013, 07:28:34 pm »

Quote from: jono88 on June 19, 2013, 07:09:46 pm

Find unit vector perpendicular to both a=2i-3j+k and b=-2i+3y+k.

I would say just cross product, but it's not part of the vce course, sorry.

Here is the VCE method

Let the perpendicular vector be v, where v=(v₁, v₂, v₃) Now,

$v\cdot a=0 \text{ and } v\cdot b=0$

$(v_1, v_2, v_3)\cdot (2, -3, 1)=0$	$(v_1, v_2, v_3)\cdot (-2, 3, 1)=0$
$2v_1 -3v_2 + v_3 =0$	$-2v_1 +3v_2 + v_3 =0$

Solving the two equations gives,
$v_2 = \frac{2 v_1}{3} \text{ and } v_3 = 0$

I can arbitrary let v₁ =3, since I will adjust the length at the very end.
$\therefore v=(3, 2, 0)$

Since the question asked for the unit vector,
$\text{Unit vector }=\pm \frac{1}{\sqrt{13}}(3,2,0)$

If you were interested in the cross product method, here is an introductory link:
https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/linear-algebra--cross-product-introduction

lzxnl · « **Reply #1855 on:** June 19, 2013, 08:38:05 pm »

Technically there are TWO unit vectors that satisfy your requirement, the other one being the negative of the vector Alwin gave.

Alwin · « **Reply #1856 on:** June 19, 2013, 08:40:40 pm »

Quote from: nliu1995 on June 19, 2013, 08:38:05 pm

Technically there are TWO unit vectors that satisfy your requirement, the other one being the negative of the vector Alwin gave.

Yeah, I just realised haha, the cross product actually comes out negative. I'll add it in now. Btw, it's $\pm$ in b^3 added in spoiler

lzxnl · « **Reply #1857 on:** June 19, 2013, 08:54:37 pm »

Well...you can't really say "the" cross product without specifying an order to the product.

Alwin · « **Reply #1858 on:** June 19, 2013, 09:03:33 pm »

Quote from: nliu1995 on June 19, 2013, 08:54:37 pm

Well...you can't really say "the" cross product without specifying an order to the product.

Haha, i was just meaning v = a x b since that's alphabetically ordered and then order is was presented in the question. This is a spesh thread, so putting the rest in the spoiler:

Spoiler

$(2, -3, 1)\times(-2, 3, 1) = (-6, -4, 0)$

Since it asks for unit vector in this direction, normalise it and $\pm$
$(\pm \frac{3}{\sqrt{13}}, \pm \frac{2}{\sqrt{13}}, 0)$

$\therefore \frac{3}{\sqrt{13}}i + \frac{2}{\sqrt{13}}j \text{ or } -\frac{3}{\sqrt{13}}i - \frac{2}{\sqrt{13}}j$

Satisfied nliu1995?

jono88 · « **Reply #1859 on:** June 20, 2013, 07:54:55 pm »

IF cos(x)-sin(x)=1/3, for domain 0<x<pi/4, find cot(2x)

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