VCE Specialist 3/4 Question Thread!

[keltingmeith]

I was just looking at the study design for specialist maths

It says it goes from 2006-2015

Say you do 1/2 specialist maths in 2015 does that mean in 2016 you will have to learn the new study design

I don't quite understand this  because if you do specialist maths 1/2 in year 11 in 2015 will they continue the same design for people going on into year 12 in 2016 or does it completely change?

While I'm not completely sure on this, and as far as I know VCAA have not commented, but I do know this:

THERE IS NO SPECIALIST MATHS 1/2 IN THE CURRENT STUDY DESIGN. GMA is not specialist, no matter WHAT your school says.

There *IS* a specialist 1/2 in the new study design, but it's not very relevant to the changes they made to specialist. So, I reckon they might get 2016 3/4ers to do it from the new study design, because you wouldn't be disadvantaged at all.

[hyunah]

thank you lzxnl
I got it now !

Capacitors are devices that store electric charge (the unit is the coulomb) and do so when an electric current
is directed onto them through a resistor connected to a battery. The charge Q(t) stored on a capacitor as a
function of time is modelled by the equation:
R (dQ/dt) + Q/C=V
where Q(0) = 0, V is the voltage of the battery, R is the resistance of the resistor and C is the capacitance of the capacitor.
For a particular circuit, V = 10 volts, R = 8.0 ohms and C = 0.1 farad.

b Find the charge, Q(t), stored on the capacitor at time t.

c Find the time taken for there to be 0.63 coulomb of charge stored on the capacitor. Show that this value is equal to the product R × C. Note that an ohm farad has the same unit (s) as the second.

d Determine the maximum amount of charge able to be stored on the capacitor given a large charging time.

e Find the charge stored as a function of time in terms of the parameters R, C and V; that is, solve the differential equation for the general case.

[Phy124]

how would I transpose y = x^2 - 2ex + e^2 +2 to make it x = something and x^2 equals something

.

[lzxnl]

thank you lzxnl
I got it now !

Capacitors are devices that store electric charge (the unit is the coulomb) and do so when an electric current
is directed onto them through a resistor connected to a battery. The charge Q(t) stored on a capacitor as a
function of time is modelled by the equation:
R (dQ/dt) + Q/C=V
where Q(0) = 0, V is the voltage of the battery, R is the resistance of the resistor and C is the capacitance of the capacitor.
For a particular circuit, V = 10 volts, R = 8.0 ohms and C = 0.1 farad.

b Find the charge, Q(t), stored on the capacitor at time t.

c Find the time taken for there to be 0.63 coulomb of charge stored on the capacitor. Show that this value is equal to the product R × C. Note that an ohm farad has the same unit (s) as the second.

d Determine the maximum amount of charge able to be stored on the capacitor given a large charging time.

e Find the charge stored as a function of time in terms of the parameters R, C and V; that is, solve the differential equation for the general case.

Let's solve this in general, noting that the initial charge is zero (this is important for the mod signs)















Now, Q(0) = 0 so the first exponential term, which is a constant, can be replaced as follows (I'm too lazy to write out everything)



Try everything else yourself

[hyunah]

thank you so much!

just wondering how did you get from

Now, Q(0) = 0 so the first exponential term, which is a constant, can be replaced as follows (I'm too lazy to write out everything)

I tried to plug in t= 0 q = 0
but im lost?

[lzxnl]

thank you so much!

just wondering how did you get from
I tried to plug in t= 0 q = 0
but im lost?

Can you see that I've replaced the first exponential term in the first line with another constant, namely -CV, and that the second line has Q(0) = 0? If you want, sub in t = 0 into the first line and set Q = 0. You'll find that -CV = e^c/RC

[hyunah]

thank you once again!

can you please help me with part c

I put 0.63 = 1 - e^-5/4t I get t = 0.795402 which is rougly 0.8 which is 0.8, but how do I show that it is equal to the product of RC?

