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anonymous
 one year ago
Can someone help with Pset 4, Part II, problem #2? On their answer for (b.), I can't work out where they're getting their Z coordinate for the normal vectors to both planes. For both, it is 1, but I have no idea why! Can anyone explain?
anonymous
 one year ago
Can someone help with Pset 4, Part II, problem #2? On their answer for (b.), I can't work out where they're getting their Z coordinate for the normal vectors to both planes. For both, it is 1, but I have no idea why! Can anyone explain?

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JoshDanziger23
 one year ago
Best ResponseYou've already chosen the best response.2jamesond, I didn't see this explained in the course but what I think is going on is this: You have \(z=f(x,y)\) so \(dz=f_x dx+f_y dy\), and so the tangent at \(P=(x_0,y_0)\) satisfies \[zz_0=f_x(x_0,y_0)(xx_0)+f_y(x_0,y_0)(yy_0).\]Rearranging and collecting all the constant terms into \(k\) gives \[f_x(x_0,y_0)x+f_y(x_0,y_0)yz=k.\]This is now the standard form for the equation for a plane (ie, \(ax+by+cz=k\)) with normal vector \(<f_x(x_0,y_0),f_y(x_0,y_0),1>\), which is the result being relied on the Pset. Best wishes Josh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see now! Thanks for pointing me to it!
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