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anonymous
 one year ago
I was working on problem #2 of problem set 4 part b, where you are supposed to find the angle of intersection of 2 surfaces at a particular point. f(x,y)=x^2y^2; g(x,y)=2+(xy)^2. point is (2,1,3). I understand that to do this, I need to find the partial derivatives for the 2 surfaces at that point for each function, and then find the normal to each of the tangent planes. The answer key does this, but it states to find the normal for the first plane, they use N1=(Fx(2,1), Fy(2,1), 1). Where Fx is derivative w/ respect to x; where does this formula come from?
anonymous
 one year ago
I was working on problem #2 of problem set 4 part b, where you are supposed to find the angle of intersection of 2 surfaces at a particular point. f(x,y)=x^2y^2; g(x,y)=2+(xy)^2. point is (2,1,3). I understand that to do this, I need to find the partial derivatives for the 2 surfaces at that point for each function, and then find the normal to each of the tangent planes. The answer key does this, but it states to find the normal for the first plane, they use N1=(Fx(2,1), Fy(2,1), 1). Where Fx is derivative w/ respect to x; where does this formula come from?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0if you have a function given by z = f(x,y), then let \(\Phi = z  f(x,y)\) and you will find that the normal to this surface is given by \(\nabla \phi = <f_x, f_y, 1>\) or \(<f_x, f_y, 1>\) if you point this vector in the other direction. try this for a simple plane, say \(\pi:\) x + 2y  z = 3 we know already that the normal is <1,2,1> but if we just try this out we see that \(\nabla (x + 2y  z  3) = <1,2,1>\) NB to be totally clear, this is not limited to xyz space, you just need the correct formulation of \(\nabla\) for say cylindrical to do it there too. working out why this works requires you to look first at the **directional derivative** and find a way to get that pointed in the right direction.

phi
 one year ago
Best ResponseYou've already chosen the best response.0They introduced an idea that is not discussed until Lecture 12, where they talk about gradients. Without using gradients, you could solve the problem this way: use \[ z  z_0 = f_x(xx_0) + f_y(yy_0) \] with \( (x_0, y_0, z_0) = (2,1,3) \) and \( f_x = 4, f_y= 2 \) you get \[ z 3 = 4(x2)  2(y1) \\ 4x 2y z = 3 \] By inspection, the normal vector is <4, 2, 1>

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0THANK YOU!! Totally makes sense now! : )

baru
 one year ago
Best ResponseYou've already chosen the best response.0If you’re wondering why normal vector is N=<fx, fy, 1> or <fx, fy, 1> (one simply points in the opposite direction as the other…so they’re both normal to the same plane) What is fx? ... its the rate of change of ‘z’ in the ‘x’ direction. In other words, if ‘x’ changes by 1 unit then ‘z’ changes by fx units. So if you think of fx as a vector, then it has the components <1, 0, fx> Similarly fy has components <0, 1, fy> We know that fx and fy together define the tangent plane… to find a normal to this plane: we take their cross product. Which is N= <1, 0, fx> X <0, 1, fy> Evaluate this and see for yourself: you get N=<fx, fy, 1> or <fx, fy, 1>
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