{"id":1148,"date":"2025-10-15T14:44:06","date_gmt":"2025-10-15T14:44:06","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1148"},"modified":"2025-10-27T21:32:51","modified_gmt":"2025-10-27T21:32:51","slug":"directional-derivatives-and-the-gradient-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/directional-derivatives-and-the-gradient-learn-it-3\/","title":{"raw":"Directional Derivatives and the Gradient: Learn It 3","rendered":"Directional Derivatives and the Gradient: Learn It 3"},"content":{"raw":"

Gradients and Level Curves<\/h2>\r\nThe gradient has a special relationship with level curves that reveals important geometric information. Let's explore this connection.\r\n\r\nRecall that if a curve is defined parametrically by [latex](x(t), y(t))[\/latex], then the vector [latex]x'(t){\\bf{i}}+y'(t){\\bf{j}}[\/latex] is tangent to the curve for every value of [latex]t[\/latex] in the domain.\r\n\r\nNow suppose [latex]z=f(x, y)[\/latex] is a differentiable function, and [latex](x_0, y_0)[\/latex] is in its domain. Assume that [latex]x_0=x(t_0)[\/latex] and [latex]y_0=y(t_0)[\/latex] for some value of [latex]t[\/latex], and consider the level curve [latex]f(x, y)=k[\/latex].\r\n\r\nDefine [latex]g(t)=f(x(t), y(t))[\/latex] and calculate [latex]g'(t)[\/latex] on the level curve. By the Chain Rule:\r\n

[latex]g'(t)=f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)[\/latex]<\/p>\r\nBut [latex]g'(t)=0[\/latex] because [latex]g(t)=k[\/latex] for all [latex]t[\/latex] (the function value is constant on a level curve). Therefore:\r\n

[latex]f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)=0[\/latex]<\/p>\r\nWe can rewrite the left side as a dot product:\r\n

[latex]f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)=\\nabla{f}(x,y)\\cdot\\langle{x}'(t),y'(t)\\rangle[\/latex]<\/p>\r\nTherefore:\r\n

[latex]\\nabla{f}(x,y)\\cdot\\langle{x}'(t),y'(t)\\rangle=0[\/latex]<\/p>\r\nThe dot product of these vectors equals zero, which means they're orthogonal. Since the vector [latex]\\langle{x}'(t),y'(t)\\rangle[\/latex] is tangent to the level curve, the gradient must be normal (perpendicular) to the level curve. This gives us an important theorem.\r\n\r\n

\r\n

gradient is normal to the level curve<\/h3>\r\nSuppose the function\u00a0[latex]z=f(x, y)[\/latex] has continuous first-order partial derivatives in an open disk centered at a point\u00a0[latex](x_0, y_0)[\/latex]. If\u00a0[latex]\\nabla{f}(x_0,y_0)\\ne0[\/latex], then\u00a0[latex]\\nabla{f}(x_0,y_0)[\/latex] is normal to the level curve of\u00a0[latex]f[\/latex] at\u00a0[latex](x_0, y_0)[\/latex].\r\n\r\n<\/section>\r\n

We can use this theorem to find tangent and normal vectors to level curves of a function.<\/p>\r\n\r\n

For the function\u00a0[latex]f(x, y)=2x^{2}-3xy+8y^{2}+2x-4y+4[\/latex], find a tangent vector to the level curve at point\u00a0[latex](-2, 1)[\/latex]. Graph the level curve corresponding to\u00a0[latex]f(x, y)=18[\/latex] and draw in\u00a0[latex]\\nabla{f}(-2,1)[\/latex] and a tangent vector.[reveal-answer q=\"345812399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"345812399\"]First, we must calculate [latex]\\nabla{f}(x,y)[\/latex]:\r\n

[latex]f_x(x,y)=4x-3y+2[\/latex] and\u00a0[latex]f_y=-3x+16y-4[\/latex] so\u00a0[latex]\\nabla{f}(x,y)=(4x-3y+2){\\bf{i}}+(-3x+16y-4){\\bf{j}}.[\/latex]<\/p>\r\nNext, we evaluate\u00a0[latex]\\nabla{f}(x,y)[\/latex] at\u00a0[latex](-2,1)[\/latex]:\r\n

