Directional Derivative: Hill problem

[helenli89]

Please see attachment for Question.
I have couple of questions about his problem.
First, what does the "due" west mean? away from the west or towards the west?
Second, how do I know if it's asending or desending? My attempt of looking at this is that if the directional derivative or the rate of change of the height of the hill is positive, it's asending; negative, it's desending. But I'm not sure if I'm correct or not. So please tell me if I'm wrong.
Thank you.

PS. My answer for b is: Northeast is in the direction of (1,1) = u. ? Dz(x) = ?z(x) . u where?z(x)=(-0.04x , -0.08y) -> ?z(-20, 5) . (1,1) = (0.08, -0.04) . (1,1) = 0.8-0.4 =0.4. She's asending at the rate of 0.4.
My answer for c is: She can traval along the level curve to be at a level path, which is that the change of the height is zero. Since the function is differenciable, by theorem, ?z(x) . y = 0 where y is a vector that's perpendicular to the gradien and y is the tangent line to the level curve. So one solution for y is (5, 20).

 

[galactus]

Find your gradient by finding \(\displaystyle f_{x}, \;\ f_{y}\)

The directional derivative is given by \(\displaystyle D_{u}f(x,y)={\nabla}f(x,y)\cdot u\)

where u is a unit vector. The above is the dot product of the gradient of f with a unit vector. This gives us the directional derivative of f in the direction of u.

We get:

\(\displaystyle {\nabla}(z)=-4xi-8yj\). If x=-20 and y=5, then we have \(\displaystyle {\nabla}(z)=80i-40j\)

a: To point due west means to head directly west. The direction vector would be u=-i. \(\displaystyle D_{u}z=[80,-40]\cdot [-1,0]=-80\)

Therefore, the climber will go down because z is going down.

b. \(\displaystyle \frac{i+j}{\sqrt{2}}\) points northeast. Find \(\displaystyle D_{u}z\), the directional derivative in the direction of northeast. This will give the rate of ascent or descent in the xy plane.

c. The climber will go along a level path that will be perpendicular to \(\displaystyle 80i-40j\). Using the unit vectors, we have \(\displaystyle \frac{(i+2j)}{\sqrt{5}}\)

forms an angle \(\displaystyle cos^{-1}(\frac{2}{\sqrt{5}})=26.565 \;\ degrees\) with the positive y-axis.

Therefore, hence, and thus, \(\displaystyle \frac{-(i+2j)}{\sqrt{5}}\) makes the same angle in the negative y-axis.

What would be the bearing?.

 

[BigGlenntheHeavy]

\(\displaystyle galactus, \ for \ a, \ I \ get \ {\nabla}(z) \ = \ .8i-.4j \ and \ D_u z \ = \ -.8\)

\(\displaystyle Am I \ wrong?\)

 

[galactus]

Oops, you are correct. I used 2 and 4. Oh well, divide by 100. The process is the same.

Sorry about that.

 

[helenli89]

Thanks for the reply. But I think it's better to just simply use the [1,1] and [-1,0] vectors instead of i, j.
And you didn't answer question b. It asks if it's desending pt asending. My answer is that since the rate change of the height is 0.4, a positive number, therefore it's asending.
I also got -0.8 for question a. And since it's negative, I say it's desending. The solution is going to posted soon. I'll post it aa soon I get a chance.

 

[galactus]

I didn't asnwer b because I wasn't going to do it all. Yes, fior part b, the climber is ascending.

 

[helenli89]

I didn't asnwer b because I wasn't going to do it all. Yes, fior part b, the climber is ascending.

It's a bit confusing when you don't say that 'from here you can find the answer' or something like this. But I guess we all need to working on our answers and questions. Anyways, here is the answer to this question.