Question

The temperature at a point $(x,y,z)$ in the sphere $x^2+y^2+z^2\le 1$ is given by $T=y^2+xz$. Find the largest and smallest values which $T$ takes (a) on the circle $y=0,x^2+z^2=1$, (b) on the surface $x^2+y^2+z^2=1$, (c) in the whole sphere.

Short answer

The maximum and minimum temperatures can be found by evaluating critical points under given constraints and checking boundary conditions.

Step-by-step solution

- Understanding the Problem

We need to find the maximum and minimum values of the function given by the temperature equation: $T=y^2+xz$. This must be done under certain constraints provided by the problem for parts (a), (b), and (c).

- Part (a): Temperature on the Circle

Given the constraint $y=0$ and $x^2+z^2=1$, we substitute $y=0$ into the temperature equation: $T=0+xz=xz$. On the circle $x^2+z^2=1$, both $x$ and $z$ vary but are bounded by the unit circle. The maximum value of $xz$ will occur at the maximum product of two numbers whose squares add up to 1. We can use the relationship $z=\sqrt{1-x^2}$ or $x=\sqrt{1-z^2}$. Substituting $z=\sqrt{1-x^2}$ into $T=xz$: $$T=x\sqrt{1-x^2}$$ The maximum and minimum can be found by taking derivatives or evaluating critical points.

- Calculating Extrema on the Circle

We find the derivative of $T=x\sqrt{1-x^2}$: $$\frac{dT}{dx}=\sqrt{1-x^2}-x\frac{x}{\sqrt{1-x^2}}=\sqrt{1-x^2}-x^2\frac{1}{\sqrt{1-x^2}}$$ Set the derivative to zero to find critical points: $$\sqrt{1-x^2}=\frac{x^2}{\sqrt{1-x^2}}$$ Solving gives potential values for $x$ corresponding to maximum/minimum temperatures. These must meet the circle condition $x^2+z^2=1$. Substitute back to evaluate $T$.

- Part (b): Temperature on the Surface

Here, not only $y$ but also $x$ and $z$ vary, all satisfying $x^2+y^2+z^2=1$. Using Lagrange multipliers: we set up: $$L(x,y,z,\lambda )=y^2+xz-\lambda (x^2+y^2+z^2-1)$$ Calculate partial derivatives and set them to zero: $$\frac{\partial L}{\partial x}=z-2x\lambda =0\frac{\partial L}{\partial y}=2y-2y\lambda =0$$$$\frac{\partial L}{\partial z}=x-2z\lambda =0\frac{\partial L}{\partial \lambda }=-(x^2+y^2+z^2-1)=0$$ Solve these equations for critical points and evaluate $T$.

- Substituting Critical Points

From the previous step's system, identify solutions for $(x,y,z)$ satisfying the constraint. Substitute these points back into the temperature function $T=y^2+xz$ to determine maximum and minimum values on the sphere's surface.

- Part (c): Temperature in the Whole Sphere

Check interior critical points using methods such as setting partial derivatives of $T=y^2+xz$ to zero. Also, ensure that points lie within the sphere. Use boundary analysis from previous parts.

- Final Values

Compare the extremum values from all parts (interior and boundary) to determine the overall maximum and minimum temperatures in each specified region (circle, surface, sphere).

Key concepts

constraint optimization

Constraint optimization deals with finding the optimal values of a function within a defined restriction or boundary. This often involves using the method of Lagrange multipliers, where the constraint is incorporated into the function. By solving the resulting system of equations, we find points that satisfy both the function's optimization and the constraint. This method is widely applied in real-world problems where restrictions are common.