{"id":626,"date":"2025-10-09T15:26:54","date_gmt":"2025-10-09T15:26:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=626"},"modified":"2025-10-27T20:45:24","modified_gmt":"2025-10-27T20:45:24","slug":"tangent-planes-and-linear-approximations-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/tangent-planes-and-linear-approximations-learn-it-1\/","title":{"raw":"Tangent Planes and Linear Approximations: Learn It 1","rendered":"Tangent Planes and Linear Approximations: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find equations of tangent planes to surfaces<\/li>\r\n \t<li>Use tangent planes for approximation<\/li>\r\n \t<li>Explain when a function of two variables is differentiable<\/li>\r\n \t<li>Use the total differential to approximate the change in a function of two variables<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Tangent Planes<\/h2>\r\n<p class=\"whitespace-normal break-words\">You've already worked with tangent lines to curves when studying functions of one variable. For [latex]y=f(x)[\/latex], the slope of the tangent line at [latex]x=a[\/latex] is [latex]m=f^{\\prime}(a)[\/latex]. Now we're extending this idea to three dimensions\u2014finding tangent planes to surfaces.<\/p>\r\n<p class=\"whitespace-normal break-words\">In two dimensions, only one line can be tangent to a curve at a point. In three dimensions, things get more interesting\u2014infinitely many lines can be tangent to a surface at a single point. If all these tangent lines lie in the same plane, that plane is the tangent plane.<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.\r\n\r\n<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>tangent plane<\/h3>\r\nLet [latex]P_{0}=(x_0, y_0, z_0)[\/latex] be a point on a surface [latex]S[\/latex], and let [latex]C[\/latex] be any curve passing through [latex]P_0[\/latex] and lying entirely in [latex]S[\/latex]. If the tangent lines to all such curves [latex]C[\/latex] at [latex]P_0[\/latex] lie in the same plane, then this plane is called the <strong>tangent plane<\/strong> to [latex]S[\/latex] at [latex]P_0[\/latex].\r\n\r\n<\/section>[caption id=\"attachment_1230\" align=\"aligncenter\" width=\"697\"]<img class=\"wp-image-1230 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22215930\/4-4-1.jpeg\" alt=\"A surface S is shown with a point P0 = (x0, y0, z0). There are two intersecting curves shown on S that pass through P0. There are tangents drawn for each of these curves at P0, and these tangent lines create a plane, namely, the tangent plane at P0.\" width=\"697\" height=\"684\" \/> Figure 1. The tangent plane to a surface [latex]\\small{S}[\/latex] at a point [latex]\\small{P_{0}}[\/latex] contains all the tangent lines to curves in [latex]\\small{S}[\/latex] that pass through [latex]\\small{P_{0}}[\/latex].[\/caption]For a tangent plane to exist at a point, the function defining the surface needs to be differentiable at that point. We'll define differentiability more carefully later in this section. For now, here's the formula for finding tangent planes.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>equation of a tangent plane<\/h3>\r\nLet [latex]S[\/latex] be a surface defined by a differentiable function [latex]z=f\\,(x,\\ y)[\/latex], and let [latex]P_0=(x_0,\\ y_0)[\/latex] be a point in the domain of [latex]f[\/latex]. Then, the equation of the tangent plane to [latex]S[\/latex] at [latex]P_0[\/latex] is given by\r\n<div style=\"text-align: center;\">[latex]z=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>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Where Does This Formula Come From?<\/h3>\r\n<p class=\"whitespace-normal break-words\">To understand why this works, let's build it from the ground up using two tangent lines to the surface [latex]S[\/latex]. First, recall what you learned about tangent lines in single-variable calculus.