• Support PF! Buy your school textbooks, materials and every day products Here!

Partial Derivates - Chain Rule

  • Thread starter sikrut
  • Start date
  • #1
49
1

Homework Statement


Parametrize the upper half of the unit circle by [itex] x = cos(t), y = sin(t), [/itex] for [itex]0\leq t \leq\pi[/itex]

Let [itex] T = f(x,y)[/itex] be the temperature at the point [itex](x,y)[/itex] on the upper half of the circle.
Suppose that:
[tex]\frac{\partial T}{\partial x} = 4x - 2y[/tex] [tex] \frac{\partial T}{\partial y} = -2x + 4y [/tex]

Using the given parametrization and the chain rule, find the derivative [itex]\frac{\partial T}{\partial t}[/itex]. Then:

(a) Using methods of single-variable calculus, find where the minimum and maximum temperatures occur on the upper half circle:
When t = [______], and this occurs at the point [itex](x,y)[/itex]=(__,__)

(b) Find the maximum and minimum values of the function: [tex]2x^2 - 2xy + 2y^2[/tex]
on the upper half of the unit circle.

Homework Equations



[tex]\frac{\partial T}{\partial t} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}[/tex]

The Attempt at a Solution



[tex]\frac{\partial T}{\partial t} = (4x - 2y)\frac{dx}{dt} + (-2x + 4y)\frac{dy}{dt}[/tex]
[tex]\frac{\partial T}{\partial t} = (4x - 2y)(-sin(t)) + (-2x + 4y)(cos(t))[/tex]
[tex]\frac{\partial T}{\partial t} = -4xsin(t) + 2ysin(t) - 2xcos(t) + 4ycos(t) [/tex]
[tex]\frac{\partial T}{\partial t} = -4cos(t)sin(t) + 2sin^2(t) - 2cos^2(t) +4sin(t)cos(t)[/tex]
[tex]\frac{\partial T}{\partial t} = 2(sin^2(t) - cos^2(t)) [/tex]
So I found [itex]\frac{dT}{dt}[/itex] like the problem asked. Then, using "single variable calculus," I found the second derivative ([itex]\frac{d^2T}{dt^2}[/itex]) and set it to zero to find the maximum and minimums.
Those turned out to be [itex]\frac{d^2T}{dt^2} = 0 = [0,pi/2,pi][/itex]

I plugged in the values, but the result comes back as in correct. What am I doing wrong?
 
Last edited:

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,303
998

Homework Statement


Parametrize the upper half of the unit circle by [itex] x = cos(t), y = sin(t), [/itex] for [itex]0\leq t \leq\pi[/itex]

Let [itex] T = f(x,y)[/itex] be the temperature at the point [itex](x,y)[/itex] on the upper half of the circle.
Suppose that:
[tex]\frac{\partial T}{\partial x} = 4x - 2y[/tex] [tex] \frac{\partial T}{\partial y} = -2x + 4y [/tex]

Using the given parametrization and the chain rule, find the derivative [itex]\frac{\partial T}{\partial t}[/itex]. Then:

(a) Using methods of single-variable calculus, find where the minimum and maximum temperatures occur on the upper half circle:
When t = [______], and this occurs at the point [itex](x,y)[/itex]=(__,__)

(b) Find the maximum and minimum values of the function: [tex]2x^2 - 2xy + 2y^2[/tex]
on the upper half of the unit circle.

Homework Equations



[tex]\frac{\partial T}{\partial t} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}[/tex]

The Attempt at a Solution



[tex]\frac{\partial T}{\partial t} = (4x - 2y)\frac{dx}{dt} + (-2x + 4y)\frac{dy}{dt}[/tex]
[tex]\frac{\partial T}{\partial t} = (4x - 2y)(-sin(t)) + (-2x + 4y)(cos(t))[/tex]
[tex]\frac{\partial T}{\partial t} = -4xsin(t) + 2ysin(t) - 2xcos(t) + 4ycos(t) [/tex]
[tex]\frac{\partial T}{\partial t} = -4cos(t)sin(t) + 2sin^2(t) - 2cos^2(t) +4sin(t)cos(t)[/tex]
[tex]\frac{\partial T}{\partial t} = 2(sin^2(t) - cos^2(t)) [/tex]
So I found [itex]\frac{dT}{dt}[/itex] like the problem asked. Then, using "single variable calculus," I found the second derivative ([itex]\frac{d^2T}{dt^2}[/itex]) and set it to zero to find the maximum and minimums.
Those turned out to be [itex]\frac{d^2T}{dt^2} = 0 = [0,pi/2,pi][/itex]

I plugged in the values, but the result comes back as in correct. What am I doing wrong?
To find a min/max for temperature, T, set its 1st derivative to zero !
 
  • #3
49
1
Ah, yes. That would be the logical thing to do wouldn't it? :blushing:

Right as I sent that question, I realized where I messed up.
[tex] \frac{d^2T}{dt^2}[/tex] is not the second derivative of [tex]\frac{dT}{dt}[/tex].
I should have derived one more time. :grumpy:
Stupid mistakes..

Thanks for your reply though. I did it both ways just to see and they both work.
 

Related Threads on Partial Derivates - Chain Rule

  • Last Post
Replies
4
Views
11K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
733
  • Last Post
Replies
1
Views
786
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
3K
Top