Understanding converting a vector field to cartesian coords

shemer77
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Homework Statement


Here is the problem and solution but I am confused as to part B
http://gyazo.com/e77d05fc67cb6ac266ff021ef88052dc


The Attempt at a Solution


I understand the first part, but I am totally lost on how they reached their cartesian answer for part B. Firstly why did they do what they did, and secondly where did 36 degrees come from?

I feel like their is some sort of equation or something I am not understanding.
 
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shemer77 said:

Homework Statement


Here is the problem and solution but I am confused as to part B
http://gyazo.com/e77d05fc67cb6ac266ff021ef88052dc

The Attempt at a Solution


I understand the first part, but I am totally lost on how they reached their cartesian answer for part B. Firstly why did they do what they did, and secondly where did 36 degrees come from?

I feel like their is some sort of equation or something I am not understanding.

36 degrees is 0.2π. It's the angle the gave you for phi. Does that help? Other than that they are just using that if u is a unit vector then the component of D along u is (D.u)u.
 
Hmm ok but why does he have .5 twice as in .5(ap.ax)ax +.5(ap.ax)ay?
 
shemer77 said:
Hmm ok but why does he have .5 twice as in .5(ap.ax)ax +.5(ap.ax)ay?

One term finds the ax component of 0.5ap and the other finds the ay. Are you sure you understood the first part?
 
hmm okay thanks I think I figured it out. All I did was use the equations x = p*cosphi, y=p*sinphi and plugged those into the original equation which was already in cartesian for me.
 
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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