VCE Specialist 3/4 Question Thread!

[lzxnl]

Why can we represent complex numbers using euler's formula.
I.e. Why does r*e^theta*i =cos(theta) +sin(theta)*i?

Explanation 1:
Let y = e^ix. Let z = cis x.
Then, dy/dx = iy, dz/dx = iz.
As y and z satisfy the same first-order differential equation with the initial condition y(0) = 1, they are equal.

Explanation 2:
Differentiate cis x e^(-ix) and you'll get 0. This means this expression is constant. The constant is 1, found by subbing in x = 0.

Explanation 3:
Take the exponential power series e^x = 1 + x + x^2/2! + x^3/3! + x^4/4!...
If you replace x with ix and collect real/imaginary terms, you'll find that the real part is 1 - x^2/2! + x^4/4!..., which is the power series expansion for cos x. Similarly, the imaginary part will be x - x^3/3! + x^5/5!... which is the power series expansion for sin x. Hence, it's only natural to define e^ix = cos x + i sin x

[deStudent]

For these questions in general http://m.imgur.com/a/eyi7M

Do you have to redraw the graphs for part (i) like the answer did? I just visualised the graph being flipped over the x-axis, I'm not sure if this a legitimate method but it worked. Kinda failed for (f) though.

Also, when they say to evaluate the integral, this means to find the signed area, not total area, right? For part (f), they found the large area then subtracted the smaller area? Wouldn't this mean they found the total area?

Thanks

[Sine]

For these questions in general http://m.imgur.com/a/eyi7M

Do you have to redraw the graphs for part (i) like the answer did? I just visualised the graph being flipped over the x-axis, I'm not sure if this a legitimate method but it worked. Kinda failed for (f) though.

Also, when they say to evaluate the integral, this means to find the signed area, not total area, right? For part (f), they found the large area then subtracted the smaller area? Wouldn't this mean they found the total area?

Thanks

I would graph it quickly just to get a quick visual and to kinda use it as a checking technique. Evaluate the integral means literally evaluate - don't think about areas or anything. Integration doesn't mean to find area it is a tool we use to find areas.

[frussell]

Could someone please help me with this...

I have completely forgotten how to seperate a regular x and y relation into two parametric equations for x and y.

[vasuk]

if you have something like this -1 < sin θ < 1
if you then took the inverse sin of all 3 sides do the inequality signs flip?
like does it become arcsin(-1) > θ > arcsin(1) ??

& how would one find the domain and range of y = cos(arcsin (x) )

[Shadowxo]

if you have something like this -1 < sin θ < 1
if you then took the inverse sin of all 3 sides do the inequality signs flip?
like does it become arcsin(-1) > θ > arcsin(1) ??

& how would one find the domain and range of y = cos(arcsin (x) )

Messing with inequalities can get quite confusing, so I'd recommend staying away from doing that. I think the general rule is, if a higher input results in a higher output and vice versa (like for arcsin), then you can do it without flipping,and for the opposite you do flip the signs, but it can get quite confusing quickly, especially if you don't test points.

For that question, the easiest way to solve it would be to let arcsin x = a
sin a =x
You want to find cos a (as arcsin x =a)
cos²a+sin²a=1
cos²a=1-sin²a
cos²a=1-x²
cos a=±√(1-x²)
-π/2≤arcsin x≤π/2, so we know cos a is positive (cos a is positive between -π/2 and π/2)
so cos a=√(1-x²)

[deStudent]

http://m.imgur.com/a/Nzx7E

For Q111) I'm lost completely. I don't understand how you can/have enough information to state that y=b? I also don't understand how x=a is the lower terminal but x is also the higher terminal?

Q112) I got this correct but I want to check if my working is legitimate. I ended up expressing my du/dx = 1/2u, is this okay? I then cancelled its' 'u' term by adding a 'u' in the numerator so the integral remains the same.

Thanks

[Gogo14]

http://m.imgur.com/a/Nzx7E

For Q111) I'm lost completely. I don't understand how you can/have enough information to state that y=b? I also don't understand how x=a is the lower terminal but x is also the higher terminal?

Q112) I got this correct but I want to check if my working is legitimate. I ended up expressing my du/dx = 1/2u, is this okay? I then cancelled its' 'u' term by adding a 'u' in the numerator so the integral remains the same.

Thanks

Q111.
You can write the equation of the original function if you are given the derivative and a coordinate, by using the equation shown in the ans (line 2). You now have y=F(x) by using the equation. Now you sub in x=4 to find the y value. Now y=F(4) and replace the x on the upper boundary with 4

[humblepie]

Would really appreciate it if someone could tell me what I've done wrong for this question:

Find the maximum value of f(x)=(sin(x))^2+cos(x)+1

The answer is 9/4 (confirmed with CAS), but for some reason I keep getting 2 (working attached below)

[cookiedream]

Would really appreciate it if someone could tell me what I've done wrong for this question:

Find the maximum value of f(x)=(sin(x))^2+cos(x)+1

The answer is 9/4 (confirmed with CAS), but for some reason I keep getting 2 (working attached below)

Hello!
Here is the graph:

You'll see that the max is 9/4 because its principles act more like a quadratics graph rather than a conventional trig graph
Therefore, you can calculate the max using the turning point form, which you have in your working out 

[Shadowxo]

Would really appreciate it if someone could tell me what I've done wrong for this question:

Find the maximum value of f(x)=(sin(x))^2+cos(x)+1

The answer is 9/4 (confirmed with CAS), but for some reason I keep getting 2 (working attached below)

Great answer by cookie, I'd just like to add that the error in your arithmetic is saying that 1/4 ≤ (cosx-1/2)^2 ≤ 9/4. The minimum would be where it equals zero, and in this case it's possible where cosx=1/2. So 0 ≤ (cosx-1/2)^2 ≤ 9/4 and you end up with the max at 9/4. This was what messed up your calculations

[zhen]

Would really appreciate it if someone could tell me what I've done wrong for this question:

Find the maximum value of f(x)=(sin(x))^2+cos(x)+1

The answer is 9/4 (confirmed with CAS), but for some reason I keep getting 2 (working attached below)

I think the mistake you made was going from -1≤cos(x)≤1 to the next step. It's supposed to be
0≤(cos(x)-1/2)^2≤9/4. Basically sub in values between -1 and 1 to find the max and min. It should be obvious that subing in -1 will get the max. But to get the min, you want it as close to or equal zero as possible, since it can't be negative, so cos(x)=1/2 makes it equal zero. From there you get -9/4≤-(cos(x)-1/2)^2≤0
Then you get 0≤-(cos(x)-1/2)^2+9/4≤9/4.
So the max is 9/4

Edit: Shadowxo beat me to it

[zhen]

Hello!
Here is the graph:

You'll see that the max is 9/4 because its principles act more like a quadratics graph rather than a conventional trig graph
Therefore, you can calculate the max using the turning point form, which you have in your working out 

This works for this question, but it won't work for all cases. For example -(cos(x)-3/2)^2+9/4, which also forms a quadratic like equation has a maximum of 2 (if my calculations are correct), rather than 9/4. The case only works for the question, as (cos(x)-1/2)^2 can equal 0, as cos(x) can equal 1/2. But if the square part can't equal zero, calculating the max using the turning point form is no longer a viable method to find the max. This is actually pretty interesting.    I only noticed this because I was going to recommend the same solution but realised that it's not always right to just assume from the turning point form that it's the max.

[humblepie]

Ah I see, thank-you so much cookiedream, Shadowxo and zhen!

[cookiedream]

This works for this question, but it won't work for all cases. For example -(cos(x)-3/2)^2+9/4, which also forms a quadratic like equation has a maximum of 2 (if my calculations are correct), rather than 9/4. The case only works for the question, as (cos(x)-1/2)^2 can equal 0, as cos(x) can equal 1/2. But if the square part can't equal zero, calculating the max using the turning point form is no longer a viable method to find the max. This is actually pretty interesting.    I only noticed this because I was going to recommend the same solution but realised that it's not always right to just assume from the turning point form that it's the max.

Oh oops haha, didn't realise that.
Thanks zhen!

Ah I see, thank-you so much cookiedream, Shadowxo and zhen!

Sorry for the mistake!
Glad I helped anyway ^^;