Hi everyone, Just wanna know how does the the unit vector become in that form: [itex]\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{x \vec{i}+y \vec{j}}{4}[/itex]
As far as I know, the unit vector or the normal vector is the vector divided by its magnitude. But that's not what I need to know, what I need to know is the manipulation that occurred. [itex]\vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{2(x \vec{i}+y\vec{j})}{\sqrt{4(x^{2}+y^{2}})}=\frac{2(x \vec{i}+y\vec{j})}{2\sqrt{(x^{2}+y^{2}})}=\frac{x \vec{i}+y\vec{j}}{\sqrt{x^{2}+y^{2}}}[/itex] That's my best. :Z
But how did it end up like this form: [itex]\frac{x \vec{i}+y \vec{j}}{4}[/itex] And I've found something similar in Thomas Calculus: Is [itex]y^{2} + z^{2}[/itex] equal to 1 or something? much like [itex]sin^{2}\theta + cos^{2}\theta = 1 [/itex]
Okay, but clearly that isn't where the gradient in the original post came from. So if you want to know what happened in post #3 (why x^{2} + y^{2} = 1) then you need to state the original problem.
Sorry, that's not the correct surface, but the surface is [itex]x^{2}+y^{2}=16[/itex]. But I think I've got the idea: [itex]\vec{n}=\frac{x\vec{i}+y \vec{j}}{\sqrt{x^{2}+y^{2}}}=\frac{x\vec{i}+y\vec{j}}{\sqrt{16}}=\frac{x\vec{i}+y\vec{j}}{4}[/itex] right?