Just did a practice trial, raised numerous questions!! (1 Viewer)

[studybuddy09]

As the title suggests, i just did a trial in preparation for the HSC (first ive done so far) and it raised a few questions, some of which ive had for a while now:

1. I can never understand how you choose which x-value goes with which side of the ratio in the division of an interval. Is there an easy way of deciding this and putting it in the formula?

2. With rates of change etc, there are three things for finding acceleration from velocity or velocity from acceleration. I know one is a = d/dx (1/2 v^2)
and i think another is a = v. dv/dx but i dont really understand how they work/ how you derive them since Maths in Focus only explains one of them. Please Explain?

3. I forget how to get a locus. It seems that when i sub x into y or y into x equations i get something wrong/ complicated. Please Help?

Will be on and off throughout the day to check but now i must study for Physics!!!


THANKS

 

[scardizzle]

1. if A(x1,y1) B(x2,y2) are divided in the interval m:n then multiply the A with n and B with m
you pretty much switch it around

2.

Lets say we are given the equation

a = v^2 + 3 and we want to find an equation for velocity

There are 2 ways to do this that i know of:

i) a = dv/dt

so v^2 + 3 = dv/dt

therefore dt/dv = 1/v^2 + 3

intergrating with respect to v gives us a velocity equation

ii) a is also = v x dv/dx

therefore v x dv/dx = v^2 + 3

therefore dv/dx = v + 3/v

intergrate with respect to x gives us a velocity eq 'n but now with x

3. when a question asks you to find the locus they're pretty much asking you to find the cartesian equation that is, get rid of parameter whether it be t or p or q etc.

this is usually done by making the parameter the subject of the equation for instance:

x/2a = p

then sub this into the y equation, obviously this is a very simple explanation and doing more questions will help you

good luck studying

(on a side note: on BOS tsk tsk tsk...)

 

[studybuddy09]

2. There are 2 ways to do this that i know of:

i) a = dv/dt

we'll use your example where a = v

so v = dv/dt

therefore dt/dv = 1/v

intergrating with respect to v gives us a velocity equation

ii) a is also = v x dv/dx

therefore v x dv/dx = v

therefore dv/dx = 1

intergrate with respect to x gives us a velocity eq 'n but now with x

This does not make any sense to me.........

Why did you let a=v? i meant a= v.dv/dx in my OP


Oh well, i guess I just have to do more questions and look over the notes when bakes told us this...


lol, just doing my SECOND trial paper for today (2U CSSA 2009) and im lovin' it except that I've forgotten positive definitite and indefinite etc so ill have to look over that

 

[scardizzle]

oh my bad, misread your post(nice use of OP ) fine let a = v^2 +3 or something and just follow those steps

 

[kwabon]

1. if A(x1,y1) B(x2,y2) are divided in the interval m:n then multiply the A with n and B with m
you pretty much switch it around

2. There are 2 ways to do this that i know of:

i) a = dv/dt

we'll use your example where a = v

got it all wrong mate.

a = v dv/dx

RHS = dx/dt . dv/dx (dx's cancels out)
= dv/dt
= a
= LHS

thats how you prove it. (y)

dont worrying too much about how to prove it, as they cannot assess you on it, just know how to use it.

 

[untouchablecuz]

Lets say you have as a function of v, and you want to find x as a function of v:

i.e.

Using



Do the same with as equations with in terms of x and t

 

[annabackwards]

For 1, use the cross method:

If you want to find P which divides point AB in the interval m:n
A(x1, y1) B(x2, y2)
...........m:n

The cross connects m to B and n to A, that way you won't get confused.
So P( mx2+nx1/m+n , my2+ny1)

 

[studybuddy09]

did another trial today and got 1. easy, so i must have just forgotten.

I think I'll need to practice 2., but thankyou for ur derivation untouchablecuz, it was most helpful

3. didnt come up but i will just have to practice it also

THANKS to you all