Suppose that the temperature on the disk x^2+y^2</=1 is given by T=x^2=2y^2=y, i

Question

Suppose that the temperature on the disk x^2+y^2</=1 is given by T=x^2=2y^2=y, indegress Celsius. Determine the hottest and coldest places on the disk, and the places they occur.

Interior

Tx=2x

Ty=4y+1

So x=0 y=-1/4

TxxTyy-Txy^2
(2)(4)-0^2= 8

Txx>0
Txy>0 so, local min of T= -1/8 at (0,-1/4)

Boundary x^2+y^2=1, so y^2=(1-x^2)
T=x^2+2y+y
T=1-y^2+2y^2+y
T=y^2+y+1 on [-1,1] for y
T'= 2y+1
y=-1/2 ; x= Sqrt(3)/2  T= 3/4 at this point.

Now I am confused on how to know the absoulte warmest temperature and spot. I believe the coldest temperate is -1/8 at (0,-1/4).
Please provide a direct response rather than an example. I have done most of the work here. Thank you

Explanation / Answer

For the boundary, transform to polar. T=x^2+2y^2+y

T = cos^2 theta + 2 sin^2 theta + sin theta

T = 1 + sin^2 theta + sin theta

Now differentiate

2 sin theta cos theta + cos theta = 0.

sin (2 theta) + cos (theta) = 0.

Solve for theta. theta is pi/2.

x=0, y=1.

T=2+1=3

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