- #1

sikrut

- 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: