Potential Function for v(x1, x2) - Find the Solution

[teng125]

Find the potential function if it exists: v (x1 , x2) = 2 (x1)^2 + 2(x2) + 2, 2(x2)+2x(1)


may i know how to do this??

pls help

 

[Office_Shredder]

Find the potential function if it exists: v (x1 , x2) = 2 (x1)^2 + 2(x2) + 2, 2(x2)+2x(1)


may i know how to do this??

pls help

The potential function is the function whose gradient = v(x1 x2) (I'll call them x and y for aesthetic purposes). So if Grad(P(x, y)) = v(x, y), that means dP/dx = 2x^2 + 2y + 2, and dP/dy = 2y + 2x.

Try going from there

 

[teng125]

so the answer is v(x,y)=(4x , 2) rite??

 

[Office_Shredder]

No... the potential function of v is going to be a scalar function P, such that P is (essentially) the integral of v. So you're looking for a scalar function whose partial derivatives are the two parts of v

 

[HallsofIvy]

No, a gradient is not a vector!

Find a single, real valued, function of 2 variables, (and x,y is much better than x1, x2!), P(x,y) such that
$$\frac{\partial P}{\partial x}= 2x^2+ 2y+ 2$$
$$\frac{\partial P}{\partial y}= 2y+ 2x.<br /> <br /> What is the anti-derivative of 2x²+ 2y+ 2 (with respect to x, treating y as a constant)? The "constant of integration" may be a function of y since you are treating y as a constant. Now, what is the derivative of that with respect to y?$$

 

[teng125]

so if i change the question to find the conservative vector field,
than my answer v(x,y)=(4x , 2) is correct rite??
just do partial derivative rite??

 

[Office_Shredder]

teng, (4x, 2) is a vector field.

You need to supply a scalar field as the answer. You're looking for a scalar field whose gradient is v(x,y).

For example, if v(x,y) was (3x^2, 3y^2) then the potential P(x,y)=x^3 + y^3 + C

Because dP/dx = 3x^2, and dP/dy = 3y^2