Author Topic: VCE Specialist 3/4 Question Thread! (Read 2601697 times) Tweet Share

allusionx · « **Reply #3390 on:** July 23, 2014, 07:34:18 am »

Anyone would like to help me? The question is starting to annoy me.(c&d)

Thanks.

Jacyan · « **Reply #3391 on:** July 24, 2014, 02:26:03 am »

Quote from: allusionx on July 23, 2014, 07:34:18 am

Anyone would like to help me? The question is starting to annoy me.(c&d)

Thanks.

Check out the attachment. The key is to use part a, and remember that y gives the depth h, and x is the radius of the circle (this is used to find the surface area A).

allusionx · « **Reply #3392 on:** July 24, 2014, 05:28:48 pm »

Thankyou very much.

Mieow · « **Reply #3393 on:** July 24, 2014, 08:53:14 pm »

Where did the negative sign at the front come from after the antidifferentiation?

Thorium · « **Reply #3394 on:** July 24, 2014, 09:02:50 pm »

Quote from: Mieow on July 24, 2014, 08:53:14 pm

Where did the negative sign at the front come from after the antidifferentiation?

The antiderivative of 1/(a+bx) is log(a+bx)/b. This is reverse chain rule. The same steps are taken to integrate the expression you attached. For your case, for the first part of the expression, b is equal to -1

Spxtcs · « **Reply #3395 on:** July 28, 2014, 08:05:50 pm »

Can someone please help me with this question, particularly part b:

A bowl is formed by rotating the parabola with equation 1/2x^2+2 about the y-axis between 2<x<4 (both signs are 'or equal to' signs)
a) Find the volume of the bowl (is it 64 $pi$ cubic units?)
b) find an expression for V(h), the volume of the liquid in the bowl if it is filled to a depth of h. State any restrictions on the value of h.

Thanks in advance.

keltingmeith · « **Reply #3396 on:** July 28, 2014, 08:42:22 pm »

a)
$\pi\int_{x=2}^{x=4}x^2dy \\ y=\frac{1}{2}x^2 + 2 \implies x^2=2y-4 \\ y=\frac{1}{2}(2)^2+2 = 4\text{ and }y=\frac{1}{2}(4)^2+2=10 \\ \therefore \pi\int_{y=4}^{y=10}2y-4dy \\ =\pi[y^2-4y]_4^{10} \\ =\pi((100-40) - (16-16)) = 60\pi$

b) The height is given by y, so $h=\frac{1}{2}x^2 + 2\text{ where }x\in [2,4] \implies h\in [4,10]$

We also know that the solid of revolution will give us the volume of the bowl - however, unlike last time when we knew what the height was, this time we don't - just that it's any number between 10 and 4. So, this gives us:

$V(h)=\pi\int_4^h 2y-4dy =\pi[y^2-4y]_4^h=\pi(h^2 - 4h) \\ \therefore V(h) = \pi(h^2 - 4h), h\in (4,10] \text{ as }h=4\text{ implies no volume}$

Spxtcs · « **Reply #3397 on:** July 28, 2014, 08:52:02 pm »

Thank you so so much!!!

hyunah · « **Reply #3398 on:** July 30, 2014, 09:58:08 pm »

A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find:
i) how much further it will travel before ocming to rest
ii) how many more seconds it takes before coming to rest

thanks and please

IndefatigableLover · « **Reply #3399 on:** July 31, 2014, 08:55:55 pm »

Verify that $y=\frac{x+1}{x-2}$ is a solution of the differential equation:

$\frac{d^2y}{dx^2}=\frac{6}{27}(y-1)^3$

keltingmeith · « **Reply #3400 on:** July 31, 2014, 09:10:20 pm »

Quote from: IndefatigableLover on July 31, 2014, 08:55:55 pm

Verify that $y=\frac{x+1}{x-2}$ is a solution of the differential equation:

$\frac{d^2y}{dx^2}=\frac{6}{27}(y-1)^3$

Before I do anything, I'm going to write y in a much neater form:
$y=\frac{x+1}{x-2} = \frac{(x-2) + 3}{x-2} = 1+\frac{3}{x-2}$

Now that that's out of the way, I'm going to have to find $\frac{d^2y}{dx^2}$ :

$\frac{dy}{dx} = \frac{-3}{(x-2)^2} \\ \frac{d^2y}{dx^2} = \frac{6}{(x-2)^3}$

Now, we sub this into both sides and hope we get the same result. The LHS is pretty obviously $\frac{6}{(x-2)^3}$ , so:

$\text{RHS }=\frac{6}{27}(1 + \frac{3}{x-2} - 1)^3 \\ =\frac{6}{27}\left(\frac{27}{(x-2)^3}\right) \\ = \frac{6}{(x-2)^3} = \text{ LHS}$
Therefore, $y=\frac{x+1}{x-2}$ is a solution to the differential equation.

