Directional derivative question

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attachment.php?attachmentid=262056&d=1389451604.jpg



I've done the first part, but I'm stuck on the second paragraph of the question. Maybe I'm being stupid, I don't even understand exactly what is meant by, 'the level curve'.

I also don't quite understand the whole concept of directional derivative. When it says, 'the gradient in the direction making an angle A with the x-axis, how should I think of this? what does the gradient mean in this context?

because when its just a simple curve/line on an xy axis, I know what the 'gradient' means, it is literally how much 'y' changes per unit x along that curve/line. But when I'm told about a 'gradient in a direction', I'm confused.
 
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hey, question dude! :smile:
question dude said:
… what is meant by, 'the level curve'.

think of the 3D graph, z = f(x,y)

you can make a 2D contour map showing the lines of equal height

those contours are the level curves :wink:
I also don't quite understand the whole concept of directional derivative. When it says, 'the gradient in the direction making an angle A with the x-axis, how should I think of this? what does the gradient mean in this context?

the directional derivative in the direction (cosθ,sinθ) is df(kcosθ,ksinθ)/dk

it's the rate at which f increases if you go along the line y/x = tanθ :smile:
 
question dude said:
attachment.php?attachmentid=262056&d=1389451604.jpg



I've done the first part, but I'm stuck on the second paragraph of the question. Maybe I'm being stupid, I don't even understand exactly what is meant by, 'the level curve'.

I also don't quite understand the whole concept of directional derivative. When it says, 'the gradient in the direction making an angle A with the x-axis, how should I think of this? what does the gradient mean in this context?

because when its just a simple curve/line on an xy axis, I know what the 'gradient' means, it is literally how much 'y' changes per unit x along that curve/line. But when I'm told about a 'gradient in a direction', I'm confused.

You are supposed to show your work, confused or not (those are PF rules). I don't see how you can have done the first part (which involves directional derivatives) but then claim you do not understand directional derivatives. I would need to see your work in order to grasp what is going on.
 
tiny-tim said:
hey, question dude! :smile:


think of the 3D graph, z = f(x,y)

you can make a 2D contour map showing the lines of equal height

those contours are the level curves :wink:


the directional derivative in the direction (cosθ,sinθ) is df(kcosθ,ksinθ)/dk

it's the rate at which f increases if you go along the line y/x = tanθ :smile:

Thanks a lot! I understand what's going on now. Before I was just plugging in the numbers without knowing the concepts properly.
 
Ray Vickson said:
You are supposed to show your work, confused or not (those are PF rules). I don't see how you can have done the first part (which involves directional derivatives) but then claim you do not understand directional derivatives. I would need to see your work in order to grasp what is going on.

Sorry about that, what I meant to get across, was that I understood the method in a sort of algorithmic way, but didn't understood the reason behind steps due to not understanding some basic stuff like 'level curve'. Anyway its sorted now.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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