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nhartung
#5
Mar31-10, 07:10 PM
P: 56
Quote Quote by Mark44 View Post
Not quite. You are confusing the derivatives of x(t), y(t), and z(t) with the value of each derivative at a specific value of t. IOW, x'(t) = 2t, but x'(0) = 0, and similarly for y'(t) vs. y'(0) and z'(t) vs. z'(0). In each case, the deriviative is a function, while the derivative evaluated at a particular value of t is just a number. x'(t) is different from x'(0), y'(t) is different from y'(0), and z'(t) is different from z'(0).
Isn't this what I did? I think I may have messed up my notation a bit but I have x'(t) = 2t then x'(0) = 0; y'(t) = 2t-1 then y'(0) = -1; z'(t) = 1 then z'(0) = 1;


Quote Quote by Mark44 View Post
The same difference exists between g'(t) and g'(0). Using the chain rule you can write g'(t) = fx(x,y,z) * x'(t) + fy(x,y,z) * y'(t) + fz(x,y,z) * z'(t). Since you aren't given any details about f(x, y, z), it's not possible to calculate the functions fx(x, y, z), fy(x, y, z), and fz(x, y, z). You are, however given the values of these partials at a particular point (1, 0, 1) that corresponds to t = 0.
So here since the point (1,0,1) corresponds to g(t) when t is 0 we can use the value of the partial at (1,0,1) as the value of fx(x,y,z)? Which would mean at (1,0,1) fx = 4; fy = 1; fz = 0.

Quote Quote by Mark44 View Post
g'(0) = fx(1,0,1) * x'(0) + fy(1,0,1) * y'(0) + fz(1,0,1) * z'(0). The only reason I have used the subscript notation (as opposed to the Leibniz notation you used) for the partial derivatives is that it's a little easier to explicitly show that they are to be evaluated at a particular point.
So if everything above is correct that gives me: 4*0 + 1*(-1) + 0*1 = -1.

Also, does that mean (if this is all correct) this method can only be used if the point the partials are evaluated at and the value of t are directly related as in this example, correct?