[Brunette15]

For part d to this question can someone please clarify why the minimum and maximum speeds occur at cos(t)=0 and cos(t)=1.

[Brunette15]

Can someone also help me out with proving this question?

[Edward Elric]

Help with this Question please

If the ellipse given by x^2+2ax+2y^2+4by+16=0 and has a centre (3,-2), the values of a and b would be?

[nhmn0301]

Help with this Question please

If the ellipse given by x^2+2ax+2y^2+4by+16=0 and has a centre (3,-2), the values of a and b would be?

x^2+2ax+2y^2+4by+16=0
(x^2 + ax ) + 2(y^2 + 2by) + 16 = 0
(x + 1/2a)^2 + 2(y+b)^2 - 1/4a^2 - 2b^2 + 16 = 0
(x + 1/2a)^2 + 2(y+b)^2 = 1/4a^2 + 2b^2 - 16
Since centre (3 , -2 ), 1/2a = -3 => hence, a = -6
                                  b = + 2
You can check this by substituting on the RHS,  which yields 1 when a = -6 and b = 2.

[tiff_tiff]

help here please !

the annual growth rate of the population of two towns P and Q are 10% and 5% respectively of their populations of their populations at anytime. If the initial population of P is 20000 and Q is 10000, find the population after three years.

Wouldn't i just find ten percent add it to the existing, and then find ten percent of that and add it for the three years. How would i set up equations to do this? differential equations? cause the topic im doing in class is differential equations?

[hyunah]

Capacitors are devices that store electric charge (the unit is the coulomb) and do so when an electric current
is directed onto them through a resistor connected to a battery. The charge Q(t) stored on a capacitor as a
function of time is modelled by the equation:
R (dQ/dt) + Q/C=V
where Q(0) = 0, V is the voltage of the battery, R is the resistance of the resistor and C is the capacitance of the capacitor.
For a particular circuit, V = 10 volts, R = 8.0 ohms and C = 0.1 farad.


c Find the time taken for there to be 0.63 coulomb of charge stored on the capacitor. Show that this value is equal to the product R × C. Note that an ohm farad has the same unit (s) as the second.
Given that we known Q(t) = 1 – e^-(5t/4)

I put 0.63 = 1 - e^-5/4t I get t = 0.795402 which is rougly 0.8 which is 0.8, but how do I show that it is equal to the product of RC? Like using the pronumerals?

Thank you in advance

[IndefatigableLover]

Solve the differential equation:



So I somehow got an answer though I wasn't sure if my way of working would be ideal... I was wondering if there was another way of working this out (maybe through substitution since I have trouble finding what to substitute...)

But yeah what I did to solve for the equation was the take out the '2' and put it outside the integral before splitting the fraction into partial fractions and integrating them there. From then onwards I made them all the same denominator and then subbed in 'y=0 when x=0' and found that the constant equalled zero resulting in my answer as

Spoiler

I'm just not sure if that's the ideal way of approaching this question... any other suggestions maybe?

[keltingmeith]

Solve the differential equation:



So I somehow got an answer though I wasn't sure if my way of working would be ideal... I was wondering if there was another way of working this out (maybe through substitution since I have trouble finding what to substitute...)

But yeah what I did to solve for the equation was the take out the '2' and put it outside the integral before splitting the fraction into partial fractions and integrating them there. From then onwards I made them all the same denominator and then subbed in 'y=0 when x=0' and found that the constant equalled zero resulting in my answer as

Spoiler

I'm just not sure if that's the ideal way of approaching this question... any other suggestions maybe?

This is the ideal way to do it - no other nice way. You could use a trig substitution - maybe - but this is a much nicer, cleaner method.

The only thing I'd fault you on is you shouldn't have modulus signs in your answer - since you have an initial condition, you should be able to gauge what that should look like without modulus signs.

EDIT: Did just try a trig substitution for interest's sake - it's not pretty. At all.