Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :).

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.