Home»Math Guides»Finding Directional Derivatives, Temperature Applied Problem
Directional Derivatives – Temperature Rate of Change moving from one point to another point
Example problem: Find the rate of change of temperature at the point (2,-1,2) in the direction towards the point (3,-3,3). The temperature at any point (x,y,z) is given by the equation: \(T(x,y,z) = 200x{e^{ – {x^2} – 3{y^2} – 9{z^2}}}\) where T is in \(^OC\) and x,y,z are in meters.
This is just asking for a directional derivative, \[{D_u}T(2, – 1,2)\]
It may be possible to find the angle between the two points here, but we can also just find the dot products of the gradients, which will be easier for this example (finding angles between two points in 3D space can be tricky to visualize).
First find the unit vector connecting (2,-1,2) to (3,-3,3). It’s just the vector connecting the points divided by the magnitude.
\[\begin{array}{l}\left\langle {3 – 2, – 3 + 1,3 – 2} \right\rangle \\ = \left\langle {1, – 2,1} \right\rangle \end{array}\]
This is a vector, now divide by the magnitude.
The magnitude is just the square root of the sum of all components squared, see below:
\[\frac{{\left\langle {1, – 2,1} \right\rangle }}{{\sqrt {{{(1)}^2} + {{( – 2)}^2} + {{(1)}^2}} }}\]
\[ = \frac{{\left\langle {1, – 2,1} \right\rangle }}{{\sqrt 6 }}\]
And we need all the partial derivatives.
We should also substitute the point (2,-1,2) into the partial derivatives.
\[\frac{{\partial T}}{{\partial x}} = – 400x{e^{ – {x^2} – 3{y^2} – 9{z^2}}}\]
\[\frac{{\partial T}}{{\partial x}}(2, – 1,2) = – 800{e^{ – 43}}\]
\[\frac{{\partial T}}{{\partial y}} = – 1200y{e^{ – {x^2} – 3{y^2} – 9{z^2}}}\]
\[\frac{{\partial T}}{{\partial y}} = 1200{e^{ – 43}}\]
\[\frac{{\partial T}}{{\partial z}} = – 3600z{e^{ – {x^2} – 3{y^2} – 9{z^2}}}\]
\[\frac{{\partial T}}{{\partial z}} = – 7200{e^{ – 43}}\]
And the directional derivative is just a dot product of the temperature gradient and the unit vector:
\[{D_u}T(2, – 1,2) = \nabla T(2, – 1,2) \cdot \frac{{\left\langle {1, – 2,1} \right\rangle }}{{\sqrt 6 }}\]
\[{D_u}T(2, – 1,2) = \left\langle { – 800{e^{ – 43}},1200{e^{ – 43}}, – 7200{e^{ – 43}}} \right\rangle \cdot \frac{{\left\langle {1, – 2,1} \right\rangle }}{{\sqrt 6 }}\]
\[{D_u}T(2, – 1,2) = ( – 800 – 2400 – 7200)\frac{{{e^{ – 43}}}}{{\sqrt 6 }}\]
\[{D_u}T(2, – 1,2) = \frac{{ – 10400{e^{ – 43}}}}{{\sqrt 6 }}\]
And this is a rate of change of temperature, so the units will be
\[\frac{{^OC}}{m}\]
Try some examples involving finding partial derivatives of multivariable functions!