Suppose that the Celsius temperature at the point (x comma y comma z )on the sphere x squared plus y squared plus z squared equals 1 is Upper T equals 500 xyz squared. Locate the highest and lowest temperatures on the sphere.

Respuesta :

Answer:

a) 62.5

b) -62.5

Step-by-step explanation:

Given:

x² + y² + z² = 1 is T = 500xyz²

Required:

Locate the highest and lowest temperatures.

Here,

S = (x² + y² + z² = 1)

T = 500 xyz²

Let's take Lagrange multiplier:

∇T = λ∇S

Thus,

500 yz² = λ(2x)........... 1

500 xz² = λ(2y).............2

500 xy² = λ(2z)..............3

Where, x² = y², 2y² = z²

Therefore,

x²+ y²+z² = 1

=> x² + x² + 2x² = 1

Solve for x:

x = ±½ = y

z = ±[tex] \frac{1}{\sqrt{2}} [/tex]

From the above calculations,

x, y, z = ±½, ±½, ±[tex]\frac{1}{\sqrt{2}} [/tex]

Calculate for the highest temperature:

Tmax = T(½, ½, [tex]\frac{1}{\sqrt{2}}) [/tex] = 500 * ½ * ½ * ½ = [tex] \frac{500}{8} = 62.5 [/tex]

Calculate for the lowest temperature:

Tmin = T(-½, ½, [tex]\frac{1}{\sqrt{2}}) [/tex] = 500 * -½ * ½ * ½ = [tex] - \frac{500}{8} = -62.5[/tex]

The highest temperatures is 62.5 and lowest temperatures is -62.5 on the sphere

Given-

The given equation is,

[tex]x^2+y^2+z^2=1[/tex]

The equation for the temperature is,

[tex]T=500(xyz)^2[/tex]

By the Lagrange multiplier

[tex]\bigtriangledown T=\lambda \bigtriangledown S[/tex]

we can rewrite the  temperature equation by the above multiplier as,

[tex]500yz^2=\lambda (2x)[/tex]

[tex]500xz^2=\lambda (2y)[/tex]

[tex]500xy^2=\lambda (2z)[/tex]

Here let,

[tex]x^2=y^2[/tex]

[tex]2y^2=z^2[/tex]

therefore,

[tex]2x^2=z^2[/tex]

Put this values in the given equation of the question we get,

[tex]x^2+x^2+2x^2=1[/tex]

[tex]4x^2=1[/tex]

[tex]x^2=\dfrac{1}{4}[/tex]

[tex]x= \pm\dfrac{1}{2}[/tex]

Therefore,

[tex]y= \pm\dfrac{1}{2}[/tex]

[tex]z= \pm\dfrac{1}{\sqrt{2} }[/tex]

Use positive value for the highest temperatures and negative values for the lowest value of temperatures in the temperature equation.

Highest temperature is,

[tex]T=500(xyz)^2[/tex]

[tex]T_h=500(\dfrac{1}{2} \times\dfrac{1}{2}\times\dfrac{1}{2} )[/tex]

[tex]T_h=\dfrac{500}{8}=62.5[/tex]

Lowest temperature is,

[tex]T_l=500(\dfrac{-1}{2} \times\dfrac{-1}{2}\times\dfrac{-1}{2} )[/tex]

[tex]T_l=-62.5[/tex]

Hence, the highest temperatures is 62.5 and lowest temperatures is -62.5 on the sphere

For more about the temperatures follow the link given below-

https://brainly.com/question/15267055