Attachment 20073

Stuck on these questions would appreciate the help very much!!!

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- Dec 12th 2010, 12:02 PMstudent31413 Interesting Questions
Attachment 20073

Stuck on these questions would appreciate the help very much!!! - Dec 12th 2010, 12:16 PMtonio
- Dec 12th 2010, 12:21 PMstudent3141
ive been trying but im tottaly stuck dont know where to begin! or how to approach it

- Dec 12th 2010, 11:04 PMmr fantastic
- Dec 12th 2010, 11:33 PMstudent3141
Ok i did question 7

8) i tried to get the mvt for both sides, do i treat the ln(x) and the RHS as seperate? and do i then juss find the mvt for both sides? cause when i try and get the mvt for the RHS i get 3/8 i think thats wrong isnt it supposed to be between 1 and 2??

9) This question i kno its implicit differentiation but what should i do with the A B C ???

I cant seem to find similar questions in text book... help needed!! - Dec 12th 2010, 11:45 PMProve It
Treat $\displaystyle \displaystyle A,B,C$ as constants.

- Dec 13th 2010, 12:09 AMstudent3141
For question 9 this is wat i did help!

2Ax + 2By(dy/dx) = 0

Dy/Dx = Ax/By

D^2y/Dx^2 = ByA - BAx(dy/dx) / (B^2).(y^2)

= BA(y-x(dy/dx)) / (B^2).(y^2)

whats going on????? did i do something wrong? - Dec 13th 2010, 12:14 AMProve It
$\displaystyle \displaystyle 2A\,x + 2By\,\frac{dy}{dx} = 0$

$\displaystyle \displaystyle 2B\,y\,\frac{dy}{dx} = -2A\,x$

$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{2A\,x}{2B\,y}$

$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{A\,x}{B\,y}$.

To find $\displaystyle \displaystyle \frac{d^2y}{dx^2}$ you need to use the Quotient Rule... Also remember that you already know what $\displaystyle \displaystyle \frac{dy}{dx}$ is so you can substitute it in... - Dec 13th 2010, 12:24 AMstudent3141
http://upload.wikimedia.org/math/a/c...7302faea9a.png

btw thanks for the help

Thats quotient rule so:

dy/dx = - Ax/By

= - (ByA - AxB(dy/dx))/(B^2).(y^2)

= BA ( y - x(dy/dx))/(B^2).(y^2)

i get stuck over here - Dec 13th 2010, 12:31 AMProve It
Be more careful with what you write, in this step you're finding $\displaystyle \displaystyle \frac{d^2y}{dx^2}$, not $\displaystyle \displaystyle \frac{dy}{dx}$.

Also, you forgot to bring the negative down into your last step.

Now, assuming you had brought the negative down...

$\displaystyle \displaystyle -\frac{AB\left(y - x\,\frac{dy}{dx}\right)}{B^2y^2} = -\frac{A\left(y - x\,\frac{dy}{dx}\right)}{B\,y^2}$

$\displaystyle \displaystyle = -\frac{A\left[y - x\left(-\frac{A\,x}{B\,y}\right)\right]}{B\,y^2}$ - Dec 13th 2010, 12:59 AMstudent3141
AHHH yess i get it now thank you so much!

Btw did you use a program to write that working out ? looks much more neater than what i was doing!

For question 8

(3/2)x - (1/4)x^2 - (5/4)

F(2)-F(1) / 2-1 = 3/4

dy/dx = 3/2 - (1/2)x

3/4 = 3/2 - (1/2)c

c= 3/2

ok so do i now juss work out the mvt of ln(x)??? - Dec 13th 2010, 03:39 AMProve It
I'm using the LaTeX compiler that is built into the forum.

See the LaTeX help subforum to learn how to use it.