VCE Specialist 3/4 Question Thread!

[SocialRhubarb]

Try using

Spoiler

      Square both sides.

[b^3]

Firstly start off by squaring both sides (just remember to take note that from the domain restriction, that and will be positive, (when we square both sides we need to be careful about introducing new solutions).

Spoiler

Then From that, you can draw out a right angled triangle with angle , you know two side lengths, so you can find the third using Pythagoras' theorem and thus find .

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Note: Take the positive root for A since we know that the length has to be positive from our initial domain restriction.

EDIT: Beaten.

[b^3]

Just careful on those brackets there, and are not included

[jono88]

How do you guys manage to solve it with such ease? :/

[SocialRhubarb]

I guess you just look for patterns and then try some things out.
I mean, a lot of problems with use the identity, and the argument (or angle) in the initial question is , while in the solution it is , so we would be looking to convert it somehow using a double angle formula. After you know what you're looking for, I guess you just play around with the equations for a bit and see what works.

[lzxnl]

Experience; the trick here is noting the 2x instead of the normal x so you need a way of getting a function of 2x.

[brightsky]

Try using

Spoiler

      Square both sides.

i would refrain from squaring both sides simply because often when you square both sides, you generate new solutions. alternative method:

cos(x)-sin(x)=1/3
sqrt(2) (cos(x)*cos(pi/4) - sin(x)sin(pi/4)) = 1/3
sqrt(2) cos(x+pi/4) = 1/3
cos(x+pi/4) = sqrt(2)/6
cos(2x+pi/2) = 2cos^2(x+pi/4) - 1 = 2(1/18) - 1 = -8/9
sin(2x) = 8/9
cos(2x) = sqrt(1-sin^2(2x)) = sqrt(1-64/81) = sqrt(17)/9 (since 2x in first quadrant)
so cot(2x) = cos(2x)/sin(2x) = sqrt(17)/8

[silverpixeli]

Hey guys,

First question:

What do I say when I don't need to include the modulus sign after integrating something like 1/x to get ln(|x|) if x is always positive?

Spoiler

(this quote from b^3 is from an older thread I found)

I get why we drop the modulus in the problem above, because x^2+9>0 for all x values, but how should I write/show this in my working? I've just been writing 'no need for modulus as <insert expressions> is always +ve', is that sufficient/do I even need to write it or can I just leave it out of my answer?

Second question:

The classic volumes of solids question with a hemispherical bowl and some water in it, this is what I've been doing

1. make an equation for a circle centred at the origin: y^2 + x^2 = r^2 where r is the radius of the bowl
2. rearrange for y^2, so I get y^2 = r^2 - x^2
3. use the solids of revolution formula:


(idk how to put terminals in but I'm integrating from to )
4. integrate and get an answer

I understand how the solids formula works and stuff, and I'm able to get an answer, but what I don't get is this:

Where does the bottom half of the circle's area go? I'm using an equation of a circle and finding the area from to , but I'm conveniently only taking the area of that section of the circle that is already above the x axis. What about the equivalent area below it? Where does it go?

Explanations or links to explanations appreciated, thanks!

[Holmes]

Is it possible to do this in yr 12 spesh without a cas calculator? Thanks!

[lzxnl]

Is it possible to do this in yr 12 spesh without a cas calculator? Thanks!

Holmes you can't even do that using more advanced methods in terms of trig, exponential, log and polynomials. Generally we don't deal with square roots in spesh unless they're square roots of quadratics of the form (a^2-(x-b)^2)^-1/2, they're square roots of linear functions or an easy substitution can be made, like with x^3*sqrt(x^4+1)

Hey guys,

First question:

What do I say when I don't need to include the modulus sign after integrating something like 1/x to get ln(|x|) if x is always positive?

(this quote from b^3 is from an older thread I found)

I get why we drop the modulus in the problem above, because x^2+9>0 for all x values, but how should I write/show this in my working? I've just been writing 'no need for modulus as <insert expressions> is always +ve', is that sufficient/do I even need to write it or can I just leave it out of my answer?

