MATH HELP NEEDED PLEASE...

[kirishi]

[deleted]

 

[pka]

Your #4 must be copied incorrectly. If f(1)=2 then \(\displaystyle \L
x^3 f(x) + \left( {f(x)} \right)^3 + f(x^3 )\) must equal 12 because \(\displaystyle \L
1^3 f(1) + \left( {f(1)} \right)^3 + f(1^3 ) = 1^3 (2) + \left( 2 \right)^3 + f(1) = 12\).

The derivative is \(\displaystyle \L
\left[ {3x^2 f(x) + x^3 f'(x)} \right] + \left[ {3\left( {f(x)} \right)^2 f'(x)} \right] + \left[ {3x^2 f'(x^3 )} \right] = 0\).
Now substitute x=1 and solve for f’(1)

 

[kirishi]

[deleted]

 

[soroban]

Hello, kirishi!

#3 is quite easy if you made a sketch . . .

3. The area of a rectangle is 4cm^2.
Find the dimensions of the rectangle so that the distance from one vertex
to the midpoint of a non adjacent side is minimum.

    A         x         B
    * - - - - - - - - - *
    | \                 |
   y|   \ z             |y
    |     \             |
    |       \           |
    * - - - - * - - - - +
    D   x/2   M         C

Given: rectangle $\displaystyle ABCD$.
. . Length: $\displaystyle x\,=\,AB\,=\,CD$
. . Width: $\displaystyle y\,=\,AD\,=\,BC$
$\displaystyle M$ is the midpoint of $\displaystyle CD$; hence, $\displaystyle DM\,=\,\frac{x}{2}$

We want to minimize distance $\displaystyle z\,=\,AM$.

Since the area is 4, we have: $\displaystyle \,xy\,=\,4\;\;\Rightarrow\;\;y\,=\,\frac{4}{x}$

From Pythagorus, we have: \(\displaystyle \L\,z^2\:=\:\left(\frac{x}{2}\right)^2\,+\,y^2\:=\:\left(\frac{x}{2}\right)^2\,+\,\left(\frac{4}{x}\right)^2\:=\:\frac{x^2}{4}\,+\,\frac{16}{x^2}\)

Then: \(\displaystyle \L\:z\;= \;\sqrt{\frac{x^4\,+\,64}{4x^2}}\;=\;\frac{(x^4\,+\,64)^{\frac{1}{2}}}{2x}\)

Can you finish it now?

 

[kirishi]

hi..thanks...i think i wud b able to finish it now...if i need more help i'll ask again...and for #4 i got -3/8 as my answer...can someone tell me if thats rite?? thanks

 

[galactus]

thanks soo much....
for question #1 what would the equation be??
so far i did this much:
v=area of a cross section x length
=area of trapezoid x length
=1/2 (5)(4) x (10)
=100

now what do i do??

See this link for a similar problem:

http://www.freemathhelp.com/forum/viewt ... ght=trough

 

[kirishi]

for #3 do i make z equal to zero then solve to find wat the points r?? so it would b:
(x^4+64)^1/2=0
x^4+64=0
x^4=-64
now wat do i do??

 

[kirishi]

for the question wid the trough and triangle i noe how to do that but if its a trapezoid wat would the equation b?? isn't it different?? i just dunt noe the equation...or how to make the equation...if i have the equation i noe wat to do after that...thanks

 

[pka]

or the question wid the trough and triangle i noe how to do that but if its a trapezoid wat would the equation b?? isn't it different?? i just dunt noe the equation...or how to make the equation...if i have the equation i noe wat to do after that...thanks

Why do you write such non-standard English?

 

[kirishi]

<Why do you right such non-standard English?>

i am used to writing like that on msn ok...and i don't think my english has anything to do with the calculus questions i am askin ok...

 

[pka]

“i don't think my english has anything to do with the calculus questions i am askin ok...”
Well it makes you look stupid! Why should we take you seriously?

 

[kirishi]

listen if you don't want to take me seriously then don't ok...and i am not the only one who writes like this...i just asked for help..and so far you don't seem to be helpin but just finding reasons to say my post is stupid...if you are going to post and say comments like this then don't even post under my topic...i'm trying to get help and all you are doing is saying stuff that is very irrelevant to the topic...

 

[Unco]

for #3 do i make z equal to zero then solve to find wat the points r?? so it would b:
(x^4+64)^1/2=0
x^4+64=0
x^4=-64
now wat do i do??

Differentiate, then set to zero.

 

[pka]

Who are you to make any demands?
“if you are going to post and say comments like this then don't even post under my topic..”
We have no obligation to help you!
All we are asking is that YOU be reasonable!
We have better things to do than try to translate your abbreviations.
We are very serious here about mathematics help!
Please help us by being serious yourself!

 

[stapel]

listen if you don't want to take me seriously then don't ok...and i am not the only one who writes like this...i just asked for help....

Yes, many kiddies use cutesy kiddie-speak -- with other kiddies. But if you're going to come to a grown-up environment and ask grown-ups to help you, it might be wise (and certainly would be polite) to speak clearly.

You want respect shown to you; perhaps you should show some of that same respect to those from whom you are demanding favors.

Thank you for your consideration.

Eliz.

 

[soroban]

Hello, kirishi!

i got wat u ment on #5.

5. Find the area of the triangle formed by the three points on the curve \(\displaystyle \L y\:=\:\left(\frac{x^2\,-\,1}{x^2\,+\,1\right)^2\)
where the slopes of the tangent lines to the curve at those points are zero.

u need the derivative.

\(\displaystyle \L\frac{dy}{dx}\;=\;2\left(\frac{x^2\,-\,1}{x^2\,+\,1}\right)\cdot\frac{(x^2+1)\cdot2x\,-\,(x^2-1)\cdot2x}{(x^2+1)^2}\;=\;\frac{8x(x^2-1)}{(x^2+1)^2}\)

We want slope 0 so we noe that: $\displaystyle \,8x(x^2-1)\,=\,0\;\;\Rightarrow\;\;x\,=\,0,\pm1$

When $\displaystyle x\,=\,0,\,y\,=\,1$
When $\displaystyle x\,=\,1,\,y\,=\,0$
When $\displaystyle x\,=\,-1,\,y\,=\,0$

Plot the three points: \(\displaystyle \,(0,1),\1,0),\-1,0)\)

We have a triangle wid base = 2 and height = 1.

$\displaystyle \text{Area }=\,\frac{1}{2}(2)(1)\:=\:1$

 

[galactus]

For the trough problem:

Let h=depth of water; b=length of top of cross-section of water.

We want $\displaystyle \frac{dh}{dt}$ when $\displaystyle h=2$

We know: $\displaystyle \frac{dV}{dt}=+5$

$\displaystyle V=h(\frac{b+2}{2})(10)$

$\displaystyle \frac{b-2}{h}=\frac{3-2}{4}$,

$\displaystyle \frac{b-2}{h}=\frac{1}{4}$

Solve for b and sub:

$\displaystyle b=\frac{h}{4}+2$

$\displaystyle h(\frac{\frac{h}{4}+4}{2})(10)=\frac{5h(h+16)}{4}=\frac{5h^{2}}{4}+20h$

$\displaystyle \frac{dV}{dt}=(\frac{5h}{2}\frac{dh}{dt}+20\frac{dh}{dt})$

When $\displaystyle h=2$

$\displaystyle 5=(\frac{5}{2}(2)\frac{dh}{dt}+20\frac{dh}{dt})$

$\displaystyle \frac{dh}{dt}=\frac{1}{5}$ m/min.