Directional derivative, help

In summary, the given equations are used to find the gradient vector of a function with respect to u, but there is not enough information to determine a unique solution. More conditions or equations are needed to fully solve the problem.
  • #1
ilyas.h
60
0

Homework Statement


[tex]D_{u}(f)(a,b) = \triangledown f(a,b)\cdot u[/tex]

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

where [tex]u = (\frac{1}{\sqrt2}, \frac{1}{\sqrt2})[/tex]

find [tex]\bigtriangledown f(a.b)[/tex]

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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  • #2
ilyas.h said:

Homework Statement


[tex]D_{u}(f)(a,b) = \triangledown f(a,b)\cdot u[/tex]

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

where [tex]u = (\frac{1}{\sqrt2}, \frac{1}{\sqrt2})[/tex]

find [tex]\bigtriangledown f(a.b)[/tex]

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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

1. What is a directional derivative?

A directional derivative is a measure of how a function changes in a specific direction. It is defined as the rate of change of the function in the direction of a unit vector.

2. How is a directional derivative calculated?

To calculate the directional derivative of a function, you first need to find the gradient vector at a given point. Then, you take the dot product of the gradient vector and the unit vector in the desired direction.

3. What is the significance of the directional derivative?

The directional derivative is important in multivariable calculus as it allows us to determine the rate of change of a function in a particular direction. It is also used in optimization problems to find the direction in which the function changes the fastest.

4. Can the directional derivative be negative?

Yes, the directional derivative can be negative. This indicates that the function is decreasing in the direction of the unit vector.

5. How is the directional derivative related to partial derivatives?

The directional derivative is essentially a generalization of the partial derivatives. The partial derivatives represent the directional derivatives in the direction of the coordinate axes. The directional derivative in any other direction can be calculated using the partial derivatives.

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