Question

Temperature on a circle Let T=f(x,y) be the temperature at the point (x,y) on the circle x=cost,y=sint,0≤t≤2π and suppose that

∂x/∂T​=8x−4y,∂T/∂y​=8y−4x.

Find where the maximum and minimum temperatures on the circle occur by examining the derivatives dT/dt and d²T/dt².

Answer 1

To find the maximum and minimum temperatures on the given circle, we need to examine the derivatives dT/dt and d²T/dt². By evaluating the derivatives and substituting the values of x and y, we can find the maximum and minimum temperatures.

Step-by-step explanation:

To find the maximum and minimum temperatures on the given circle, we need to examine the derivatives \(\frac{{dT}}{{dt}}\) and \(\frac{{d^2T}}{{dt^2}}\). Since the circle is defined parametrically as \(x = \cos(t)\) and \(y = \sin(t)\), we can express the temperature \(T\) as a function of \(t\) by substituting \(x\) and \(y\) in the given partial derivative equations:

\(\frac{{dT}}{{dt}} = \frac{{\partial T}}{{\partial x}} \cdot \frac{{dx}}{{dt}} + \frac{{\partial T}}{{\partial y}} \cdot \frac{{dy}}{{dt}}\)

\(\frac{{d^2T}}{{dt^2}} = \frac{{\partial^2 T}}{{\partial x^2}} \cdot \left(\frac{{dx}}{{dt}}\right)^2 + 2 \cdot \frac{{\partial^2 T}}{{\partial x \partial y}} \cdot \frac{{dx}}{{dt}} \cdot \frac{{dy}}{{dt}} + \frac{{\partial^2 T}}{{\partial y^2}} \cdot \left(\frac{{dy}}{{dt}}\right)^2\)

By evaluating the derivatives and substituting the values of \(x\) and \(y\), we can find the maximum and minimum temperatures.