Can Different Choices of w Lead to Continuously Differentiable u?

In summary, the conversation discusses the differentiability of u, a function of two real variables r and theta, and its relationship to w, which is defined in terms of sines and cosines. The experts come to the conclusion that u is continuously differentiable as long as w is continuously differentiable, and provide examples of functions for u that are and are not continuously differentiable.
  • #1
sara_87
763
0

Homework Statement


Given:
w=9(cos([tex]\theta[/tex])+sin([tex]\theta[/tex])) and u=r*w.
1) Is u continuously differentiable?
2)Is it possible to get u continuously differentiable with a different w?

Homework Equations





The Attempt at a Solution



1) u is continuously differentiable since w is in terms of sines and cosines which are continuously differentiable functions. (Is that right?)
2) I'm stuck on this one.

Thank you
 
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  • #2
What's r? A constant? A function of theta? A variable? I'm inclined to think that r and theta are polar co-ordinates, is that correct? If so what are you differentiating with respect to?

What level of proof is required? I.e. do you have to do things from first principles? Can you appeal to known results: sin and cos are certainly continuously differentiable, but can you use that?
 
  • #3
As a function of two real variables r and theta, u is continuously differentiable. But if you think of u as a function on the x-y plane described by the polar coordinates (r,theta) you MAY have a problem. Think about what happens at r=0. Do you?
 
  • #4
u(x,y)=rw(theta) where r and theta are polar coordinates.When we differentiate, it will be with respect to x and y.
When r=0 then u=0 which is continuously differentiable..??
 
  • #5
u=0 is the value of the function at r=0. I would be more worried about the values of the derivatives of u near r=0. Try writing it as a function of x and y.
 
  • #6
Do you mean I should write w in terms of x and y?
 
  • #7
sara_87 said:
Do you mean I should write w in terms of x and y?

Sure. And then write u as a function of x and y.
 
  • #8
w=9cos(theta)+9sin(theta)
=9(x/r)+9(y/r)
so u=rw=9(x+y)
the derivative (whether its with respect to x or y) is 9 which is continuously differentiable.
 
  • #9
sara_87 said:
w=9cos(theta)+9sin(theta)
=9(x/r)+9(y/r)
so u=rw=9(x+y)
the derivative (whether its with respect to x or y) is 9 which is continuously differentiable.

Seems fine to me.
 
  • #10
Thanks.
What about the next one: Is it possible to get u continuously differentiable with a different w?
 
  • #11
sara_87 said:
Thanks.
What about the next one: Is it possible to get u continuously differentiable with a different w?

You shouldn't have any trouble coming up with others that are. It's more interesting to think about ones that aren't. How about u=r?
 
  • #12
u=r is continuously differentiable.
and we can have other functions like u=rsin(2theta) etc.. right?
 
  • #13
sara_87 said:
u=r is continuously differentiable.
and we can have other functions like u=rsin(2theta) etc.. right?

u=r is not even differentiable at 0. Think harder about it. It's a lot like absolute value, f(x)=|x|.
 
  • #14
u=r=sqrt(x^2+y^2)
the derivative is u'=(1/2)(2x or 2y)(x^2+y^2)^(-1/2)
the derivative at 0 is 0.


But can we not have other functions for u that are continuously differentiable?
 
  • #15
sara_87 said:
u=r=sqrt(x^2+y^2)
the derivative is u'=(1/2)(2x or 2y)(x^2+y^2)^(-1/2)
the derivative at 0 is 0.


But can we not have other functions for u that are continuously differentiable?

The derivative at zero is not zero. It's undefined. To see whether it's continuously differentiable you have to look at limits. Try limit du/dx as x->0+ and x->0- and y=0. To find other functions that ARE continuously differentiable, write any function that is continuously differentiable at 0 in x-y coordinates and convert to polar coordinates. Like u=x^2+y^2=r^2.
 
  • #16
oh right i see, we have to take the limit of the derivative.
ok so say u=r^2=x^2+y^2
then du/dx=2x and the limit of 2x as x tends to 0 is 0 so it is continuously differentiable...right?
 
  • #17
sara_87 said:
oh right i see, we have to take the limit of the derivative.
ok so say u=r^2=x^2+y^2
then du/dx=2x and the limit of 2x as x tends to 0 is 0 so it is continuously differentiable...right?

Right. u=r^2 is continuously differentiable. u=r is not.
 
  • #18
thank you.
 

What does it mean for a function to be continuously differentiable?

Continuously differentiable means that a function has a continuous derivative, meaning that the slope of the function is defined at every point in its domain.

How is continuously differentiable different from just differentiable?

Being continuously differentiable is a stronger condition than just being differentiable. A function can be differentiable, meaning it has a derivative at every point, but not necessarily have a continuous derivative.

What is the difference between continuously differentiable and infinitely differentiable?

Continuously differentiable means that the function has a continuous derivative, while infinitely differentiable means that the function has derivatives of all orders, meaning its derivatives are also differentiable.

Can a function be continuously differentiable but not differentiable?

No, a function cannot be continuously differentiable and not differentiable. If a function is continuously differentiable, it must also be differentiable at every point in its domain.

Why is continuously differentiable important in mathematics?

Continuously differentiable functions are important in mathematics because they are used to model many real-world phenomena, and they have many useful properties for solving problems in calculus and other areas of mathematics.

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