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Definition of a Tangent Line<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]y = f(x)[\/latex] is differentiable at [latex]x = x_0[\/latex], then the equation of the tangent line to the curve at [latex]x=x_0[\/latex] is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]y = f(x_0)+f'(x_0)(x-x_0)[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Now, consider two curves on our surface [latex]S[\/latex]. The first curve is where [latex]S[\/latex] intersects the vertical plane [latex]x=x_0[\/latex]. The tangent line to this curve at our point is [latex]z=f(x_0, y_0)+f_y(x_0, y_0)(y-y_0)[\/latex]. The second curve is where [latex]S[\/latex] intersects the vertical plane [latex]y=y_0[\/latex]. Its tangent line is [latex]z=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">These two tangent lines determine our tangent plane. To find the equation of the plane, we need a normal vector. Let's find vectors parallel to each tangent line:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Parallel to the first tangent line: [latex]{\\bf{a}}={\\bf{j}}+f_y(x_0, y_0){\\bf{k}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Parallel to the second tangent line: [latex]{\\bf{b}}={\\bf{i}}+f_x(x_0, y_0){\\bf{k}}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The cross product of these vectors gives us a normal vector to the plane:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{a}}\\times{\\bf{b}}} &amp; =\\hfill &amp; {({\\bf{j}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{k}})\\times({\\bf{i}}+f_x\\,(x_0,\\ y_0)\\,{\\bf{k}})} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\begin{vmatrix}{\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}}\\\\0&amp;1&amp;f_y\\,(x_0,\\ y_0)\\\\1&amp;0&amp;f_x(x_0,\\ y_0)\\end{vmatrix}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {f_x\\,(x_0,\\ y_0)\\,{\\bf{i}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{j}}-{\\bf{k}}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\nThis vector is perpendicular to both tangent lines, so it's perpendicular to the tangent plane. We can now use this normal vector [latex]{\\bf{n}}[\/latex] and the point [latex]P_0=(x_0, y_0, f(x_0, y_0))[\/latex] to write the equation of the plane:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{n}}\\cdot ((x-x_0)\\,{\\bf{i}}+(y-y_0)\\,{\\bf{j}}+(z-f\\,(x_0,\\ y_0))\\,{\\bf{k}})} &amp; =\\hfill &amp; {0} \\hfill \\\\ \\hfill {(f_x\\,(x_0,\\ y_0)\\,{\\bf{i}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{j}}-{\\bf{k}})\\cdot ((x-x_0)\\,{\\bf{i}}+(y-y_0)\\,{\\bf{j}}+(z-f\\,(x_0,\\ y_0))\\,{\\bf{k}})} &amp; =\\hfill &amp; {0} \\hfill \\\\ \\hfill {f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)-(z-f\\,(x_0,\\ y_0))} &amp; =\\hfill &amp; {0} \\hfill \\\\ \\hfill \\end{array} [\/latex]<\/div>\r\n&nbsp;\r\n<p class=\"whitespace-normal break-words\">Solving for [latex]z[\/latex] gives us the tangent plane equation.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the equation of the tangent plane to the surface defined by the function\u00a0[latex]f\\,(x,\\ y)=2x^{2}-3xy+8y^{2}+2x-4y+4[\/latex] at point [latex](2,-1)[\/latex].[reveal-answer q=\"fs-id1367793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367793933124\"]\r\n<p style=\"text-align: left;\">First, we must calculate [latex]f_x\\,(x,\\ y)[\/latex] and [latex]f_y\\,(x,\\ y)[\/latex], then use our definition with [latex]x_0=2[\/latex] and [latex]y_0=-1[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f_x\\,(x,\\ y)} &amp; =\\hfill &amp; {4x-3y+2}\\hfill \\\\ \\hfill {f_y\\,(x,\\ y)} &amp; =\\hfill &amp; {-3x+16y-4} \\hfill \\\\ \\hfill {f\\,(2,\\ -1)} &amp; =\\hfill &amp; {2(2)^{2}-3(2)(-1)+8(-1)^{2}+2(2)-4(-1)+4=34.}\\hfill \\\\ \\hfill {f_x\\,(2,\\ -1)} &amp; =\\hfill &amp; {4(2)-3(-1)+2=13} \\hfill \\\\ \\hfill {f_y\\,(2,\\ -1)} &amp; =\\hfill &amp; {-3(2)+16(-1)-4=-26.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThen our equation becomes\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {z} &amp; =\\hfill &amp; {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 {z} &amp; =\\hfill &amp; {34+13(x-2)-26(y-(-1))}\\hfill \\\\ \\hfill {z} &amp; =\\hfill &amp; {34+13x-26-26y-26}\\hfill \\\\ \\hfill {z} &amp; =\\hfill &amp; {13x-26y-18.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n(See the following figure).\r\n<div>\r\n\r\n[caption id=\"attachment_1232\" align=\"aligncenter\" width=\"764\"]<img class=\"size-full wp-image-1232\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22220223\/4-4-2.jpeg\" alt=\"A curved surface f(x, y) = 2x2 \u2013 3xy + 8y2 + 2x + 4y + 4 with tangent plane z = 13x \u2013 26y \u2013 18 at point (2, \u20131, 34).\" width=\"764\" height=\"474\" \/> Figure 2.\u00a0Calculating the equation of a tangent plane to a given surface at a given point.