[latex]\\nabla{f}(-2,1)=(4(-2)-3(1)+2){\\bf{i}}+(-3(-2)+16(1)-4){\\bf{j}}=-9{\\bf{i}}+18{\\bf{j}}.[\/latex]<\/p>\r\nThis vector is orthogonal to the curve at point [latex](-2, 1)[\/latex]. We can obtain a tangent vector by reversing the components and multiplying either one by [latex]-1[\/latex]. Thus, for example, [latex]-18{\\bf{i}}-9{\\bf{j}}[\/latex] is a tangent vector (see the following graph).\r\n\r\n[caption id=\"attachment_1272\" align=\"aligncenter\" width=\"304\"]\"A Figure 5. A rotated ellipse with equation [latex]\\small{f(x,y)=18}[\/latex]. At the point [latex]\\small{(-2,1)}[\/latex] on the ellipse, there are drawn two arrows, one tangent vector and one normal vector. The normal vector is marked [latex]\\small{\\nabla{f(-2,1)}}[\/latex] and is perpendicular to the tangent vector.[\/caption][\/hidden-answer]<\/section>","rendered":"

Gradients and Level Curves<\/h2>\n

The gradient has a special relationship with level curves that reveals important geometric information. Let’s explore this connection.<\/p>\n

Recall that if a curve is defined parametrically by [latex](x(t), y(t))[\/latex], then the vector [latex]x'(t){\\bf{i}}+y'(t){\\bf{j}}[\/latex] is tangent to the curve for every value of [latex]t[\/latex] in the domain.<\/p>\n

Now suppose [latex]z=f(x, y)[\/latex] is a differentiable function, and [latex](x_0, y_0)[\/latex] is in its domain. Assume that [latex]x_0=x(t_0)[\/latex] and [latex]y_0=y(t_0)[\/latex] for some value of [latex]t[\/latex], and consider the level curve [latex]f(x, y)=k[\/latex].<\/p>\n

Define [latex]g(t)=f(x(t), y(t))[\/latex] and calculate [latex]g'(t)[\/latex] on the level curve. By the Chain Rule:<\/p>\n

[latex]g'(t)=f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)[\/latex]<\/p>\n

But [latex]g'(t)=0[\/latex] because [latex]g(t)=k[\/latex] for all [latex]t[\/latex] (the function value is constant on a level curve). Therefore:<\/p>\n

[latex]f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)=0[\/latex]<\/p>\n

We can rewrite the left side as a dot product:<\/p>\n

[latex]f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)=\\nabla{f}(x,y)\\cdot\\langle{x}'(t),y'(t)\\rangle[\/latex]<\/p>\n

Therefore:<\/p>\n

[latex]\\nabla{f}(x,y)\\cdot\\langle{x}'(t),y'(t)\\rangle=0[\/latex]<\/p>\n

The dot product of these vectors equals zero, which means they’re orthogonal. Since the vector [latex]\\langle{x}'(t),y'(t)\\rangle[\/latex] is tangent to the level curve, the gradient must be normal (perpendicular) to the level curve. This gives us an important theorem.<\/p>\n

\n

gradient is normal to the level curve<\/h3>\n

Suppose the function\u00a0[latex]z=f(x, y)[\/latex] has continuous first-order partial derivatives in an open disk centered at a point\u00a0[latex](x_0, y_0)[\/latex]. If\u00a0[latex]\\nabla{f}(x_0,y_0)\\ne0[\/latex], then\u00a0[latex]\\nabla{f}(x_0,y_0)[\/latex] is normal to the level curve of\u00a0[latex]f[\/latex] at\u00a0[latex](x_0, y_0)[\/latex].<\/p>\n<\/section>\n

We can use this theorem to find tangent and normal vectors to level curves of a function.<\/p>\n

For the function\u00a0[latex]f(x, y)=2x^{2}-3xy+8y^{2}+2x-4y+4[\/latex], find a tangent vector to the level curve at point\u00a0[latex](-2, 1)[\/latex]. Graph the level curve corresponding to\u00a0[latex]f(x, y)=18[\/latex] and draw in\u00a0[latex]\\nabla{f}(-2,1)[\/latex] and a tangent vector.<\/p>\n