VCE Specialist 3/4 Question Thread!

[Professor Polonsky]

I forgot me partial fractions

Show, using partial fractions that

[Phy124]

I forgot me partial fractions

Show, using partial fractions that

[zhe0001]

da hello

struggling with a problem, da prease help

Expand the following,



The book says the answer is: but I never get it right.

Attached is my working out

Thanks!

MOD EDIT: Moved from Methods Question thread to Specialist Question thread

[09Ti08]

This is what I get, I know it's not what you want, but is there another question before this?

[aestheticatar]

I keep getting this wrong! Help me out guys!

Thanks ALOT in advance

[Conic]

For part a:

It states the acceleration is constant, which means we can use the constant acceleration formulas. We know the time (20s), the initial velocity (4m/s). We need to find the displacement before we find the acceleration. Since it returns to the origin, it needs to travel the same distance as it did when moving at a constant velocity, so the displacement is . Now we can find the acceleration:









For part b:

Now we need to find the time at which it turns around, ie where v=0:







So at t=18.25s it turns around. The time at wich it is travelling back is the total time (32s) minus the time when it turns around, which is 13.25s, or .

[aestheticatar]

Not sure how to start as I don't understand the question. Someone care to explain and help me with a?

Thanks again!

[b^3]

We have a chemical reaction taking place, of which the amount of mass consumed/formed of each compound are related.
"It takes 1 kg of and 3 kg of to form 4 kg of ".
So that means that every time we gain a kg of , the mass of decreases by kg. Since we started with 2 kg of we subtract times the mass of , as is the mass of compound at time . What we have left is dependent on the amount of compound we have, so the mass of compound is given by
.
A similar situation goes for compound . Every time we gain a kg of , the mass of decreases by kg, and starting with 3 kg means we have
.
Now the rate of increase of is proportional to the product of and .



Hope that helps

EDIT: For part b, flip and the RHS and try integrating, using partial fractions.

[Kanye East]

Need help with this question ASAP! It's to do with matrices
(see attachment below)

[Professor Polonsky]

19a + 4b = 5
6a + b = 0
18a + 3b = 0
7a + 2b = 5

[random_person]

2013 MAV Specialist Mathematics Exam 1 Question 9 (4 Marks)

Find all the solutions to:

over the domain .

Thanks in advance.

[lzxnl]

(1 + cos 2t)/sin(2t) = sqrt 3
2cos^2 t / 2sin t cos t = sqrt 3
tan t = sqrt(1/3)
Solve.

Or:
1 + cos 2t = sqrt3 sin 2t
sqrt3 sin 2t - cos 2t = 1
Let the LHS be expressed in the form r sin(2t - a)
r sin(2t-a) = r(sin 2t cos a - sin a cos 2t) = sqrt 3 sin 2t - cos 2t
Comparing coefficients
r sin a = 1
r cos a = sqrt3
r^2 = 4, r =2
2 sin a = 1, 2 cos a = sqrt 3 => a = pi/6

so LHS = 2sin(2t-pi/6) = 1
Solve as normal

[random_person]

Cheers

[Phy124]

(1 + cos 2t)/sin(2t) = sqrt 3

Note that if you do this you will eliminate one of the solutions you're looking for.

is a solution to the original equation. However, as you'll notice, is undefined for all and so are the alternate forms from here on out, which is where the problem arises.

Adjusting the expression as to form a single sine or cosine term is probably a more ideal solution. (as you did in your alternate method)

so LHS = 2sin(2t-pi/6) = sqrt3

I believe this should be

[lzxnl]

Note that if you do this you will eliminate one of the solutions you're looking for.

is a solution to the original equation. However, as you'll notice, is undefined for all and so are the alternate forms from here on out, which is where the problem arises.

Adjusting the expression as to form a single sine or cosine term is probably a more ideal solution. (as you did in your alternate method)I believe this should be

For the last bit, point taken. My mind has been malfunctioning at a regular rate recently.

As for the first point, I see your point there. Sigh. Dividing by zero since 2013. As you can probably see, I wasn't really thinking when I was typing this and it looked cool to use a tangent half-angle formula somewhere. Oh well.