Differentiate an equation to find the minimum peak of a variable

[souky101]

Summary:: Differentiate an equation to get the minimum peak

I have this equation:

6.3504 = 23.04 ( 1-141978.24XY)^2 + 141978.24X^2
I want to differentiate it with respect to X to get the minimum peak of Y
.
dy/dx =0
.
See attached file

Thanks for help

[Moderator's note: moved from a technical forum.]

 

[Office_Shredder]

What work have you done to try to solve this? Did you attempt to differentiate the equation?

 

[haruspex]

Homework Statement:: 6.3504 = 23.04 ( 1-141978.24XY)^2 + 141978.24X^2
find the minimum peak of y
Relevant Equations:: I have this equation:
6.3504 = 23.04 ( 1-141978.24XY)^2 + 141978.24X^2

I want to differentiate it with respect to X to get the minimum peak of y
dx/dy=0
see attached photo
.
Thanks for help

differentiate the equation with respect to X to get the minimum peak of y
dx/dy=0

I think you want a local minimum of Y, not a "minimum peak".
And that means dy/dx=0, not dx/dy=0.

For convenience, let's get rid of those numbers and write the equation as
$D= B( 1-Axy)^2 + Cx^2$
So differentiate wrt x. What do you get?

 

[jedishrfu]

Did you try to solve for x ie create a y = f(x) format from there it’s a simple matter to get the dy/dx?

Show your work.

 

[souky101]

Thanks for your reply
.
Yes , we can put it in this form:
D=B(1-Axy)^2 + Ax^2

The graph of this equation depends on the constants D and B - but if we differentiate this equation with respect to x , dy/dx =0 , then the constant D will disappear though the graph depends on it ??
.
See the attached graph

 

[haruspex]

the constant D will disappear

It will come back, trust me.
Do the differentiation and post what you get.

 

[souky101]

D=B(1-Axy)^2 + Ax^2
dy/dx=0

0=2B(1-Axy)(-Ax(dy/dx) -A y(dx/dx)) +2Ax
=2B(1-Axy)(-Ay)+2Ax
=2B-2BAxy(-Ay) +2Ax
=2B+2BA^2xy^2 +2Ax

 

[haruspex]

=2B(1-Axy)(-Ay)+2Ax
=2B-2BAxy(-Ay) +2Ax

Check that step.

 

[souky101]

2B(-Ay)+2BA^2xy^2+2Ax =0

 

[souky101]

2B(-Ay)+2BA^2xy^2+2Ax =0
-By+BAxy^2+x=0

 

[haruspex]

2B(-Ay)+2BA^2xy^2+2Ax =0
-By+BAxy^2+x=0

Ok. You can use that in combination with your original equation, but it is not going to be pretty.

 

[souky101]

This equation is function of x and y
Now how to get the minimum value of y ?
and this equation is independent of the constant D though the plotted graph shape depends on the constant D ?.

 

[haruspex]

This equation is function of x and y
Now how to get the minimum value of y ?
and this equation is independent of the constant D though the plotted graph shape depends on the constant D ?.

As I posted, combine it with your original equation. Use the equation in post#13 to eliminate x from the equation in post #6.
You should end up with a quadratic in y².

 

[souky101]

Thanks for your help - I will try
.
Thanks again

 

[souky101]

May I ask you what the name of the text editor you are using to write mathematical equations ?
.
Thanks for help

 

[haruspex]

May I ask you what the name of the text editor you are using to write mathematical equations ?
.
Thanks for help

LaTeX.
Click the LaTeX Guide button just below the text entry window.

 

[souky101]

.
I did the substitution but the resultant equation still contains the two variables x and y

D=B(1-Axy)^2 +Ax^2 ... (1)
-By+ABxy^2 +x =0 .....(2)

Substitute (2) in (1) :
(1-Axy) = x/By

D=B(x/By)^2 + Ax^2
.
This equation is still function of 2 variables x and y !

 

[haruspex]

.
I did the substitution but the resultant equation still contains the two variables x and y

D=B(1-Axy)^2 +Ax^2 ... (1)
-By+ABxy^2 +x =0 .....(2)

Substitute (2) in (1) :
(1-Axy) = x/By

D=B(x/By)^2 + Ax^2
.
This equation is still function of 2 variables x and y !

I didn't think I would need to explain that you first have to get eqn 2 into the form x=f(y).