VCE Methods Question Thread!

[keltingmeith]

5^n+1-5^n-1/5^n+5^n-2

Need help with this question

First, clean up the bits hanging around:


And... You're basically done. You could rewrite the fraction as follows, though:

[faso]

First, clean up the bits hanging around:


And... You're basically done. You could rewrite the fraction as follows, though:

Sorry,the division was for the whole expression what I meant was :
5^n+1-5^n-1 divide by 5^n+5^n-2

[keltingmeith]

Sorry,the division was for the whole expression what I meant was :
5^n+1-5^n-1 divide by 5^n+5^n-2

Oh, okay. Well then:



So, clean up the top and bottom:



And we see that it cleans up to 0.

[faso]

Oh, okay. Well then:



So, clean up the top and bottom:



And we see that it cleans up to 0.

Sorry again......
What I meant was
5^(n+1)-5^(n-1)divide by 5^(n)+5^(n-2)

Forgot the brackets.
And What program are you using ?

[keltingmeith]

Sorry again......
What I meant was
5^(n+1)-5^(n-1)divide by 5^(n)+5^(n-2)

Forgot the brackets.
And What program are you using ?

OHHH okay, thought the last two were a bit easy. So,



So first, clean up the exponents so they're all to the n:



Factorise that 5^n out:



Simplify:



This was all written using the LaTeX type-setting language, which the forum supports through tex tags.

[knightrider]

How do they cancel out the in the working out attached?
Why can they do this and does this always work?

[lzxnl]

The log function is 1 to 1. Therefore, as only one x value returns a y value, if you have log x = log y, then x = y
Alternatively, think of it as the base of the log to the power of both sides. So if ln x  = ln y, then e^(ln x) = e^(ln y) => x = y

[faso]

OHHH okay, thought the last two were a bit easy. So,



So first, clean up the exponents so they're all to the n:



Factorise that 5^n out:



Simplify:



This was all written using the LaTeX type-setting language, which the forum supports through tex tags.

What happened to the negative in the 5^n-2

[knightrider]

The log function is 1 to 1. Therefore, as only one x value returns a y value, if you have log x = log y, then x = y
Alternatively, think of it as the base of the log to the power of both sides. So if ln x  = ln y, then e^(ln x) = e^(ln y) => x = y

Thanks Lzxnl 

so whenever you have the same base logs on both sides of an equation you can equate them right?

[cosine]

Why can't we divide by zero?

[keltingmeith]

What happened to the negative in the 5^n-2

Whoops, that time I actually forgot to do that. .__. Let's go again from where I messed up:

Why can't we divide by zero?

Good question! Let's assume that n/0 equals some number, k, where n can be any number. From this, we get:



This, of course, means that n=0 (as anything multiplied by 0 equals 0). However, earlier we said that n could be any number - however, we've just shown that the quotient only exists if n is 0. So, we've reached a contradiction, which shows that we cannot divide by 0.

Another method is to look at the graph of 1/x. The limit as x approaches 0 from the negative side and the limit as x approaches 0 from the positive side produces two different results. Since , then cannot exist.

[Apink!]

Hi there! If someone could help me, that would be awesome!

This is an integration question:
A long trough whose cross-section is parabolic is 3/2 meters wide at the top and 2 meters deep. Find the depth of water when it is half- full.
I don't understand how I would approach this question because I don't understand what's happening in the question (i.e what is actually happening)

Also, how would you do this question?
Find the exact area of the region bounded by the graphs of equations:
 y²= x and x-y=2.

I think I can't seem to solve this because I don't know which is the "top" and "bottom" function I need to identify to find the area under the curve. Just how would you do this if the curves don't seem to be so clear on which one is top and bottom?

Thanks for helping me. I can't ask these questions to my maths teacher at school because she simply refuses to help me questions that she hasn't taught yet (we are still doing the functions bit, and if I give her a calculus question, she won't answer me because she didn't teach it to us )

[IntelxD]

Just what my opinion is, I have always wondered why the answer cannot equal zero, I mean i proved it above right?
Also literally speaking division can be read as how many times does 0 go into 'n', assuming it is any number, zero would go into it 0 times, right? Some deep stuff here..

Nope.

[keltingmeith]

Just what my opinion is, I have always wondered why the answer cannot equal zero, I mean i proved it above right?
Also literally speaking division can be read as how many times does 0 go into 'n', assuming it is any number, zero would go into it 0 times, right? Some deep stuff here..

Except what both of us have done aren't really proofs - rather, they're arguments. They're really not rigorous enough to be proofs.

I'd disagree with what you said in that second paragraph, though - if you want to put 0 into n, you'd have to put an infinite amount of 0s in until you finally reached n (and even then, you'd never really reach n). Rather, the statement makes no sense, which lends to the idea that n/0 doesn't have a defined result.

Hi there! If someone could help me, that would be awesome!

This is an integration question:
A long trough whose cross-section is parabolic is 3/2 meters wide at the top and 2 meters deep. Find the depth of water when it is half- full.
I don't understand how I would approach this question because I don't understand what's happening in the question (i.e what is actually happening)

Try drawing it out - I've seen this question before, and to memory it did have a diagram attached?

Also, how would you do this question?
Find the exact area of the region bounded by the graphs of equations:
 y²= x and x-y=2.

I think I can't seem to solve this because I don't know which is the "top" and "bottom" function I need to identify to find the area under the curve. Just how would you do this if the curves don't seem to be so clear on which one is top and bottom?

Thanks for helping me. I can't ask these questions to my maths teacher at school because she simply refuses to help me questions that she hasn't taught yet (we are still doing the functions bit, and if I give her a calculus question, she won't answer me because she didn't teach it to us )

It's not really an issue that you identify which is the top or bottom function - integrate the function both ways, and you should get the same magnitude, but one will be negative and one will be positive. You want the positive answer - do you think you can see why, though?

[Apink!]

Hi Eulerfan
The question is from Cambridge essentials and unfortunately it doesn't have a diagram attached.
and for the second question, is it because area can't be negative?

Thanks!