{"id":1122,"date":"2025-10-15T14:31:03","date_gmt":"2025-10-15T14:31:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1122"},"modified":"2025-10-27T20:49:52","modified_gmt":"2025-10-27T20:49:52","slug":"tangent-planes-and-linear-approximations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/tangent-planes-and-linear-approximations-learn-it-2\/","title":{"raw":"Tangent Planes and Linear Approximations: Learn It 2","rendered":"Tangent Planes and Linear Approximations: Learn It 2"},"content":{"raw":"

Linear Approximations<\/h2>\r\n

You've seen linear approximations before in single-variable calculus. The idea is straightforward: use a tangent line to estimate function values near a known point.<\/p>\r\n\r\n

\r\n

Linear Approximations and Differentials<\/strong><\/p>\r\n

The formula for the linear approximation of a function [latex]f(x)[\/latex] at the point [latex]x=a[\/latex] is:<\/p>\r\n

[latex]y\\approx f(a)+f'(a)(x-a)[\/latex]<\/p>\r\n\r\n<\/section>\r\n

Here's what this looks like graphically:<\/p>\r\n\r\n\r\n[caption id=\"attachment_1236\" align=\"aligncenter\" width=\"342\"]\"A Figure 4. Linear approximation of a function in one variable.[\/caption]\r\n

The tangent line gives you a good approximation to [latex]f(x)[\/latex] for values of [latex]x[\/latex] reasonably close to [latex]x=a[\/latex]. When working with functions of two variables, we use the same idea\u2014just replace the tangent line with a tangent plane.<\/p>\r\n\r\n

\r\n

linear approximation of a function of two variables<\/h3>\r\nGiven a function [latex]z=f(x, y)[\/latex] with continuous partial derivatives at the point [latex](x_0, y_0)[\/latex], the linear approximation of [latex]f[\/latex] at [latex](x_0, y_0)[\/latex] is:\r\n
[latex]L(x, y)=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0).[\/latex]<\/div>\r\n<\/section>Notice this is exactly the equation of the tangent plane to the surface [latex]z=f(x, y)[\/latex] at [latex](x_0, y_0)[\/latex]. The linear approximation works because if you know the exact value of [latex]f(x, y)[\/latex] at [latex](x_0, y_0)[\/latex], then for points [latex](x, y)[\/latex] close to [latex](x_0, y_0)[\/latex], the tangent plane gives you a value close to the actual value of [latex]f(x, y)[\/latex] (Figure 5).\r\n\r\n[caption id=\"attachment_1238\" align=\"aligncenter\" width=\"650\"]\"A Figure 5. Using a tangent plane for linear approximation at a point.[\/caption]\r\n\r\n
Given the function [latex]f(x, y)=\\sqrt{41-4x^{2}-y^{2}}[\/latex], approximate [latex]f(2.1, 2.9)[\/latex] using point [latex](2, 3)[\/latex] for [latex](x_0, y_0)[\/latex]. What is the approximate value of [latex]f(2.1, 2.9)[\/latex] to four decimal places?[reveal-answer q=\"fs-id1467793993124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793993124\"]\r\n

To apply our definition, we first must calculate [latex]f(x_0, y_0)[\/latex], [latex]f_x(x_0, y_0)[\/latex], and [latex]f_y(x_0, y_0)[\/latex] using [latex]x_0=2[\/latex] and [latex]y_0=3[\/latex]:<\/p>\r\n\r\n

