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  • 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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  1. JoshDanziger23
    • one year ago
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    jamesond, 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 \[z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).\]Rearranging and collecting all the constant terms into \(k\) gives \[f_x(x_0,y_0)x+f_y(x_0,y_0)y-z=k.\]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

  2. anonymous
    • one year ago
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    Oh I see now! Thanks for pointing me to it!

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