Question

Let T = f(x, y) be the temperature at the point (r, y) on the circle r = cost, y-inf.0≤ t≤2 and suppose that

ar 8-4y 8y-42

. Find where the maximum and minimum temperatures on the circle occur by examining the derivatives dT> dt and dT/dt

r=4x^ 2 -4xy+4y^ 2 . . Suppose that Find the maximum and minimum values of T7 on the circle

Answer 1

$\frac{\partial }{\partial x}\left(T\right)=8x-4y\\ \frac{\partial }{\partial y}\left(T\right)=8y-4x\\ x = \cos t \\ y= \sin t\\ x^2+y^2=1\\ \sin ^2\left(x\right)+\cos ^2\left(x\right)=1\\ g = x^2 +y^2\\ d∇g = 2x i +2yj\\ =∇+= \lambda ∇ g \\ ∇+= (8x-4y)i + (8y -4x)j \\ ∇h= (8x-4y)i+(8y-4x)j + \lambda (2x i +2y j)\\ ∇h=0\\ 8 x-4y + 2x \lambda = 0....(1)\\ dy -4x + 2y \lambda = 0.....(2)\\ 2x λ=4y-8x\\ λ= \frac{4y-8x}{2x}\\ 2x λ+ 8x = 4y\\ x(2 λ+8)=4y\\ x= \frac{4y}{2λ+8}\\ Aslo\\ 2y λ+8y = 4x\\ y(2λ+8) =4x\\ y= \frac{4x}{2λ+8}\\ \left(\frac{4y}{2λ+8}\right)^2+\left(\frac{4x}{2λ+8}\right)^2=1\\ \left(\frac{1}{2λ+8}\right)^2\left(16y^2+16x^2\right)=1\\ Thus \\ \left(\frac{4}{2λ+8}\right)^2=1\\ \frac{4}{\left(λ+4\right)^2}=1\\ 4=\left(λ+4\right)^2\\ λ=-2,\:λ=-6\\ When \space λ= -2, x = y \\ When \space λ= -6, x = -y \\ \implies x = ± \frac{1}{\sqrt{2}}\\ Maxima \implies x = - \frac{1}{\sqrt{2}}, y = \frac{1}{\sqrt{2}}\\ Manima \implies x = \frac{1}{\sqrt{2}}, y = \frac{1}{\sqrt{2}}\\ Min_T= 4(\frac{1}{\sqrt{2}})^2-4(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})+4(\frac{1}{\sqrt{2}})=2\\ Max_T= 4(-\frac{1}{\sqrt{2}})^2-4(\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})+4(\frac{1}{\sqrt{2}})=6$