Is My Directional Derivative Calculation Correct?

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SUMMARY

The discussion centers on calculating the directional derivative of the function f(x,y,z) = z/(2x + y) at the point (0,1,1) in the direction of the vector d = 2i - 2j - k. The initial calculation yielded a vector result of -3i - 6k, which is incorrect as the directional derivative is a scalar. The correct approach involves using the gradient, which was derived as -2z/(2x+y)^2 i + -z/(2x+y)^2 j + 2x+y/(2x+y)^2 k. The final scalar result for the directional derivative is confirmed to be -1.

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  • Knowledge of scalar functions and their properties
  • Basic vector operations and unit vectors
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Im doing the following question:

calculate the directional derivative of the function f(x,y,z) = z/(2x + y) at the point (0,1,1) in the direction d = 2i - 2j - k

could someone please check my answer is correct as i calculated -3i -6k

Also how do i find the unit vector in the direction of the greatest rate of change of the function f(x,y,z)

thanks
 
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Think again. The directional derivative of a scalar function is not a vector. Hint: look at the concept of gradient.
 
the gradient i found was -2z/(2x+y)^2 i + -z/(2x+y)^2 j + 2x+y/(2x+y)^2 k

i thought the directional derivative was then s.grad

what should the answer be?
 
It IS s.grad. But s.grad is a scalar.
 
ok sorry. i think iv got the answer now. -1 ?
 
What is your vector for grad? You'd better check arithmetic...
 
I've got to go now. But as for your second question, if derivative is s.grad, what direction should s point to maximize the derivative?
 
how about -3?
 
-3? I like it.
 

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