Question

Heat-Seeking Path In Exercises 57 and \(58,\) find the path of a heat-seeking particle placed at point \(P\) on a metal plate with a temperature field \(T(x, y)\). $$ T(x, y)=400-2 x^{2}-y^{2}, \quad P(10,10) $$

Short answer

The path of the heat-seeking particle initially at point \((10,10)\) on the metal plate will proceed in the direction of the vector \((-40, -20)\).

Step-by-step solution

Solve for the gradient of the temperature field

First, obtain the gradient of the function \(T(x,y)\). The gradient is a vector that gives the direction of the greatest rate of increase of the function. The gradient of a function \(f(x, y)\) is given by \(\nabla f = (df/dx, df/dy)\). For the function \(T(x,y) = 400 - 2x^2 - y^2\), this gives us \(\nabla T = (-4x, -2y)\).

Compute the gradient at the initial point

Now, calculate the gradient of \(T\) at the point \((10,10)\) by inserting the coordinates of the point into the gradient function: \(\nabla T(10,10) = (-4*10, -2*10) = (-40, -20)\). Therefore, the path the particle will follow at the initial point will be towards the direction of the vector \((-40, -20)\), which indicates the steepest ascent in the temperature field at the point \((10,10)\).

Construct the directional path

This means the initial path of the heat-seeking particle once placed on the metal plate at point \((10,10)\) will be in the direction of the vector \((-40, -20)\). However, the complete path is more complex and will involve solving a system of differential equations, as it will continuously shift in the direction of greatest ascent. Thus, \((-40, -20)\) only gives us the initial direction of movement for the heat-seeking particle.