hi yaz!!!!!

4)
a)
         i. y'= 3x^2 - 12
         ii. here we wanna find out where the tangents are horizontal soooo... we set it equal to zero!
                    3x^2 - 12 = 0
                    3x^2 = 12
                      x^2 = 4
                      x = +2, -2

             so le'ts plug these guys back in2 the original equation:
                 (2)^3 - 12(2) + 7 = -9 --------> (2, -9)
                 (-2)^3 - 12(-2) + 7 = 23 ------> (-2, 23)

        iii. here we wanna find out when the graph is decreasing and increasing soo we choose points!
                let's try -1000000000000
                  3(-1000000000000)^2 - 12 = some positive number!!!  so increasing!!!

                let's try 0
                   3(0)^2 - 12 = some negative number!!!  so decreasing!!!

                let's try 1000000000000
                   3(1000000000000)^2 - 12 = some positive number!!!  so increasing!!!

               sooooo.... the graph is increasing from (-infinity, -2) U (2, infinity)
                                             and decreasing from (-2, 2)

b) i. since u know this one i'll go hed and skip it!

    ii. we know the total surface area = 2040
         so let's come up w/ an equation for it!!!!!

          we know the base and lit of the box are squares, so the surface area of a square is x^2 but we have two of these so let's go hed and multiply that by 2 so it looks like this 2x^2

           if we cut open the box and spread out the pieces we will end up w/ 4 rectangles and 2 squares.  so the surface area of a rectangle is x*(height) and we have four of these so let's multiply that by 4 so it looks like this 4xh.

now we have an equation!!!!  surface area = 2x^2 + 4xh = 2040

oh noes!!!!  but we wanna maximize the volume and only got the surface figured out... bummer.

we know the volume of the box = x^2 * h. 

now lets solve for h in the surface euqation:
510/x - 1/2 * x = h

and plug that in in2 our volume equation:
volume = x^2 (510/x - 1/2 * x)
            = 510x - 1/2 * x^3

iii) let's find the derivative of the volume!!!!
            =510x - x^3 * 1/2
            = 510 - 3/2 * x^2

now we wanna find out the maximum volume of the box, but how do we do that???  well, let's find out the largest point possible!!!!  set the derivate of the volume equal to zero

             510 - 3/2 x^2 = 0
                         x = +(340)^(1/2) or -(340)^(1/2)
                                  but we know the volume can't be negative!!!  silly!!!  so let's only keep positive answers!!
               
finally, we must prove this is the largest value, so in order to do this, we wanna c how the graph behaves.  let's find out where it deacreses and increases by choosing points

    lets try 5
                   510 - 3/2 * (5)^2 = some positive number!!!  increasing baby!!!!
    lets try 2000000000000000
                    510 - 3/2 * (2000000000000000)^2 = some negative number!!!  oh dear!!!

   finally, we know the graph is increasing from zero to (340)^(1/2) and decreasing from (340)^(1/2) all the way to infinity land!!!  so x = (340)^(1/2) must be the maximum point!!!

good day!

Last edited by Flowers4Carlos (2005-08-29 17:52:54)

Hi again...Ok I have some more questions for you!!!!!! These are mixed diffrentiation and intergration and are pretty hard........ i will state whther to diffrentiate or intergrate..

Diffrentiate the following with respect to x:

(a)y=5x+1

(b)f(x)=6x² -ln2x

(c)y=2(x+cos2PIEx)³

(d)f(x)=x^4e^-³x

(e)y=sin²x/2x+3


Evaluate the following integrals:

(a) ∫(2x+1/x+2)dx

(b)^4∫^°(1+2/ √t)dt
{the small zero is at the bottom of the curvy line}

(c) ^PIE/4∫^°4sin2θdθ
{the small zero is at the bottom of the curvy line}

(d) ∫x(x-3)^4dx       Use the substitution u=x-3



HAHAH lets see how quick and accurate you can respond to these questions!!!!!!

Let the differential operator with respect to x be d

Differentiation:

A) y=5x+1
-------------------------------------
dy/dx = d(5x+1)=5



B) f(x)=6x² -ln2x
-------------------------------------
df/dx = d(6x² - ln(2x))=d(6x²)-d(ln(2x))=12x-(1/x)



C) y=2(x+cos(2πx) )³
-------------------------------------
k=2π
y=2(x+cos(kx) )³

t=x+cos(kx)
y=2t³

dy/dx=dy/dt * dt/dx

dy/dt=d(2t³)=6t²=6[x+cos(kx)]²
dt/dx=d( x+cos(kx) ) = dx+d(cos(kx))=1-ksin(kx)

dy/dx=6[x+cos(kx)]²*[1-ksin(kx)]=6[x+cos(2πx)]²*[1-2πsin(2πx)]     //do you really want me to expand this? :)
                                                                                                                //we better leave the expression as is



