Maximum value of directional derivative (Duf)?

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SUMMARY

The maximum value of the directional derivative at a point is indeed the length of the gradient vector ∇f, also known as grad. f. In this discussion, the gradient vector is given as ∇f = (56/5) i - (3/5) j. The calculated length of this vector is √(3145)/5, which aligns with the answer provided in the referenced textbook. The confusion arises from distinguishing between the vector itself and its magnitude.

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JC3187
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Hi guys,

I am confused from what I know the max. value of directional derivative at a point is the length of the gradient vector ∇f or grad. f?

Why does the answer in my book of a question say that
Max. val of Duf = (√3145)/5
when ∇f = (56/5) i- (3/5) j
?

Thanks
 
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JC3187 said:
Hi guys,

I am confused from what I know the max. value of directional derivative at a point is the length of the gradient vector ∇f or grad. f?

Why does the answer in my book of a question say that
Max. val of Duf = (√3145)/5
when ∇f = (56/5) i- (3/5) j
?

Thanks

\|\nabla f\| = \sqrt{\left(\frac{56}{5}\right)^2 + \left(-\frac{3}{5}\right)^2} = \dots
 
It is the difference between "a vector" and "length of a vector".
 

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