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

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The discussion focuses on estimating the directional derivative Duf at the point (1,1,1) using the function values f(1,1,1)=3 and f(1.1,1.2,1.1)=3.1. A unit vector u was calculated as <0.41, 0.82, 0.41>, representing the direction of change. The challenge lies in determining the gradient vector del f, which was approximated using the differences in function values to estimate partial derivatives. The final calculation for the directional derivative resulted in a value of approximately 1.23. The conversation emphasizes the importance of correctly interpreting notations and understanding the limit definition of directional derivatives.
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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 = (\partialf/\partialx, \partialf/\partialy)

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>/\sqrt{.1^2+.2^2+.1^2} = <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
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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