IndefatigableLover · « **Reply #3401 on:** July 31, 2014, 09:17:16 pm »

Quote from: EulerFan101 on July 31, 2014, 09:10:20 pm

Before I do anything, I'm going to write y in a much neater form:
$y=\frac{x+1}{x-2} = \frac{(x-2) + 3}{x-2} = 1+\frac{3}{x-2}$

Now that that's out of the way, I'm going to have to find $\frac{d^2y}{dx^2}$ :

$\frac{dy}{dx} = \frac{-3}{(x-2)^2} \\ \frac{d^2y}{dx^2} = \frac{6}{(x-2)^3}$

Now, we sub this into both sides and hope we get the same result. The LHS is pretty obviously $\frac{6}{(x-2)^3}$ , so:

$\text{RHS }=\frac{6}{27}(1 + \frac{3}{x-2} - 1)^3 \\ =\frac{6}{27}\left(\frac{27}{(x-2)^3}\right) \\ = \frac{6}{(x-2)^3} = \text{ LHS}$
Therefore, $y=\frac{x+1}{x-2}$ is a solution to the differential equation.

Thanks for that EulerFan101! I got a tad confused when I saw the 'y' since it wasn't like any of the other examples in the book so I thought it was an error but apparently not! Thanks for that

keltingmeith · « **Reply #3402 on:** July 31, 2014, 09:26:16 pm »

Quote from: IndefatigableLover on July 31, 2014, 09:17:16 pm

Thanks for that EulerFan101! I got a tad confused when I saw the 'y' since it wasn't like any of the other examples in the book so I thought it was an error but apparently not! Thanks for that

When they ask you in your 3/4 exam to answer a diff-eq, they are always in a form like that. In fact, in specialist, you will learn how to solve equations of the form $\frac{dy}{dx} = f(y)$ , and from 2016 specialist will even include solving questions of form $\frac{dy}{dx} = f(x)g(y)$ , but you'll have to wait for uni maths to solve the second one.

lzxnl · « **Reply #3403 on:** July 31, 2014, 11:17:36 pm »

Quote from: hyunah on July 30, 2014, 09:58:08 pm

A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find:
i) how much further it will travel before ocming to rest
ii) how many more seconds it takes before coming to rest

thanks and please

Define t = 0 before bus moves
60 metres in 10 seconds => average velocity 6 m/s, must occur half-way in the 10 s time period, so at t = 5s, v = 6 m/s
60 m in 15 s => average velocity 4 m/s, must occur half-way in 15 s time period so at 7.5 s, v = 4 m/s
This 7.5 s is after the initial 10 s time period, so t = 17.5 s at v = 4 m/s
In 17.5 - 5 = 12.5 s, the velocity has dropped from 6 to 4 m/s, so the deceleration is -2/12.5 m/s^2
So clearly another 25 seconds are needed for the bus to stop completely
Now you can try the rest yourself

knightrider · « **Reply #3404 on:** July 31, 2014, 11:26:08 pm »

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?

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2601697 times) Tweet Share

allusionx

Re: Specialist 3/4 Question Thread!

Jacyan

Re: Specialist 3/4 Question Thread!

allusionx

Re: Specialist 3/4 Question Thread!

Mieow

Re: Specialist 3/4 Question Thread!

Thorium

Re: Specialist 3/4 Question Thread!

Spxtcs

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

Spxtcs

Re: Specialist 3/4 Question Thread!

hyunah

Re: Specialist 3/4 Question Thread!

IndefatigableLover

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

IndefatigableLover

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

lzxnl

Re: Specialist 3/4 Question Thread!

knightrider

Re: Specialist 3/4 Question Thread!

Recent Posts

User:
Password: