I have 2 constants and one variable (x).

I need an easy ease - smooth transition function (something like "S" shape) between 2 constant values (c1 & c2) with a variable (x, from 0 to 10). So when the x is 0, target value will be "c1". When the x is 10, target value will be "c2".

And the proper smoothed target values in between 0 & 10 values of x variable.

c1: first value of the constant

c2: last value of the constant

y: our target value that we want to determine

m: Middle point of the constants that when the x is 5.

Which function or method do you suggest for put this datastogether? Sigmoid-like interpolation?

I couldn't bring this equation together with 2 constants, mid point and x variable.

Any help will be appreciated....

Best regards... ]]>

The final question of my problem set due in 5 hours eludes me!!

z0 is a complex different from 2'

z^3-3z=z0

z1=(z0+2)/(z0-2)

z2=z1^(1/2) so the square root

z3=(z2+1)/(z2-1)

z4=z3^(1/3) so the cubic root

z5=z4+1/z4

Prove that z5^3-3z5 = z0

I've tried to express z5 in terms of z0 and see if it could be simplified but that didn't really work out. I've been trying to maybe see if something can be done by expressing z exponentially but I'm very stuck.

Thank you for any help!! ]]>

. . . . .[tex]\displaystyle \int_0^{\pi}\, \left[\, \sin(2x)\, \cdot\, \cos(3x)\, \cdot\, \sin(4x)\, \cdot\, \cos(5x)\, \cdot\, \sin(6x)\, \right]\, dx[/tex]

I tried to group the terms, then to transform in sum but didn't work.

Some ideas?

Im not sure if Im applying the right rules to the equation.

We have to find the first level derivative.

The equation given is:

y = x^2 * 2^x^2

Hope that makes sense?

Its x to the power of 2 times 2 to the power of x to the power of 2. If that helps?

So the way I understood it, I have to apply certain rules to the equation to gain the result.

In this case - (x^n) = nx^n-1

So my way of thinking:

y = x^2 * 2^x^2 becomes

2x * 2 * 2x

Is this right? Or completely wrong (more likely). ]]>

First post here so I hope I'm doing it right.

Following comparison gives a three-dimensional butterfly figure:

Code:

`x=sin(t)*(exp(cos(t))-2*cos(4*t)-(sin(t/12))^5) `

y=cos(t)*(exp(cos(t))-2*cos(4*t)-(sin(t/12))^5)

z=0

So I figured out that following integral gives me the labour:

I guess C will be replaced by 12*Pi, but what should be on the dr?

I hope you guys can help.

Thank you very much ]]>

{ x^2,x>1}

Now,after this the problem requires me to find the antiderivative of f(x) which is F(x)(the primitive function).

F(x)={x^2-x,x<=1}

{x^3/3,x>1}

If the integral of f(x) is F(x) then the derivative of F(x) is f(x)(the original function).I stated this because I didn't want to create a confusion.

The problem also requires me to check if the values of the primitive function F(x) is F(4)=-8.I tried to plug the values into the function F but I couldn't get the -8 value.How do I solve this problem?Also the problem wants me to find F(-1) and the answer to this is -27.How can F(-1)=-27? ]]>

I

really having trouble with this one.

first thing I tried was just integrating by parts as it is, which I couldn't make work.

I tried changing the expression to:

I

which simplified to:

I

which then integrating by parts on this new expression didn't have much luck either.

is there any hints on how I should approach this question?

thanks ]]>

In a book I found that for f(x,t):

dff.png

But I could not remember this rule of partial differentiation. Any help?

Thanks

If f(x,y)=(xy)

Thanks ]]>

R(T) = exp(-∫

I have the answer from notes, however cannot work it through logically myself to get the provided answer.

Ans noted:

exp(-3.5x10

exemplo.jpg

Intuitively, I thought of this parameterization as a way to describe the graph mathematically, for that I divided it into 3 regions of the domain and for each one I found a corresponding function, for example, in the first region, I found a first degree equation (a and b would be the parameters?) and so on. Does it have anything to do with it? If not, what would this parameterization be? Thanks in advance!

a) Show that F satisfies the partial differential equation

∂F/∂x + 2x•∂F/∂y = 0

b) Given that F(0,y) = sin y for all y, find a formula for F(x,y).

for a) I take LHS and subbed F for

∂/∂x(*f *(x^{2 }− y)) + 2x•∂/∂y (*f *(x^{2 }− y))

= 2x• *f ' *(x^{2 }− y) - 2x• *f ' *(x^{2 }− y) = 0

so LHS = RHS

for b) I'm a bit confusedso LHS = RHS

if F( 0, y ) = sin(y) thenF( 0, y) =

now the answer in the book is F(x,y) = sin ( y - x

thanks ]]>