Multivariable Vector: Gradient @ particular speed  Find Rateby MJSemper Tags: gradient, multivariable, rate, speed, vector 

#1
Apr312, 03:45 PM

PF Gold
P: 8

1. The problem statement, all variables and given/known data
Given: Concentration of Fluid = F(x,y,z) = 2x^2 + 4y^4 + 2*x^2*z^2 at point (1,1,1) Found Grad(F(x,y,z)) = <8,16,4> If you start to move in the direction of Grad(F) at a speed of 8, how fast is the concentration changing? 2. Relevant equations Already found the gradient ... = < 4x+4xz^2, 16y^3, 4*x^2*z > ...at point (1,1,1) = < 8,16,4 > 3. The attempt at a solution I know that the gradient is the direction of greatest change, and its magnitude is the particular rate thereof. What I'm stuck at is how the "speed of 8" plays in there. Is it a scalar value to the magnitude of the gradient vector? I know grad(F) = sqrt(336) do I take the Unit Vector of the gradient and multiply it by the scalar 8? 



#2
Apr412, 06:21 AM

P: 937

I guess you should calculate [tex] \frac{dF}{dt} = \nabla F \cdot \frac{d\mathbf{x}}{dt} [/tex]



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