# Partial Derivates - Chain Rule

## 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?

Last edited:

SammyS
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## 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 !

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. :grumpy:
Stupid mistakes..

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