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Question

http://www.chegg.com/homework-help/calculus-early-transcendentals-2nd-edition-chapter-14.4-problem-11e-solution-9781429231848

in step 1.

why do they move Z over to one side to start?

How do they get -1? somone mentioned gradient in the comments but gradient has not been covered yet. I don't think they would require us to use gradient on this problem.

in step 3.

conceptually why do they set (I should probably know this). yes I know its parrelel, but I don't remember this form.

why do they move Z over to one side to start? How do they get -1? somone mentioned gradient in the comments but gradient has not been covered yet. I don't think they would require us to use gradient on this problem. conceptually why do they set (I should probably know this). yes I know its parrelel, but I don't remember this form.

Explanation / Answer

I cannot see the answer. But to some extent by what you asked:

In step 3, if that was done, that means you have parallel vectors. If two vectors are parallel, they get related by means of one scalar multiple. That multiple in this case is Lambda.

As far as gradient is concerned, it's nothing but the three dimensional slope of a potential function. I know I sound too technical, but yes that is what it really means. In simple language, to determine the gradient, you need to differentiate the function in x,y,z partially with respect to x, y and z. The x differential becomes the i-th vector component. Similarly y differential becomes the j-th component and so on.

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