VCE Methods Question Thread!

[paper-back]

A price of string 10 metres long is cut into two prices to form two squares
If one peice of string has length x meters show that the combined area of the 2 squares is given by
A=1/8(x^2-10x+50)

Having difficulty with this one, any help would be appreciated
Thanks

[b^3]

Let's let the first piece of string have length m, then the second is of length m. Since we know that they form two squares, the side length of these squares will be a quarter of the length of the string for that square. So for the first the side length is while for the second the side length is m. Now the area of the square is the side length squared, adding those results together should give you the above.

Spoiler

[paper-back]

Thank you b^3!

[Einstein]

how do you solve

2(x-1/2) - x+3/4 = 9x-5/4

Since year 7 ive never got this "subtract 7 from both sides"  "times both sides by AB" etc, now going it higher maths im getting ssome questions wrong by not following this rule, could some show me how it is done with the above, and also with this question as well

1/y = 1/a - 1/b for y

thank you

[Zealous]

how do you solve
2(x-1/2) - x+3/4 = 9x-5/4

This is very standard algebra - your Year 11 Methods book should have a few exercises on this sort of thing. I suggest do as many questions as you can so you get fast and confident at it.



Expand the brackets:


Combine like terms on the LHS:


Bring the 'x' from the LHS to the RHS, bring the 5/4 from the RHS to the LHS:


Simplify both sides:




Divide both sides by 8:

1/y = 1/a - 1/b for y

You've already asked this question recently and b^3 gave an excellent explanation on how to solve it here:

Re: Methods [3/4] Question Thread!

Please make sure you check back a few pages to make sure you haven't missed any responses to the questions you've asked.

[achre]

how do you solve

2(x-1/2) - x+3/4 = 9x-5/4

Since year 7 ive never got this "subtract 7 from both sides"  "times both sides by AB" etc, now going it higher maths im getting ssome questions wrong by not following this rule, could some show me how it is done with the above, and also with this question as well

1/y = 1/a - 1/b for y

thank you

beaten by a fair bit

It kinda clicked for me when I saw someone do it backwards. For example:







etc. In solving any equation, you just go from the last step back to the first, by "undoing" the operations via the opposite operation (where there's multiplication, you divide). Though, I suppose the explanation is also fairly straightforward - the equals sign means that both sides are the exact same quantity. So if you take 2 from one side, you have to take 2 from the other to keep them equal, and just repeatedly simplify one side until you have the desired expression in terms of another.
For your question:

Spoiler

Now I add 1/4 to each side. -(1/4)+(1/4)=0, so it's gone from the RHS. -(5/4)+(1/4)=1, so our expression is now . I do the same with the 9x (9x-9x=0, x-9x=-8x). Then I divide each side by -8 (-8x/-8=x, -1/-8=1/8) to get my final expression .

For question 2, to get y in terms of a and b, you just take the reciprocal (literally just flip the fraction of 1/y upside down to y/1, and (1/a-1/b)/1 to 1/(1/a-1/b), which simplifies to )

[Einstein]

With this question



why cant i just flip the reciprocal from the start?

edit: btw how do you actually write with latex, do you actually write {alignedat} etc.. or is their some software? i just copied the above from b^3 text.

[Phy124]

With this question



why cant i just flip the reciprocal from the start?

edit: btw how do you actually write with latex, do you actually write {alignedat} etc.. or is their some software? i just copied the above from b^3 text.

You can.





Regarding latex yes we type it out (I do anyway, some might type it up elsewhere like lyx and copy it across) but what you have typed is only necessary if you're trying to align something, you'll notice that when does it he is usually aligning the text along the equals signs, to do this you put an ampersand before each equal sign.

e.g.

Code: [Select]

\begin{alignedat}{1}y&=\frac{1}{\frac{1}{a}-\frac{1}{b}}
\\ &=\frac{1}{\frac{b-a}{ab}}
\end{alignedat}


Rather than having:

Code: [Select]

y=\frac{1}{\frac{1}{a}-\frac{1}{b}}
\\=\frac{1}{\frac{b-a}{ab}}



Websites like this are good for understanding latex basics but you'll probably have to google tutorials for things like alignment.

