If your answer to the question posed at the end of the Section 13.3 #40 example related to the directional derivative, or if you have not yet heard about directional derivatives but your answer somehow related to a derivative as a rate of change relative to change along the line y = (4/3)x or change relative to change in the direction of a vector, say <3,4>, then good show.  Here is the DPGraph Picture again that the question related to.  Maple Picture

Here are some notes on the directional derivative and more examples involving the gradient and the directional derivative..

Maximizing the Directional Derivative

DPGraph Picture of the Surface     DPGraph Picture with the Point

DPGraph Picture of the surface, the point, and a plane containing the point and a line in the xy-coordinate plane with direction vector u where the x- and y-components of u equal the x- and y-components of the direction vector for the directional derivative.  See the analysis below for more details.

Find the maximum value of the directional derivative of the given function at the indicated point.

Animation   QT Version

Heat Seeking Particle Examples

T(x,y) = 100 - x² - 2y²        Starting Point (2,4)

Let the path be described by  r(t) = < x(t) , y(t) >

Then  r'(t) = < dx/dt , dy/dt >  gives the direction of motion.

Since this is a heat seeking particle it must move at all times in the direction of maximum increase in temperature.  This must be the direction of the gradient of the temperature function T(x,y).

grad T(x,y) = < -2x , -4y >          Powerpoint Enhancement

Click here to see a Maple worksheet on this example.

Click on the picture to see a larger picture of the analytical solution constrained so that the path looks orthogonal to the equal temperature lines.

dx/dt = -2x       x(0) = 2

(1/x)dx = -2dt

ln | x | = -2t + c

| x | = e^{-2t + c} = e^-2t e^c = c₁e^-2t

x = c₂e^-2t

x = 2e^-2t   since x(0) = 2 yields c₂ = 2

In a similar fashion we would find that from

dy/dt = -4y and y(0) = 4 we get

y = 4e^-4t

Thus the path is given by r(t) = < 2e^-2t , 4e^-4t >.

Since y = (2e^-2t)² we see that the path is also given by y = x².

The picture on the right shows the path.  Animation

In general then it must be the case that grad T = k r'(t) where k is some constant.  Let us assume a parameterization of the path such that k = 1 so that grad T = r'(t).  Note that we are not claiming that r'(t) gives the velocity of the heat seeking particle, only that it gives the direction.

From grad T = r'(t) we equate vector components to get 

dx/dt = -3, dy/dt = -1, and dz/dt = -2z.  

Let 0 be the value of t at the starting point so that 

x(0) = 2, y(0) = 2, and z(0) = 5.  

We have three differential equations to solve and their solutions 

will yield x = -3t + 2, y = -t + 2, and z = 5e^-2t.

Thus r(t) = < -3t + 2 , -t + 2 , 5e^-2t >

Click here to see an animated picture of a portion of the path.

Click here to see a Maple worksheet on the problems above.