Solve for f: gradf = (2xy + sin(x)i + (x2 + 2)j

  • Thread starter -EquinoX-
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In summary, the conversation involves finding the function f given gradf = (2xy + sin(x)i + (x2 + 2)j and discussing why the answer may be marked as wrong on a particular website. The solution is confirmed to be correct and it is suggested that the website may not accept the answer due to formatting issues. When asked to find the exact change in f between two specific points, it is recommended to plug the numbers into the function.
  • #1
-EquinoX-
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  • #2
If [tex] \nabla f = (2xy + \sin x)\bold{i} + (x^2 + 2)\bold{j}[/tex], your answer is correct.
 
  • #3
but then why webassign say it's wrong...
 
  • #4
I checked again and It still looks right to me. Maybe you read the question wrong, or entered it wrong.
 
  • #5
pretty sure it's not wrong... hmmm
 
  • #6
okay.. I think the answer is right, it's just the way I format the answer in webassign and it doesn't accepts it.. also if I am asked:

Find the exact change in f between (0, 0) and (1, π/2).

do I just plug the numbers in?
 

Related to Solve for f: gradf = (2xy + sin(x)i + (x2 + 2)j

What is the meaning of "gradf" in the equation?

"gradf" stands for the gradient of the function f. It is a vector that represents the direction and magnitude of the steepest increase of the function at a given point.

What are the variables in the equation and their roles?

The variables in this equation are x and y. They are independent variables that determine the value of the function f. In this case, x and y are used to calculate the gradient of f.

What does the "i" and "j" represent in the equation?

The "i" and "j" represent the unit vectors in the x and y directions, respectively. They are used to indicate the direction of the gradient vector in the x and y directions.

What is the purpose of the sine function in the equation?

The sine function is used to calculate the component of the gradient vector in the x direction. It is a trigonometric function that relates the angle and side lengths of a right triangle.

How can this equation be solved for f?

This equation cannot be directly solved for f because it is a gradient vector equation. Instead, it can be used to calculate the gradient of f at a specific point (x,y). To solve for f, the equation would need to be rearranged and integrated, assuming f is a function of x and y.

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