Second question:

The classic volumes of solids question with a hemispherical bowl and some water in it, this is what I've been doing

1. make an equation for a circle centred at the origin: y^2 + x^2 = r^2 where r is the radius of the bowl
2. rearrange for y^2, so I get y^2 = r^2 - x^2
3. use the solids of revolution formula:


(idk how to put terminals in but I'm integrating from to )
4. integrate and get an answer

I understand how the solids formula works and stuff, and I'm able to get an answer, but what I don't get is this:

Where does the bottom half of the circle's area go? I'm using an equation of a circle and finding the area from to , but I'm conveniently only taking the area of that section of the circle that is already above the x axis. What about the equivalent area below it? Where does it go?

Explanations or links to explanations appreciated, thanks!

For your first point, just say x^2+9>0 and then omit the modulus signs. That'll be enough. You have to say that though; otherwise the examiner doesn't know if you know that the antiderivative of 1/x is ln|x|+c

For your second point...you have to be really careful with these things. First of all consider what volume you're looking for. Draw out your hemispherical bowl. You have y^2=r^2-x^2. But at the same time, your bounds are with respect to y NOT x. Due to the symmetry of the bowl, it doesn't matter which variable you integrate with respect to. You can still integrate with respect to x. Your function is now y=+sqrt(r^2-x^2) (yes only a semicircle) and you want to find the volume when you rotate this around the x axis between two particular bounds.
If you revisit the shape of your hemisphere, imagine that the top of the water surface divides the hemisphere into two parts. The volume that you want is the bit on the same end as the diameter of the sphere. Going back to revolving y=+sqrt(r^2-x^2) around the x axis, this means that the volume you want is actually from x=0 to x=r-depth.
Now, when you revolve this semicircle around the x-axis, you get the full sphere. Think about it. There's no problem of not getting the other half of the sphere. So integrate pi*(r^2-x^2) from 0 to r-depth and you're done!

These volume questions need you to visualize what you're actually doing. A diagram is great.

[Holmes]

Euler's approximation method, I get confused when the derivative has both an x and a y variable. How do we find the y variable?
Eg:

Euler's method with a stepsize of is used to solve , given the initial condition . Find the approximation obtained for .

Also:
Euler's method with a stepsize of 0.5 is used to solve the differential equation  , with the initial value of . Find the approximation obtained for y(2).

Many many thanks.

[lzxnl]

Basically it's the same as the normal Euler method, except this time when you see a y in the derivative function you plug in the previous value of y just like you plug in previous values of x where you see x.

[Holmes]

Basically it's the same as the normal Euler method, except this time when you see a y in the derivative function you plug in the previous value of y just like you plug in previous values of x where you see x.

Thanks! That made the first question surprisingly simple, but for some reason I can't get this question to work out:

Euler's method with a stepsize of is used to solve the differential equation  , with the initial value of . Find the approximation obtained for .

Many many thanks.

[lzxnl]

Well let's solve it first to see what sort of answer we want.
dy/dx=x/y => y dy/dx = x
Integrating both sides and doubling
y^2=x^2+c
y(1)=p so p^2=1+c
c=p^2-1
y^2=x^2+p^2-1
x=2, y^2=3+p^2

Hmm ok.

Now for Euler's method.
So we have y(1.5)=p+0.5*1/p
y(2)=(p+0.5/p)+0.5*1.5/(p+0.5/p)

Simplifying yields p+(1.25p+0.25)/(p^2+0.5p)

Which seems funny, but subbing in p = 1 yields y = 2 by both methods, while subbing in p=2 yields y=sqrt 7 or roughly 2.64 while Euler method gives 2.55, so it's not TOO far off.

[Holmes]

y(2)=(p+0.5/p)+0.5*1.5/(p+0.5/p)

Simplifying yields p+(1.25p+0.25)/(p^2+0.5p)

It's probably really easy, but how do we simplify that top expression to get to the one you've got?