# directional derivatives

• Oct 21st 2009, 09:13 PM
zpwnchen
directional derivatives
Suppose that you are climbing a hill whose shape is given by http://webwork.asu.edu/webwork2_file...719c509491.png, and that you are at the point (90, 40, 300).
1) In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? Answer : $\displaystyle <-12.6/ \sqrt{174.76},-4/ \sqrt{174.76}>$

2) If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?

How to do 2) ?
• Oct 22nd 2009, 09:18 AM
galactus
First, find your derivatives:

$\displaystyle f_{x}=-.14x, \;\ f_{y}=-.10y$

Sub in your given coordinates.

$\displaystyle \nabla f(90,40)=-12.6i-4j$

The max value of the directional derivative is $\displaystyle ||{\nabla}f(90,40)||=\sqrt{(\frac{-63}{5})^{2}+(-4)^{2}}=\frac{\sqrt{4369}}{5}\approx 13.22$

A unit vector in this direction is:

$\displaystyle \frac{-12.6}{13.22}i-\frac{4}{13.22}j$

This is in decimal form, but is equivalent to your answer.

$\displaystyle \sqrt{174.76}=13.22$
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