Locate the maximum and minimum temperatures on the ellipse by examining dT/dt and d2T/dt2. Suppose that T = xy - 2. Find the maximum and minimum values of T on the ellipse.

Steps to Solve

1. Calculate ( \frac{dT}{dt} )

Using the parametric equations of the ellipse ( x = 2\sqrt{2} \cos t ) and ( y = \sqrt{2} \sin t ), we can find ( T ) implicitly defined by ( \frac{dT}{dt} = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt} ).

Given ( \frac{\partial T}{\partial x} = y ) and ( \frac{\partial T}{\partial y} = x ), we first compute ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ):

[ \frac{dx}{dt} = -2\sqrt{2} \sin t ] [ \frac{dy}{dt} = \sqrt{2} \cos t ]

Substituting the derivatives and ( \frac{\partial T}{\partial x} ) and ( \frac{\partial T}{\partial y} ):

$$ \frac{dT}{dt} = y \cdot (-2\sqrt{2} \sin t) + x \cdot (\sqrt{2} \cos t) $$

2. Substitute ( x ) and ( y )

Now, substitute ( x ) and ( y ) in the equation:

[ y = \sqrt{2} \sin t \quad \text{and} \quad x = 2\sqrt{2} \cos t ]

Then,

$$ \frac{dT}{dt} = (\sqrt{2} \sin t)(-2\sqrt{2} \sin t) + (2\sqrt{2} \cos t)(\sqrt{2} \cos t) $$

Simplifying gives:

$$ \frac{dT}{dt} = -4 \sin^2 t + 2\cdot2 \cos^2 t = -4 \sin^2 t + 4 \cos^2 t $$

3. Set ( \frac{dT}{dt} = 0 )

To find critical points, set the derivative equal to zero:

$$ -4 \sin^2 t + 4 \cos^2 t = 0 $$

This simplifies to:

$$ \sin^2 t = \cos^2 t $$

Thus,

$$ \tan^2 t = 1 \quad \Rightarrow \quad t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} $$

4. Calculate ( \frac{d^2 T}{dt^2} )

Next, compute the second derivative to identify the nature of these critical points. First, differentiate ( \frac{dT}{dt} ):

$$ \frac{d^2T}{dt^2} = \frac{d}{dt}(-4 \sin^2 t + 4 \cos^2 t) = -8 \sin t \cos t $$

5. Evaluate ( \frac{d^2 T}{dt^2} ) at critical points

Evaluate ( \frac{d^2T}{dt^2} ) at each critical point:

• For ( t = \frac{\pi}{4} ) and ( t = \frac{5\pi}{4} ):

$$ \frac{d^2T}{dt^2} = -8 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = -8 \cdot \frac{1}{2} = -4 \quad (\text{local max}) $$

• For ( t = \frac{3\pi}{4} ) and ( t = \frac{7\pi}{4} ):

$$ \frac{d^2T}{dt^2} = -8 \cdot \frac{\sqrt{2}}{2} \cdot -\frac{\sqrt{2}}{2} = 8 \cdot \frac{1}{2} = 4 \quad (\text{local min}) $$

6. Find temperatures ( T ) at critical points

Now, substitute back to find temperatures at critical points:

• At ( t = \frac{\pi}{4} ) and ( t = \frac{5\pi}{4} ):

$$ T = xy - 2 = (2\sqrt{2} \cdot \sqrt{2}) - 2 = 4 - 2 = 2 \text{ (max)} $$

• At ( t = \frac{3\pi}{4} ) and ( t = \frac{7\pi}{4} ):

$$ T = xy - 2 = (2\sqrt{2} \cdot -\sqrt{2}) - 2 = -4 - 2 = -6 \text{ (min)} $$