# 3 Interesting Questions

• Dec 12th 2010, 01:02 PM
student3141
3 Interesting Questions
Attachment 20073

Stuck on these questions would appreciate the help very much!!!
• Dec 12th 2010, 01:16 PM
tonio
Quote:

Originally Posted by student3141
Attachment 20073

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

This looks like homerwork and you show no self work at all...what've you done so far?

Tonio
• Dec 12th 2010, 01:21 PM
student3141
ive been trying but im tottaly stuck dont know where to begin! or how to approach it
• Dec 13th 2010, 12:04 AM
mr fantastic
Quote:

Originally Posted by student3141
ive been trying but im tottaly stuck dont know where to begin! or how to approach it

Have you reviewed your class notes and textbook for similar examples that you can follow?
• Dec 13th 2010, 12:33 AM
student3141
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 13th 2010, 12:45 AM
Prove It
Treat $\displaystyle A,B,C$ as constants.
• Dec 13th 2010, 01:09 AM
student3141
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, 01:14 AM
Prove It
$\displaystyle 2A\,x + 2By\,\frac{dy}{dx} = 0$

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

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

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

To find $\displaystyle \frac{d^2y}{dx^2}$ you need to use the Quotient Rule... Also remember that you already know what $\displaystyle \frac{dy}{dx}$ is so you can substitute it in...
• Dec 13th 2010, 01:24 AM
student3141

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, 01:31 AM
Prove It
Be more careful with what you write, in this step you're finding $\displaystyle \frac{d^2y}{dx^2}$, not $\displaystyle \frac{dy}{dx}$.

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

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

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

$\displaystyle = -\frac{A\left[y - x\left(-\frac{A\,x}{B\,y}\right)\right]}{B\,y^2}$
• Dec 13th 2010, 01:59 AM
student3141
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, 04:39 AM
Prove It
I'm using the LaTeX compiler that is built into the forum.

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