How Do You Find the Gradient Vector from a Directional Derivative?

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To find the gradient vector from a directional derivative, the equation D_{u}(f)(a,b) = ∇f(a,b)·u is used, where u is a unit vector. The discussion reveals that the dot product leads to the equation df/dx + df/dy = 6, indicating a relationship between the partial derivatives. However, it is noted that there are infinitely many gradient vectors that satisfy this condition, requiring additional information for a unique solution. Without more constraints, the solution remains incomplete. The final conclusion emphasizes that the current findings are as far as one can go without further data.
ilyas.h
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Homework Statement


D_{u}(f)(a,b) = \triangledown f(a,b)\cdot u

D_{(\frac{1}{\sqrt2}, \frac{1}{\sqrt2})}(f)(a,b) = 3 \sqrt{2}

where u = (\frac{1}{\sqrt2}, \frac{1}{\sqrt2})

find \bigtriangledown f(a.b)

Homework Equations

The Attempt at a Solution



first you change grad f into it's partial derivative form and then take the dot product:

(df/dx, df/dy).(1/root2, 1/root2) = 3root2

you'll find that:

df/dx + df/dy = 6

where would I go from here? quite confused.
 
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ilyas.h said:

Homework Statement


D_{u}(f)(a,b) = \triangledown f(a,b)\cdot u

D_{(\frac{1}{\sqrt2}, \frac{1}{\sqrt2})}(f)(a,b) = 3 \sqrt{2}

where u = (\frac{1}{\sqrt2}, \frac{1}{\sqrt2})

find \bigtriangledown f(a.b)

Homework Equations

The Attempt at a Solution



first you change grad f into it's partial derivative form and then take the dot product:

(df/dx, df/dy).(1/root2, 1/root2) = 3root2

you'll find that:

df/dx + df/dy = 6

where would I go from here? quite confused.

There are infinitely many different gradient vectors that satisfy the given condition. You need more conditions in order to get a unique answer.
 
Ray Vickson said:
There are infinitely many different gradient vectors that satisfy the given condition. You need more conditions in order to get a unique answer.

so is my solution complete? if not, where would I obtain more equations to make the conditions more robust?

you could say that:

grad f(6-y, y) . (1/root2, 1/root2) = 3root2
 
ilyas.h said:
so is my solution complete? if not, where would I obtain more equations to make the conditions more robust?

you could say that:

grad f(6-b, b) . (1/root2, 1/root2) = 3root2

You cannot pull more information out of the air; somebody has to give it to you. If they do not give you more information, you have gone as far as you can go.
 
Ray Vickson said:
You cannot pull more information out of the air; somebody has to give it to you. If they do not give you more information, you have gone as far as you can go.

great, so df/dx + df/dy = 6 is the final solution?

sorry for bugging you, this question is quite important to me.
 
ilyas.h said:
great, so df/dx + df/dy = 6 is the final solution?

sorry for bugging you, this question is quite important to me.

Asked and answered.
 
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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