How Can You Estimate the Directional Derivative Using Given Function Values?

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SUMMARY

The discussion focuses on estimating the directional derivative of the function f at the point (1,1,1) using given function values f(1,1,1)=3 and f(1.1,1.2,1.1)=3.1. The unit vector u is calculated as <0.41, 0.82, 0.41>, which is derived from the differences in the function's input values. The gradient vector del f is estimated as <1, 1/2, 1>, leading to a directional derivative Duf of approximately 1.23. The method employed utilizes the definition of the directional derivative and approximates changes in function values.

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Homework Statement


Assume f(1,1,1)=3 and f(1.1,1.2,1.1)=3.1

a) Which directional derivative Duf at (1,1,1) can be estimated from this information? Give vector u

b) Estimate the directional derivative in part a

Homework Equations



Duf = del f (dot product) vector u
del f = ([itex]\partial[/itex]f/[itex]\partial[/itex]x, [itex]\partial[/itex]f/[itex]\partial[/itex]y)

The Attempt at a Solution



So far I've been able to get unit vector
u = <1.1-1, 1.2-1, 1.1-1>/[itex]\sqrt{.1^2+.2^2+.1^2}[/itex] = <0.41, 0.82, 0.41>

I've been rolling it around in my head but I can't think of a way to obtain del f.
How would I get any thing resembling the partial derivatives of the unknown function, if all I know are points? I understand del f is the vector pointing in the direction of greatest change.
Edit: I have made another push at an answer.
I figured del f= <partial f/ partial x, partial f/ partial y, partial f/partial x>
using partial f = 3.1-1=.1
partial x = .1
partial y= .2
partial z= .1

del f= <.1/.1, .1/.2, .1/.1>=<1, 1/2, 1>

so the directional derivative would be
<1, 1/2, 1>dot<.41, .82, .41> =1.23
 
Last edited:
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What I would have used is that
$$D_{\vec u}f(\vec x)=\lim_{t\to 0}\frac{f(\vec x+t\vec u)-f(\vec x)}{t} \approx\frac{f(\vec x+t\vec u)-f(\vec x)}{t}$$ when t is small. But your approach should work too. You should however avoid notations that suggest that ∂f, ∂x etc. are numbers.
 

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