Differentiate an equation to find the minimum peak of a variable

In summary: Then you substitute that into eqn 1 to get a single equation in y only. As I said a few posts back, it won't be pretty.
  • #1
souky101
21
1
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.]
 

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  • #2
What work have you done to try to solve this? Did you attempt to differentiate the equation?
 
  • #3
souky101 said:
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?
 
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  • #4
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.
 
  • #5
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
 

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Last edited:
  • #6
souky101 said:
the constant D will disappear
It will come back, trust me.
Do the differentiation and post what you get.
 
  • #7
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
 
  • #8
souky101 said:
=2B(1-Axy)(-Ay)+2Ax
=2B-2BAxy(-Ay) +2Ax
Check that step.
 
  • #9
2B(-Ay)+2BA^2xy^2+2Ax =0
 
  • #10
2B(-Ay)+2BA^2xy^2+2Ax =0
-By+BAxy^2+x=0
 
  • #11
souky101 said:
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.
 
  • #12
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 ?.
 
  • #13
souky101 said:
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.
 
  • #14
Thanks for your help - I will try
.
Thanks again
 
  • #15
May I ask you what the name of the text editor you are using to write mathematical equations ?
.
Thanks for help
 
  • #16
souky101 said:
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.
 
  • #17
.
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 !
 
  • #18
souky101 said:
.
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).
 

Related to Differentiate an equation to find the minimum peak of a variable

What is differentiation?

Differentiation is a mathematical process that involves finding the rate of change of a function with respect to its independent variable. It is essentially the opposite of integration.

Why do we need to differentiate equations?

Differentiation allows us to analyze the behavior of functions and understand how they change over time or in response to different inputs. It is a fundamental tool in many fields of science, including physics, engineering, and economics.

How do you differentiate an equation?

To differentiate an equation, you need to use the rules of differentiation, which include the power rule, product rule, quotient rule, and chain rule. These rules allow you to find the derivative of a function, which represents the rate of change of that function.

What is the difference between differentiation and integration?

Differentiation and integration are inverse operations. Differentiation finds the rate of change of a function, while integration finds the area under a curve. In other words, differentiation is used to find the slope of a curve, while integration is used to find the area under a curve.

What are some real-world applications of differentiation?

Differentiation has many practical applications, such as calculating velocity and acceleration in physics, optimizing production processes in engineering, and determining marginal cost and revenue in economics. It is also used in fields like biology, chemistry, and medicine to model and analyze complex systems.

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