Partial Derivates - Chain Rule

[sikrut]

Homework Statement

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

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

Using the given parametrization and the chain rule, find the derivative \frac{\partial T}{\partial t}. 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 (x,y)=(__,__)

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

Homework Equations

\frac{\partial T}{\partial t} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}

The Attempt at a Solution

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

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

 

[SammyS]

Homework Statement

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

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

Using the given parametrization and the chain rule, find the derivative \frac{\partial T}{\partial t}. 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 (x,y)=(__,__)

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

Homework Equations

\frac{\partial T}{\partial t} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}

The Attempt at a Solution

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

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 !

 

[sikrut]

Ah, yes. That would be the logical thing to do wouldn't it?

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

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