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nhartung
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Homework Statement
Suppose the differentiable function f(x,y,z) has the partial derivatives fx(1,0,1) = 4, fy(1,0,1) = 1 and fz(1,0,1) = 0. Find g'(0) if g(t) = f(t2 + 1, t2-t, t+1).
Homework Equations
The Attempt at a Solution
Ok I'm given the solution for this and I'm trying to work through it but I'm confused.
I understand we have g(t) = f(x(t),y(t),z(t)) where x = t2+1; y = t2-t; z = t+1.
So i thought I would just use the chain rule to find g'(t) and plug in 0 for t. But I check the solutions sheet and he uses a very different (easier) method that I don't understand. ( I just realized that we aren't given a function f in terms of x y or z so that's why this wouldn't work and we need another method. Still I don't understand this other method.)
First he says when t=0 (x,y,z)|t=0 = (1,0,0) (I think he may have made a mistake, shouldn't it be (1,0,1) ?) And then he gets a point P = (1,-1,0) No idea were this point comes from...
Next he sets up the chain rule equation: [tex]\frac{dg}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}[/tex]
Now I think he somehow uses that point and the partial derivatives given above to solve the equation and he gets: = 1*0 + 1*(-1) + (0*1) = -1 (depending on whether P was correct above may change the answer here I think)
So could someone please explain the method used here?