[Einstein]

ok so ill just show you my working out and please tell me if im missing anything

    [/tex] \frac{1}{y} = \frac{1}{a} - \frac{1}{b}
ab(\frac{1}{y}) = b - a   i have multiplied 1/a by b and 1/b by a, and the LHS by ab, btw do i need it to be \frac{b}{ab}-\frac{a}{ab} or am i correct in what i have done?

continuing on \frac{ab}{y} =  b - a

y = \frac{ab}{a-b}

quick question: why have some people got it as = \frac{b}{ab} - \frac{a}{ab} and not just ab(\frac{1}{y}) = b-a [/tex]
jeez that was confusing haha!

edit: can someone understand what ive done here, latex didnt seem to work.

[IndefatigableLover]

ok so ill just show you my working out and please tell me if im missing anything

\frac{1}{y} = \frac{1}{a} - \frac{1}{b}
ab(\frac{1}{y}) = b - a   i have multiplied 1/a by b and 1/b by a, and the LHS by ab, btw do i need it to be \frac{b}{ab}-\frac{a}{ab} or am i correct in what i have done?

continuing on \frac{ab}{y} =  b - a

y = \frac{ab}{a-b}

quick question: why have some people got it as = \frac{b}{ab} - \frac{a}{ab} and not just ab(\frac{1}{y}) = b-a

jeez that was confusing haha!

edit: can someone understand what ive done here, latex didnt seem to work.

isn't working because you need to type within the setting box (well that's what I call it anyway LOL) but when people type in you need first press the button and type within that and it should come out alright!

Code: [Select]

[tex]\frac{1}{y} = \frac{1}{a} - \frac{1}{b}[/tex]
^What you just typed in your post but remember to place it within the 'tex boxes'

[alchemy]

Can someone explain the way of expanding cubics in the attached image? Some of the notation I don't understand in this context.
Yes, I'm into overkills.

[Mieow]

Can someone please help me with these two questions:

A rectangular block of ice is melting at a rate of 2 cm^3/s. The block has a square base and its height is three times the
base length. Find the rate at which the base length is changing when it is 8 cm

An inverted cone full of sand is being emptied via a hole in the bottom at a rate of 6 cm^3/h. The rate at which the height of the sand in the cone is changing, given the radius is always equal to the height, is:

[Billion]

V(t)=20+0.02t+5sin((pi/6)t), where V is the value of the stock in dollars and t is the number of months after Jan 06.

Can someone explain in detail how the inflation rate is 0.1% per month.

Thanks.

[Orb]

V(t)=20+0.02t+5sin((pi/6)t), where V is the value of the stock in dollars and t is the number of months after Jan 06.

Can someone explain in detail how the inflation rate is 0.1% per month.

Thanks.

Inflation rate is the percentage that the value rises in accordance to time.
So inflation rate is dV/dt
= 0.02 + (5pi/6) x cos ((pi/6)t)

That's how I thought that you'd approach the question, but I'm not quite sure how you obtain the 0.1% per month value
 

[Zealous]

A rectangular block of ice is melting at a rate of 2 cm^3/s. The block has a square base and its height is three times the
base length. Find the rate at which the base length is changing when it is 8 cm

1: Write down the rate we know:

(note the negative because the block is melting!)

2: Write down what we are trying to find:

(where L is the side length of the square base)

3: Set up your chain rule equation:



4: We now need to find dL/dV, we can do that using the information provided:





5: Bring this back into the chain rule equation:



6: Substitute L=8 into dL/dt:



Hence when the base length of the block is 8cm, the rate in which it is changing is

An inverted cone full of sand is being emptied via a hole in the bottom at a rate of 6 cm^3/h. The rate at which the height of the sand in the cone is changing, given the radius is always equal to the height, is:

1: Write down the rate we know:

(note the negative because it is being emptied!)

2: Write down what we are trying to find:



3: Set up your chain rule equation:



4: We now need to find dh/dV, we can do that using the information provided:





5: Bring this back into the chain rule equation:



______________________

I used the exact same procedure for both questions - hopefully this gives you an idea of the steps you can take to solve these questions. I might write up a more in depth guide on Related Rates for everyone as I'm seeing a lot of these questions popping up and I found it challenging myself when I started doing them.