Author Topic: VCE Specialist 3/4 Question Thread! (Read 2545306 times) Tweet Share

pi · « **Reply #1110 on:** January 10, 2013, 02:11:17 pm »

-3sin(x/3) isn't right o.O

(did you mean inverse sin?)

polar · « **Reply #1111 on:** January 10, 2013, 02:28:21 pm »

Quote from: Homer on January 10, 2013, 01:59:49 pm

when anti differentiating -3/square root(9-x^2)

3cos-1(x/3) and -3sin(x/3) be right? the book only has 3cos-1(x/3)

yeah, technically both are fine since $\arcsin(x) = \frac{\pi}{2} - \arccos(x)$ , you forgot a +c though which is where the difference would be

$\begin{aligned} 3\arccos\left(\frac{x}{3}}\right) + c_1 &= 3\left(\frac{\pi}{2} - \arcsin(x)\right) + c_1<br />\\ &= \frac{3\pi}{2} - 3\arcsin\left(\frac{x}{3}\right) + c_1<br />\\ &= -3\arcsin\left(\frac{x}{3}\right) + \left(\frac{3\pi}{2} + c_1\right)<br />\\ &= -3\arcsin\left(\frac{x}{3}\right) + c_2, \; \text{where \;} c_2 = \frac{3\pi}{2} + c_1\end{aligned}$

b^3 · « **Reply #1112 on:** January 10, 2013, 02:30:03 pm »

Both $3\cos^{-1}(\frac{x}{3})$ and $-3\sin^{-1}(\frac{x}{3})$ will work, but if you graph them they will be a vertical translation of each other (thats why if we were to find the graph with an initial condition the +C would be needed). Most of the time if it has the negative out the front we make it into a cos, and if it doesn't have the negative we make it into the sin, but its still not wrong if you do it the other way but its best to stick to convention.

EDIT: slightly beaten to it

Homer · « **Reply #1113 on:** January 10, 2013, 03:11:56 pm »

oops yes pi i did mean sin-1. So, suppose that question comes up on the exam, would both of them be acceptable with the +c's?
thanks guys

Homer · « **Reply #1114 on:** January 11, 2013, 03:26:46 pm »

could someone show me how they would work this question out, i get the answer right but I have a feeling that I use a dodgy method to work it out haha

Hancock · « **Reply #1115 on:** January 11, 2013, 03:37:33 pm »

$\int^{}_{}\frac{2}{49+25x^2} \ dx$

$2\int^{}_{}\frac{1}{49+25x^2} \ dx$

$\frac{2}{25}\int^{}_{}\frac{1}{\frac{49}{25}+x^2} \ dx$

$\frac{2}{25}\int^{}_{}\frac{1}{(\frac{7^2}{5^2})+x^2} \ dx$

$\frac{2}{25}*\frac{5}{7}\int^{}_{}\frac{\frac{7}{5}}{(\frac{7}{5})^2+x^2} \ dx$

$\frac{2}{35}\int^{}_{}\frac{\frac{7}{5}}{(\frac{7}{5})^2+x^2} \ dx$

$\frac{2}{35}arctan(\frac{x}{\frac{7}{5}}}) + C$

$\frac{2}{35}\arctan(\frac{5x}{7}) + C$

zvezda · « **Reply #1116 on:** January 13, 2013, 07:34:01 pm »

hey guys, need some help with Ex 4G q8ai (finding the value of b; you will know what im talking about once you read the q) and q8b which is just getting on my nerves. For 8b, im managing to get the answer (well not really), except all of the terms in my answer are doubled. Been through my working many a time and still have no clue where ive gone wrong. Help would be much appreciated. cheers

pi · « **Reply #1117 on:** January 13, 2013, 08:22:09 pm »

^Which book?

Possible to screenshot and upload a pic?

Homer · « **Reply #1118 on:** January 13, 2013, 08:58:35 pm »

Maths quest doesn't go upto 4g, so it should be essentials, but 4g 8ai in essentials has nothing to do with b so idk haha :/

zvezda · « **Reply #1119 on:** January 13, 2013, 10:12:05 pm »

Wow my bad, forgot to specify lol. Its the essentials text book:
8a. Find the square roots of 1+i by:
i) cartesian methods
ii) de moivre's theorem (noted this cos part b is a "hence" q)

8b. Hence find the exact values of cos(pi/8) and sin(pi/8)

Jenny_2108 · « **Reply #1120 on:** January 14, 2013, 02:28:44 am »

Quote from: stankovic123 on January 13, 2013, 10:12:05 pm

Wow my bad, forgot to specify lol. Its the essentials text book:
8a. Find the square roots of 1+i by:
i) cartesian methods
ii) de moivre's theorem (noted this cos part b is a "hence" q)

8b. Hence find the exact values of cos(pi/8) and sin(pi/8)

8ai) Let square root of 1+i is z=a+bi => z^2=1+i => (a+bi)^2=1+i
You expand LHS then equate to RHS
ii) Find polar form of 1+i and let z = rcis(theta) is the sqrt of 1+i
Then use de moivre's theorem to solve

8b) After you solve 8aii, I guess the ans will have something with cis(pi/8), expand it to cos(pi/8) and sin(pi/8) then solve
Maybe you can try 1st and see how you go

Homer · « **Reply #1121 on:** January 14, 2013, 10:36:55 am »

8i)
$(a+bi) ^ 2 = 1+i$

$a ^ 2 + 2bai - b ^ 2 = 1 + i$

$a ^ 2 - b ^ 2=1$

$2 b a i =i$

$b= \frac {1} {2a}$

$a ^ 2 - \frac {1} {4a^2} =1$

$4a ^ 4 - 4 a ^ 2 - 1=0$

$a= ± \sqrt { \frac {1+ \sqrt {2} } {2}}$

$b = ± \sqrt { \frac {-1+ \sqrt {2} } {2} }$

8ii) $\left | z \right | = \sqrt {2}$

Arg(z) = $\tan^{-1}(1)=\frac{\pi }{4}$

z= $\sqrt{2}$ cis $\frac{\pi }{4}$

$\sqrt {z}$ = $2 ^ { \frac {1} {4} } cis( \frac {\pi } {8})$

also

$\sqrt {z}$ = $2 ^ { \frac{1} {4}}cis( \frac {-7\pi } {8})$

Jenny_2108 · « **Reply #1122 on:** January 14, 2013, 02:36:24 pm »

So from 8a you already found 2 solutions
i) cartesian form:

Quote from: Homer on January 14, 2013, 10:36:55 am

$a= ± \sqrt { \frac {1+ \sqrt {2} } {2}}$

$b = ± \sqrt { \frac {-1+ \sqrt {2} } {2} }$

They ask about sqrt of 1+i, not a and b thus you have to conclude $\sqrt {1+i}= ± \sqrt { \frac {1+ \sqrt {2} } {2}} ± \sqrt { \frac {-1+ \sqrt {2} } {2} } i$

ii) in polar form

Quote from: Homer on January 14, 2013, 10:36:55 am

$\sqrt {z}$ = $2 ^ { \frac {1} {4} } cis( \frac {\pi } {8})$

$\sqrt {z}$ = $2 ^ { \frac{1} {4}}cis( \frac {-7\pi } {8})$

8b) Find cos(pi/8) and sin (pi/8)

Angle pi/8 in the 1st quadrant thus sin and cos values are positive

Using ans from part a:

$\sqrt {z}$ = $2 ^ { \frac {1} {4} } cis( \frac {\pi } {8}) = \sqrt { \frac {1+ \sqrt {2} } {2}} + \sqrt { \frac {-1+ \sqrt {2} } {2} } i$

From here you can do it by yourself

zvezda · « **Reply #1123 on:** January 14, 2013, 10:09:40 pm »

Im appreciating the help but i think you guys have misunderstood my question. I know what the overall process is, its just that something isnt clicking algebraically. how did you manage to get the b value? Cos im sure there is an algebraic process after subbing in the a value. As for part b, i know the process, it's just that in my answer, all the terms are double that of those in the answer. So i know theres something wrong with my algebra, but im struggling to find it.

Jenny_2108 · « **Reply #1124 on:** January 15, 2013, 03:54:49 am »

Quote from: stankovic123 on January 14, 2013, 10:09:40 pm

Im appreciating the help but i think you guys have misunderstood my question. I know what the overall process is, its just that something isnt clicking algebraically. how did you manage to get the b value?

you already found a, so just sub value of a back into equation $a^2 -b^2=1$ to find b value

Quote from: stankovic123 on January 14, 2013, 10:09:40 pm

As for part b, i know the process, it's just that in my answer, all the terms are double that of those in the answer. So i know theres something wrong with my algebra, but im struggling to find it.

$2 ^ { \frac {1} {4} } cis( \frac {\pi } {8}) = \sqrt { \frac {1+ \sqrt {2} } {2}} + \sqrt { \frac {-1+ \sqrt {2} } {2} } i$

$\implies 2 ^ { \frac {1} {4} } cos( \frac {\pi } {8}) = \sqrt { \frac {1+ \sqrt {2} } {2}}$

$\implies cos( \frac {\pi } {8}) = \frac{({ \frac {1+ \sqrt {2} } {2}})^\frac{1}{2}}{2 ^ {\frac {1} {4}} }$

$\implies cos( \frac {\pi } {8}) = (\frac{1+ \sqrt {2} } {2\sqrt{2}})^{\frac{1}{2}}$

$\implies cos( \frac {\pi } {8}) = (\frac{(1+ \sqrt {2})\times \sqrt{2} } {4}) ^{\frac{1}{2}}$

$\implies cos( \frac {\pi } {8}) =\frac{\sqrt {{2+\sqrt{2}}}}{2}$

To find sin(pi/8), you do similarly. I'm so lazy to type LaTex again

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