quick question

[Natters]

http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/specialist/pastexams/2010/2010specmath2-w.pdf
vcaa exam 2 2010, question 22 of the multiple choice
why is there a +v₀² instead of a +c?

[dc302]

F=ma

f/m = dv/dt

dt/dx f/m = (dv/dt) (dt/dx)

1/v (f/m) = dv/dx

(f/m)dx = v dv

integrate LHS from x0 to x1, integrate RHS from v0 to v1.

As you can see, once you integrate the RHS, v becomes (1/2)v^2, and then you sub in v0 and v1, so that is where all the square roots and stuff come from.

[Natters]

i get all of it except for the V₀² but i think i get it now
for some reason this vcaa exam seems easy as f compared to all the company ones

[Zebra]

i still dont understand why v0^2 is true.

[HarveyD]

howd you go on the exam natters?

[dc302]

i still dont understand why v0^2 is true.

Ok continuing from where I left off:

(f/m)dx = v dv

integrate LHS from x0 to x1, RHS from v0 to v1

we get

int(f/m)dx = int v dv

int(f/m)dx = (1/2) v1 ^2  - (1/2) v0 ^2

so (1/2) v1 ^2 = int(f/m)dx + (1/2)v0 ^2

v1 ^2 = 2int(f/m)dx + v0 ^2

Note that m (mass) is a constant, so we can take it out

v1 ^2 = (2/m)int(f)dx + v0 ^2

v1 = sqrt[  (2/m)int(f)dx + v0 ^2  ]


edit: if someone could write this up with latex, that would be nice

[Natters]

this 2010 practice exam im getting about 60% on which is good enough, im assuming ur talking about this one

[Natters]

i didnt use v dv dx though i used d/dx (.5v²)
ended up with +c instead of the Vo2

int(f/m)dx = int v dv

int(f/m)dx = (1/2) v1 ^2  - (1/2) v0 ^2
dont get that line mainly

dw about it though, if it was on the 2010 it won't be on this years

[dopplereffect]

You need to understand how The Special Equation works. It's an extension of the Fundamental Theorem of Calculus and I'm fairly certain you have to use it in this equation.

Fundamental Theorem: If F(x) is an antiderivative of f(x)

Then F(b) - F(a) = int(f(x),a,b)

The Special Equation is F(b) = int(f(x),a,b) + F(a)

This is very useful when you are given an equation that can't be integrated but you have to find y when x is a value.

EG. VCAA 2007 Exam 2 - MC Q11

If dy/dx = sqrt(sin(x)) and y=1 when x=0 then the value of y when x=pi/3 can be found by evaluating...

Using The Special Equation then y = int(sqrt(sin(x),0,pi/3) + 1

I haven't seen many questions where you have to use it but it is useful.

It's not really called The Special Equation, that's what Derrick Ha calls it and how I remember it

[Natters]

didnt you just write the whole definite integral to find an area thing (though it doesnt always have to be an area)?
i dont see anything there i didnt already know lol dw about it tho

[dc302]

didnt you just write the whole definite integral to find an area thing (though it doesnt always have to be an area)?
i dont see anything there i didnt already know lol dw about it tho

Well you only get +c if you're doing an indefinite integral with no initial conditions. However, here you are given conditions so naturally you need to find c. (which will end up as v0^2 or whatever)

Anyway good luck for exams!

edit: and no, I'm not finding the area, although you can think of it like that if you want. It's probably to do with a fundamental flaw in understanding about how the +c and stuff works but it's not gonna be big in the exam.

[Natters]

polish a turd it's still a turd thats what it is and thanks =p

btw for a "show this = that"
say its like.... show umm something simple for an example
(2x)^4=16x^4

is it sufficient to show that (2x)^4=(4x^2)^2 and that 16x^4=(4x^2)^2?
like change both sides to something halfway, or do i have to just commit to changing one side?

[Zebra]

i tried but i still don't understand why you wouldn't use

d(1/2 v^2) /dx?

[Zebra]

or is this question to do with v^2=u^2 + 2 a s ?

[dc302]

polish a turd it's still a turd thats what it is and thanks =p

btw for a "show this = that"
say its like.... show umm something simple for an example
(2x)^4=16x^4

is it sufficient to show that (2x)^4=(4x^2)^2 and that 16x^4=(4x^2)^2?
like change both sides to something halfway, or do i have to just commit to changing one side?

Yes that's enough, I'm pretty sure.

i tried but i still don't understand why you wouldn't use

d(1/2 v^2) /dx?

I think you don't understand because you have simply memorised some formula/rule whereas I'm actually doing it without such things. You can use that derived result if you want to. And no it's not about the 2nd equation. But do you now understand why it's vo^2 and not c? (because using the derived result or not doesn't change whether its +c or not).