Multivariable Vector: Gradient @ particular speed - Find Rate

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SUMMARY

The discussion focuses on calculating the rate of change of fluid concentration, represented by the function F(x,y,z) = 2x^2 + 4y^4 + 2*x^2*z^2, at the point (-1,1,1). The gradient of the function at this point is determined to be Grad(F) = <-8,16,4>. To find the concentration change while moving in the direction of the gradient at a speed of 8, participants confirm that the rate of change can be calculated using the formula \(\frac{dF}{dt} = \nabla F \cdot \frac{d\mathbf{x}}{dt}\), where \(\frac{d\mathbf{x}}{dt}\) is the unit vector of the gradient scaled by the speed.

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MJSemper
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Homework Statement


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?

Homework Equations



Already found the gradient ... = < 4x+4xz^2, 16y^3, 4*x^2*z >
...at point (-1,1,1) = < -8,16,4 >

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?
 
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I guess you should calculate \frac{dF}{dt} = \nabla F \cdot \frac{d\mathbf{x}}{dt}
 

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