{"id":1367,"date":"2025-10-17T14:30:18","date_gmt":"2025-10-17T14:30:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1367"},"modified":"2025-11-06T13:28:33","modified_gmt":"2025-11-06T13:28:33","slug":"conservative-vector-fields-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/conservative-vector-fields-learn-it-3\/","title":{"raw":"Conservative Vector Fields: Learn It 3","rendered":"Conservative Vector Fields: Learn It 3"},"content":{"raw":"
Conservative Vector Fields and Potential Functions<\/h2>\r\n
The Fundamental Theorem for Line Integrals tells us that if [latex]\\mathbf{F}[\/latex] is conservative, then calculating [latex]\\displaystyle\\int_{C}\\mathbf{F}\\cdot d\\mathbf{r}[\/latex] has two steps: first, find a potential function [latex]f[\/latex] for [latex]\\mathbf{F}[\/latex], and second, calculate [latex]f(P_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of [latex]C[\/latex] and [latex]P_0[\/latex] is the starting point.<\/p>\r\n
To use this theorem, we need to answer a key question: Given a conservative vector field [latex]\\mathbf{F}[\/latex], how do we find a function [latex]f[\/latex] such that [latex]\\nabla f=\\mathbf{F}[\/latex]? Let's start with an example to see how this works.<\/p>\r\n\r\nFind a potential function for [latex]{\\bf{F}}(x,y)=\\langle2xy^3,3x^2y^2+\\cos{(y)}\\rangle[\/latex], thereby showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.[reveal-answer q=\"342587498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342587498\"]\r\n
Suppose that [latex]f(x, y)[\/latex] is a potential function for\u00a0[latex]{\\bf{F}}[\/latex]. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex], and therefore<\/p>\r\n
[latex]f_x=2xy^3\\text{ and }f_y=3x^2y^2+\\cos{y}[\/latex].<\/p>\r\n
Integrating the equation [latex]f_x=2xy^3[\/latex] with respect to [latex]x[\/latex] yields the equation<\/p>\r\n
[latex]f(x,y)=x^2y^2+h(y)[\/latex].<\/p>\r\n
Notice that since we are integrating a two-variable function with respect to [latex]x[\/latex], we must add a constant of integration that is a constant with respect to [latex]x[\/latex], but may still be a function of [latex]y[\/latex]. The equation [latex]f(x,y)=x^2y^2+h(y)[\/latex] can be confirmed by taking the partial derivative with respect to [latex]x[\/latex]:<\/p>\r\n
This implies that [latex]h^\\prime(y)=\\cos{(y)}[\/latex], so [latex]h(y)=\\sin{y}+C[\/latex]. Therefore,\u00a0any<\/em>\u00a0function of the form [latex]f(x,y)=x^2y^3+\\sin{(y)}+C[\/latex] is a potential function. Taking, in particular, [latex]C=0[\/latex] gives the potential function [latex]f(x,y)=x^2y^3+\\sin{(y)}[\/latex].<\/p>\r\n
To verify that [latex]f[\/latex] is a potential function, note that [latex]\\nabla{f}=\\langle2xy^3,3x^2y^2+\\cos{y}\\rangle={\\bf{F}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
The logic of the previous example extends to finding the potential function for any conservative vector field in [latex]\\mathbb{R}^2[\/latex]. Here's the strategy:<\/p>\r\n\r\nHow to: Find a Potential Function for a Conservative Vector Field [latex]\\mathbf{F}(x,y)=\\langle P(x,y),Q(x,y)\\rangle[\/latex]\u00a0<\/strong>\r\n\r\n \t
Integrate [latex]P[\/latex] with respect to [latex]x[\/latex].<\/strong> This results in a function of the form [latex]g(x, y)+h(y)[\/latex], where [latex]h(y)[\/latex] is unknown.<\/li>\r\n \t
Take the partial derivative of [latex]g(x, y)+h(y)[\/latex] with respect to [latex]y[\/latex].<\/strong> This results in the function [latex]g_y(x, y)+h'(y)[\/latex].<\/li>\r\n \t
Use the equation [latex]g_y(x, y)+h'(y)=Q(x ,y)[\/latex] to find [latex]h'(y)[\/latex].<\/strong><\/li>\r\n \t
Integrate [latex]h'(y)[\/latex] to find [latex]h(y)[\/latex].<\/strong><\/li>\r\n \t
Write the potential function.<\/strong> Any function of the form [latex]f(x,y)=g(x,y)+h(y)+C[\/latex], where [latex]C[\/latex] is a constant, is a potential function for [latex]\\mathbf{F}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n
We can adapt this strategy to find potential functions for vector fields in [latex]\\mathbb{R}^3[\/latex], as shown in the next example.<\/p>\r\n\r\nFind a potential function for [latex]{\\bf{F}}(x,y,z)=\\langle2xy,x^2+2yz^3,3y^2z^2+2z\\rangle[\/latex], thereby showing that [latex]{\\bf{F}}[\/latex] is conservative.[reveal-answer q=\"764523874\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"764523874\"]\r\n
Suppose that [latex]f[\/latex] is a potential function. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex] and therefore [latex]f_x=2xy[\/latex]. Integrating this equation with respect to [latex]x[\/latex] yields the equation [latex]f(x, y, z)=x^{2}y+g(y, z)[\/latex] for some function [latex]g[\/latex]. Notice that, in this case, the constant of integration with respect to [latex]x[\/latex] is a function of [latex]y[\/latex] and\u00a0[latex]z[\/latex].<\/p>\r\n
Since [latex]f[\/latex] is a potential function,<\/p>\r\n
Integrating this function with respect to [latex]y[\/latex] yields<\/p>\r\n
[latex]g(y, z)=y^{2}z^{3}+h(z)[\/latex]<\/p>\r\n
for some function [latex]h(z)[\/latex] of [latex]z[\/latex] alone. (Notice that, because we know that [latex]g[\/latex] is a function of only [latex]y[\/latex] and [latex]z[\/latex], we do not need to write [latex]g(y, z)=y^{2}z^{3}+h(x,z)[\/latex].) Therefore,<\/p>\r\n
To verify that [latex]f[\/latex] is a potential function, note that [latex]\\nabla{f}=\\langle2xy,x^2+2yz^3,3y^2z^2+2z\\rangle[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
We can apply this process to a gravitational force. Recall that if an object has unit mass and is located at the origin, then the gravitational force in [latex]\\mathbb{R}^2[\/latex] that the object exerts on another object of unit mass at the point [latex](x, y)[\/latex] is given by vector field<\/p>\r\n
where [latex]G[\/latex] is the universal gravitational constant. In the next example, we build a potential function for [latex]\\mathbf{F}[\/latex], thus confirming what we already know: that gravity is conservative.<\/p>\r\n\r\nFind a potential function [latex]f[\/latex] for [latex]{\\bf{F}}(x,y)=-G\\left\\langle\\frac{x}{(x^2+y^2)^{3\/2}},\\frac{y}{(x^2+y^2)^{3\/2}}\\right\\rangle[\/latex].[reveal-answer q=\"384572983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384572983\"]\r\n
Suppose that [latex]f[\/latex] is a potential function. Then, [latex]\\nabla{f}={\\bf{F}}[\/latex] and therefore<\/p>\r\n
To integrate this function with respect to [latex]x[\/latex], we can use [latex]u[\/latex]-substitution. If [latex]u=x^{2}+y^{2}[\/latex] then [latex]\\frac{du}2=x \\ dx[\/latex], so<\/p>\r\n
[latex]\\begin{aligned}\r\n\\displaystyle\\int\\frac{-Gx}{(x^2+y^2)^{3\/2}}&=\\displaystyle\\int\\frac{-G}{2u^{3\/2}} \\ du \\\\\r\n&=\\frac{G}{\\sqrt{u}}+h(y) \\\\\r\n&=\\frac{G}{\\sqrt{x^2+y^2}}+h(y)\r\n\\end{aligned}[\/latex]<\/p>\r\n
for some function [latex]h(y)[\/latex]. Therefore,<\/p>\r\n
Since [latex]f(x,y)=\\frac{G}{\\sqrt{x^2+y^2}}+h(y)[\/latex],\u00a0[latex]f_y[\/latex] also equals [latex]\\frac{-Gy}{(x^2+y^2)^{3\/2}}+h^\\prime(y)[\/latex].<\/p>\r\n
which implies that [latex]h^\\prime(y)=0[\/latex]. Thus, we can take [latex]h(y)[\/latex] to be any constant; in particular, we can let [latex]h(y)=0[\/latex]. The function<\/p>\r\n
is a potential function for the gravitational field\u00a0[latex]{\\bf{F}}[\/latex]. To confirm that [latex]f[\/latex] is a potential function, note that<\/p>\r\n
Conservative Vector Fields and Potential Functions<\/h2>\n
The Fundamental Theorem for Line Integrals tells us that if [latex]\\mathbf{F}[\/latex] is conservative, then calculating [latex]\\displaystyle\\int_{C}\\mathbf{F}\\cdot d\\mathbf{r}[\/latex] has two steps: first, find a potential function [latex]f[\/latex] for [latex]\\mathbf{F}[\/latex], and second, calculate [latex]f(P_1)-f(P_0)[\/latex], where [latex]P_1[\/latex] is the endpoint of [latex]C[\/latex] and [latex]P_0[\/latex] is the starting point.<\/p>\n
To use this theorem, we need to answer a key question: Given a conservative vector field [latex]\\mathbf{F}[\/latex], how do we find a function [latex]f[\/latex] such that [latex]\\nabla f=\\mathbf{F}[\/latex]? Let’s start with an example to see how this works.<\/p>\nFind a potential function for [latex]{\\bf{F}}(x,y)=\\langle2xy^3,3x^2y^2+\\cos{(y)}\\rangle[\/latex], thereby showing that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\n