D) f(x)=x^4e^-³x            //I really don't understand the nesting here. so I'll solve it as x^[4e^(-3x)]
-------------------------------------
This one's a bit complex, but you can check the formula by yourself

u,v,w are functions
d[u^v^w] = (u-¹)(u^v^w)(v^(w-1))[ ln(u).[ v.u.d(w).ln(v)+w.u.d(v) ] + v.d(u) ]
u=x;  d(u)=1
v=4e;  d(v)=0
w=(-3x);  d(w)=-3

df/dx=d[u^v^w]=(x-¹).[x^(4e^(-3x))].[e^(-3x)].[4-12xln(x)]     //I hope I didn't forgot anything...:)



E) y=sin²x/2x+3          //a bit confusing too.
-------------------------------------
y=(sin²x)/(2x)+3         //was this what you wanted to write?

d[u/v]=[vd(u)-ud(v)]/v²
u=sin²x; d(u)=d(sin²x)=2sin(x)cos(x)
v=2x; d(v)=2

dy/dx=d[u/v]=[sin(x)cos(x)]/(x)-sin(x)/(2x²)



Integration:

A)∫(2x+1/x+2)dx
-------------------------------------
∫(2x+1/x+2)dx=∫(2x)dx+∫(1/x)dx+∫2dx=x²+ln(x)+2x+C



B)^4∫^°(1+2/ √t)dt
-------------------------------------
did you mean:
4
∫( 1+2/√t)dt
0


∫( 1+2/√t)dt=∫1dt+∫(2/√t)dt=t+2∫[1/ (t^(1/2)) ]dt=t+2[(t^(-1/2+1))/(-1/2+1)]+C=t+2[2(t^1/2)]=t+2(2√t)=t+4√t+C

b
∫u'du=U(b)-U(a)
a

let U(t)=t+4√t+C

4
∫( 1+2/√t)dt = U(4)-U(0)
0

U(4)=4+4√4+C=12+C
U(0)=0+4√0+C=C
U(4)-U(0)=12+C-C=12

4
∫( 1+2/√t)dt = 12
0



C)^PIE/4∫^°4sin2θdθ
-------------------------------------
did you mean:
π/4
∫ 4sin(2θ)dθ
0


let w=2θ
dw/dθ=2 <=> dw=2dθ

∫4sin(2θ)dθ=2∫2sin(2θ)dθ=2∫sin(2θ)2dθ=2∫sin(w)dw=2[-cos(w)]=2[-cos(2θ)]=-2cos(2θ)+C

U(θ)=-2cos(2θ)+C

π/4
∫ 4sin(2θ)dθ =U(π/4)-U(0)
0

U(π/4)=C
U(0)=-2+C
U(π/4)-U(0)=C-[-2+C]=2


π/4
∫ 4sin(2θ)dθ =2
0



D)∫x(x-3)^4dx
-------------------------------------
∫x(x-3)^4dx
u=x-3
x=u+3

du/dx=d[x-3]=1; du=dx

∫x(x-3)^4dx=∫[(u+3)(u)^4]du=∫(u^5+3u^4)du=∫[u^5]du+3∫[u^4]du=(u^6)/6+3(u^5)/5+C=
=(5u^6+18u^5)/30+C
var.substituition
(5u^6+18u^5)/30+C=(u^5)(5u+18)/30+C=[ (x-3)^5 ].[5(x-3)+18]/30+C=[ (x-3)^5 ].[5x-15+18]/30+C=
=[ (x-3)^5 ].[5x+3]/30+C

∫x(x-3)^4dx=[ (x-3)^5 ].[5x+3]/30+C



finally finished!!!
Please watch out for any mistakes here... I was typing for almost half an hour....ufff!  :)!!!
I really enjoyed all the exercises! thanks!

-------------------------------------------------------------------------------------------------------------------
The problems are not too hard ;) Its faster and easier working on the paper...! And if we had LATEX It would be easier...
I hope mathisfun has LATEX on the "to do" list :)

Last edited by kylekatarn (2005-09-01 23:34:35)

kylekatarn!!!!!!!!!!  it looks like u made a simple mistake in question c integration:

4∫sin2θdθ = -2cos2θ + c

what is the name of the formula u used for question d first part???  i've never seen it!!!!

Although the result is the same I fixed the expression.
Thanks again.!

I'll post the proof to that derivation formula asap.

Last edited by kylekatarn (2005-09-01 23:40:30)

all numbers are special and deserve to be celebrated whenever we want =P