Find gradient at a point and the directional derivative Multivariable calculus

In summary, the problem is to calculate the gradient of the function f(x,y)=x^3+2y^3 at point P(1,1) and the directional derivative at P in the direction u of the given vector A -> i-j. This can be done by first finding the partial derivatives of f(x,y), which are f_x=3x^2 and f_y=6y^2. Then, the formula for the directional derivative is the dot product of the gradient at P with a unit vector in the given direction. The issue the person is facing is not knowing how to complete the calculation, specifically finding the unit vector in the direction of A -> i-j.
  • #1
Nugget3
1
0
Calculate gradient of f
f(x,y)=x^3+2y^3
at point P (1,1)
and the directional derivative at P
in the direction u of the given vector A -> i-j

I tried to attempt this but i honestly don't know where to start. I began to take the partial derivatives of f. I got f'=3x^2dx+6y^2dy, however that answer doesn't seem correct to me.
 
Physics news on Phys.org
  • #2
Nugget3 said:
Calculate gradient of f
f(x,y)=x^3+2y^3
at point P (1,1)
and the directional derivative at P
in the direction u of the given vector A -> i-j

I tried to attempt this but i honestly don't know where to start. I began to take the partial derivatives of f. I got f'=3x^2dx+6y^2dy, however that answer doesn't seem correct to me.

What you have calculated is the total differential of f(x,y), which includes both partials ##f_x## and ##f_y##. What is the formula for the directional derivative? Show us what happens when you try to calculate it.
 
  • #3
it is the dot product of the gradient at P with a unit vector in the given direction
 
  • #4
LCKurtz said:
What you have calculated is the total differential of f(x,y), which includes both partials ##f_x## and ##f_y##. What is the formula for the directional derivative? Show us what happens when you try to calculate it.

hedipaldi said:
it is the dot product of the gradient at P with a unit vector in the given direction

You ignored the part I highlighted in red.
 
  • #5
try to calculate what?don't you need to calculate the directional derivative?
 
  • #6
hedipaldi said:
try to calculate what?don't you need to calculate the directional derivative?

Yes. You posted a question because you were apparently stuck on the problem. You have now stated that you need to take the gradient at P dotted into a unit vector. So please do it and show us what exactly you are stuck on.
 

FAQ: Find gradient at a point and the directional derivative Multivariable calculus

1. What is a gradient?

The gradient is a mathematical concept that represents the directional rate of change of a function. It is a vector that points in the direction of the steepest increase of a function and its magnitude represents the rate of change in that direction.

2. How do you find the gradient at a point?

To find the gradient at a point, you need to take the partial derivatives of the function with respect to each variable and then evaluate them at the given point. The resulting values will be the components of the gradient vector.

3. What is the significance of the gradient?

The gradient is significant because it provides important information about the behavior of a function. It tells us the direction of the steepest increase or decrease of the function, and the magnitude of the gradient represents the rate of change in that direction. This information is useful in various applications, such as optimization problems and modeling physical phenomena.

4. What is a directional derivative?

A directional derivative is a measure of the instantaneous rate of change of a function in a specific direction. It is calculated by taking the dot product of the gradient vector and a unit vector in the desired direction. This gives us the slope of the tangent line to the function in that direction.

5. How do you calculate the directional derivative?

The directional derivative can be calculated by taking the dot product of the gradient vector and a unit vector in the desired direction. Alternatively, it can also be calculated using the formula:
Dvf(x,y) = ∇f(x,y) · v = (fx(x,y), fy(x,y)) · (v1, v2) = fx(x,y)v1 + fy(x,y)v2
where ∇f(x,y) represents the gradient vector, v is the unit vector in the desired direction, and fx(x,y) and fy(x,y) are the partial derivatives of the function at the given point.

Similar threads

Back
Top