# Directional derivative (muiltivariable calculus)

I have an encountered an example in my text book which I don't fully see the intuition of. I will write out the part of the example I'm struggling with:

A hiker is standing beside a stream on the side of a mountain examining her map of the region. The height of the land (in meters) at any point $(x,y)$ is given by the function

$$h(x,y) = \frac{20000}{3 + x^2 + 2y^2}$$

where $x$ and $y$ (in kilometers) denote the coordinates of the point on the hiker's map. The biker is at the point $(3,2)$

At what angle to the path of the stream (on the map) should the hiker set out if she wishes to climb at a $15^o$ inclination to the horizontal?

ANSWER (from textbook)

Suppose the hiker moves away from $(3,2)$ in the direction of the unit vector $\vec{u}$. She will be ascending at an inclination of $15^o$ if the directional derivative of $h$ in the direction of $\vec{u}$ is $1000 tan(15^{o}) \approx 268$ (The 1000 compensates for the fact that the vertical units are meters while the horizontal units are kilometers.). If $\theta$ is the angle between $\vec{u}$ and the upstream direction, then

$$500cos(\theta) = |\bigtriangledown h(3,2)|cos(\theta) \approx 268$$

OK, so what I have some problems seeing here is is the part where it is stated that we know that the ascending will be at an inclination of $15^{o}$ if the directional derivative of $h$ in the direction of $\vec{u}$ is $1000 tan(15^{o}) \approx 268$

Am I correct when I interpret the $15^{o}$ angle here as the angle between $\bigtriangleup y$ and $\bigtriangleup x$ in the $(x,y)$ plane? If so, why would this angle then give us the desired inclination to the horizontal? Or am I totally off here?

I would truly appreciate it if anyone could make me see this more intuitively :).

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## 1 Answer

Am I correct when I interpret the $15^{o}$ angle here as the angle between $\Delta y$ and $\Delta x$ in the $(x,y)$ plane?

I don't think so. The $\theta=15^{o}$ angle refers to the climbing inclination, so $\tan(\theta) = \frac{\Delta z}{|u|}$ where $u$ is the displacement in the $(x,y)$ plane (so $|u| = \sqrt{\Delta x^2+\Delta y^2}$) and $\Delta z = f[(x,y)+u]-f[(x,y)]$ is the difference in "altitude". And $\frac{\Delta z}{|u|}$ is, when $|u|\to 0$, the directional derivative.

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Thanks a lot! Of course - this makes sense :). I really appreciate it! –  Kristian Jun 18 '12 at 12:26