[latex]\\begin{array}{ccc}\\hfill {f(x_0, y_0)} & =\\hfill & {f(2, 3)=\\sqrt{41-4(2)^{2}-(3)^{2}}=\\sqrt{41-16-9}=\\sqrt{16}=4}\\hfill \\\\ \\hfill {f_x(x, y)} & =\\hfill & {-\\frac{4x}{\\sqrt{41-4x^{2}-y^{2}}}\\ so\\ f_x(x_0, y_0)=-\\frac{4(2)}{\\sqrt{41-4(2)^{2}-(3)^{2}}}=-2}\\hfill \\\\ \\hfill {f_y(x, y)} & =\\hfill & {-\\frac{y}{\\sqrt{41-4x^{2}-y^{2}}}\\ so\\ f_y(x_0, y_0)=-\\frac{3}{\\sqrt{41-4(2)^{2}-(3)^{2}}}=-\\frac{3}{4}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nNow we substitute these values into our definition equation:\r\n
[latex]\\begin{array}{ccc}\\hfill {L(x, y)} & =\\hfill & {f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)}\\hfill \\\\ \\hfill & =\\hfill & {4-2(x-2)-\\frac{3}{4}(y-3)} \\hfill \\\\ \\hfill & =\\hfill & {\\frac{41}{4}-2x-\\frac{3}{4}y.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\nLast, we substitute [latex]x=2.1[\/latex] and [latex]y=2.9[\/latex] into [latex]L(x, y)[\/latex]:\r\n
[latex]L(2.1, 2.9)=\\frac{41}{4}-2(2.1)-\\frac{3}{4}(2.9)=10.25-4.2-2.175=3.875.[\/latex]<\/div>\r\n \r\n\r\nThe approximate value of [latex]f(2.1, 2.9([\/latex] to four decimal places is\r\n
[latex]f(2.1, 2.9)=\\sqrt{41-4(2.1)^{2}-(2.9)^{2}}=\\sqrt{14.95}\\approx 3.8665[\/latex],<\/div>\r\n \r\n\r\nwhich corresponds to a [latex]0.2\\%[\/latex] error in approximation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Linear Approximations<\/h2>\n

You’ve seen linear approximations before in single-variable calculus. The idea is straightforward: use a tangent line to estimate function values near a known point.<\/p>\n

\n

Linear Approximations and Differentials<\/strong><\/p>\n

The formula for the linear approximation of a function [latex]f(x)[\/latex] at the point [latex]x=a[\/latex] is:<\/p>\n

[latex]y\\approx f(a)+f'(a)(x-a)[\/latex]<\/p>\n<\/section>\n

Here’s what this looks like graphically:<\/p>\n

\"A
Figure 4. Linear approximation of a function in one variable.<\/figcaption><\/figure>\n

The tangent line gives you a good approximation to [latex]f(x)[\/latex] for values of [latex]x[\/latex] reasonably close to [latex]x=a[\/latex]. When working with functions of two variables, we use the same idea\u2014just replace the tangent line with a tangent plane.<\/p>\n

\n

linear approximation of a function of two variables<\/h3>\n

Given a function [latex]z=f(x, y)[\/latex] with continuous partial derivatives at the point [latex](x_0, y_0)[\/latex], the linear approximation of [latex]f[\/latex] at [latex](x_0, y_0)[\/latex] is:<\/p>\n

[latex]L(x, y)=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0).[\/latex]<\/div>\n<\/section>\n

Notice this is exactly the equation of the tangent plane to the surface [latex]z=f(x, y)[\/latex] at [latex](x_0, y_0)[\/latex]. The linear approximation works because if you know the exact value of [latex]f(x, y)[\/latex] at [latex](x_0, y_0)[\/latex], then for points [latex](x, y)[\/latex] close to [latex](x_0, y_0)[\/latex], the tangent plane gives you a value close to the actual value of [latex]f(x, y)[\/latex] (Figure 5).<\/p>\n

\"A
Figure 5. Using a tangent plane for linear approximation at a point.<\/figcaption><\/figure>\n
Given the function [latex]f(x, y)=\\sqrt{41-4x^{2}-y^{2}}[\/latex], approximate [latex]f(2.1, 2.9)[\/latex] using point [latex](2, 3)[\/latex] for [latex](x_0, y_0)[\/latex]. What is the approximate value of [latex]f(2.1, 2.9)[\/latex] to four decimal places?<\/p>\n