Solve Diff Eqns for Polar Functions & Critical Points

  • Thread starter vinverth
  • Start date
Use algebra and trigonometry to solve for these points.In summary, the conversation is about a student seeking help with two math questions, one involving finding the differential equations for polar functions and the other involving locating critical points. The conversation includes the student's attempts at solving the problems and their request for assistance.
  • #1
vinverth
3
0
Hello everyone..Find it embarrassing enough on asking a question on my very first post but I've been an avid reader of the forums for the past couple of months and been finding what i need for all my assignments here.So a big Thank You to all who've helped.I'm A EE grad and have a math course in my final semester so am a complete noob when it comes to grad math courses,a little consideration here while posting replies or even answers.So here i have a couple of q's whose answers or at least a decent start I've been searching all over the web.

1.Find the differential equations for the polar functions r,ө of the following two-dimensional systems.

(a) x'=x+y
y'=x-y

2.Locate the critical points of the following systems.

(a) x'=x-y²
y'=x²-y²
These are both separate questions.Answers to anyone pleasezzz..
(b) x'=sin(y)
y'=cos(x)Thank You again to everyone and please bail me out guys!
 
Physics news on Phys.org
  • #2
vinverth said:
1.Find the differential equations for the polar functions r,ө of the following two-dimensional systems.

(a) x'=x+y
y'=x-y

Polar coordinates are defined by

[tex]x = r \cos\theta,~y=r\sin\theta.[/tex]

You should rewrite the equations using these and solve for [tex]r', \theta'[/tex].

2.Locate the critical points of the following systems.

(a) x'=x-y²
y'=x²-y²
These are both separate questions.Answers to anyone pleasezzz..
(b) x'=sin(y)
y'=cos(x)

Critical points are the points where the derivatives of functions either don't exist or are zero.
 
  • #3
I already tried that approach but it got me nowhere..what confuses me is the fact that the equations are already in their first derived form for both q's wrt x and y.
i tried to plug in x=rcosө and y=rsinө which gives me x'=rcosө+rsinө.
(rcosө)'=rcosө+rsinө
How do i proceed frm here..

and for the second q..since x'=sin(y) and y'=cos(x)..do i just check for what values x' and y' are 0 and figure out their critical points.
Any kind of help or a start is appreciated..and I've been trying to solve them in every possible way via research on internet.but maybe its just my fundamentals..too weak at them..like i said before i hail from a diff background
 
  • #4
vinverth said:
I already tried that approach but it got me nowhere..what confuses me is the fact that the equations are already in their first derived form for both q's wrt x and y.
i tried to plug in x=rcosө and y=rsinө which gives me x'=rcosө+rsinө.
(rcosө)'=rcosө+rsinө
How do i proceed frm here..

Use the product rule for derivatives

[tex] (fg)' = f' g + f g'.[/tex]

and for the second q..since x'=sin(y) and y'=cos(x)..do i just check for what values x' and y' are 0 and figure out their critical points.

The (x,y) values for which either x'=0 or y'=0 are critical points.
 

FAQ: Solve Diff Eqns for Polar Functions & Critical Points

What are polar functions and how are they different from Cartesian functions?

Polar functions are a type of mathematical function that uses polar coordinates, which consist of a radius and an angle, to describe points on a graph. They are different from Cartesian functions, which use x and y coordinates, in that they describe points in terms of distance and direction from the origin rather than horizontal and vertical distances.

How do you solve differential equations for polar functions?

To solve differential equations for polar functions, you first need to convert the polar coordinates to Cartesian coordinates. Then, you can use standard methods for solving differential equations, such as separation of variables or using a substitution. Once you have the solution in Cartesian form, you can convert it back to polar coordinates to obtain the final solution.

What is a critical point in a polar function?

A critical point in a polar function is a point where the derivative of the function is equal to zero. In other words, it is a point where the slope of the tangent line is undefined or flat. Critical points are important because they can help us determine the behavior of the function, such as where it has maximum or minimum values.

How do you find critical points in a polar function?

To find critical points in a polar function, you first need to find the derivative of the function. Then, set the derivative equal to zero and solve for the values of the angle that make the derivative equal to zero. These angles correspond to the critical points of the function. You can then plug these angles back into the original function to find the corresponding radius values at the critical points.

What are some real-life applications of solving differential equations for polar functions?

Differential equations for polar functions have many applications in the real world, such as in physics, engineering, and astronomy. For example, they can be used to model the motion of celestial bodies, the growth of populations, and the vibrations of a guitar string. They are also useful in designing and analyzing circular and spiral-shaped objects, such as gears and springs.

Back
Top