[\/caption]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the equation of the tangent plane to the surface defined by the function [latex]f\\,(x,y)=\\sin{(2x)}\\cos{(3y)}[\/latex] at point [latex](\\pi \/3,\\pi \/4)[\/latex].[reveal-answer q=\"fs-id1377793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1377793933124\"]\r\n<p style=\"text-align: left;\">First, calculate [latex]f_x\\,(x,\\ y)[\/latex] and [latex]f_y\\,(x,\\ y)[\/latex], then use our definition with [latex]x_0=\\pi \/3[\/latex] and [latex]y_0=\\pi \/4[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f_x\\,(x,\\ y)} &amp; =\\hfill &amp; {2\\cos{(2x)}\\cos{(3y)}}\\hfill \\\\ \\hfill {f_y\\,(x,y)} &amp; =\\hfill &amp; {-3\\sin{(2x)}\\sin{(3y)}}\\hfill \\\\ \\hfill {f,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} &amp; =\\hfill &amp; {\\sin{(2(\\frac{\\pi}{3}))}\\cos{(3\\,(\\frac{\\pi}{4}))}=(\\frac{\\sqrt{3}}{2})(-\\frac{\\sqrt{2}}{2})=-\\frac{\\sqrt{6}}{4}}\\hfill \\\\ \\hfill {f_x\\,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} &amp; =\\hfill &amp; {2\\cos{(2(\\frac{\\pi}{3}))}\\cos{(3,(\\frac{\\pi}{4}))}=2\\,(-\\frac{1}{2})(-\\frac{\\sqrt{2}}{2})=\\frac{\\sqrt{2}}{2}}\\hfill \\\\ \\hfill {f_y\\,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} &amp; =\\hfill &amp; {-3\\sin{(2\\,(\\frac{\\pi}{3}))}\\sin{(3\\,(\\frac{\\pi}{4}))}=-3\\,(\\frac{\\sqrt{3}}{2})(\\frac{\\sqrt{2}}{2})=-\\frac{3\\sqrt{6}}{4}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n\r\nThen our definition becomes\r\n\r\n[latex]\\begin{alignat}{2} \\hspace{6cm}z&amp;=f(x_0,y_0)+f_x(x_0,y_0) (x-x_0)+f_y(x_0,y_0) (y-y_0)\\\\\r\n\r\nz&amp;=-\\frac{\\sqrt6}{4}+\\frac{\\sqrt2}{2}\\left(x-\\frac{\\pi}3\\right)-\\frac{3\\sqrt6}4\\left(y-\\frac{\\pi}4\\right)&amp;\\quad\\\\\r\n\r\nz&amp;=\\frac{\\sqrt2}{2}x-\\frac{3\\sqrt6}{4}y-\\frac{\\sqrt6}4-\\frac{\\pi\\sqrt2}6+\\frac{3\\pi\\sqrt6}{16}.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n<\/div>\r\n<div><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">[\/hidden-answer]<\/span><\/div>\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">When Tangent Planes Don't Exist<\/h3>\r\n<p class=\"whitespace-normal break-words\">Not every surface has a tangent plane at every point. Consider this function:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]f\\,(x,\\ y)=\\begin{cases}\\frac{xy}{\\sqrt{x^{2}+y^{2}}}\\ \\ (x,\\ y)\\neq(0,\\ 0) \\\\ 0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (x,\\ y)=(0,\\ 0).\\end{cases}[\/latex]<\/div>\r\nHere's what its graph looks like:\r\n\r\n[caption id=\"attachment_1234\" align=\"aligncenter\" width=\"581\"]<img class=\"size-full wp-image-1234\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22220328\/4-4-3.jpeg\" alt=\"A curved surface that passes through (0, 0, 0) and that folds up on either side of the z axis.\" width=\"581\" height=\"418\" \/> Figure 3.\u00a0Graph of a function that does not have a tangent plane at the origin.[\/caption]\r\n<p class=\"whitespace-normal break-words\">When either [latex]x=0[\/latex] or [latex]y=0[\/latex], we have [latex]f(x, y)=0[\/latex]. The function doesn't change along either axis, so [latex]f_x(x, 0)=f_y(0, y)=0[\/latex]. As [latex]x[\/latex] or [latex]y[\/latex] approach zero along the axes, these partial derivatives stay zero. Using our formula, we'd get [latex]z=0[\/latex] as the equation of the tangent plane.<\/p>\r\n<p class=\"whitespace-normal break-words\">But watch what happens when we approach the origin from a different direction. Let's approach along the line [latex]y=x[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]f\\,(x,\\ x)=\\dfrac{x(x)}{\\sqrt{x^{2}+(x)^{2}}}=\\dfrac{x^{2}}{\\sqrt{2x^{2}}}=\\dfrac{|x|}{\\sqrt{2}}.[\/latex]<\/div>\r\n<p class=\"whitespace-normal break-words\">When [latex]x&gt;0[\/latex], this curve has slope [latex]\\sqrt{2}\/2[\/latex]. When [latex]x&lt;0[\/latex], the slope is [latex]-\\sqrt{2}\/2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This creates a problem. Our definition of tangent plane assumes all tangent lines through a point lie in the same plane. That's clearly not happening here\u2014we're getting different slopes depending on which direction we approach from. When we discuss differentiability later, we'll see that this function isn't differentiable at the origin, which is why a tangent plane doesn't exist there.<\/p>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find equations of tangent planes to surfaces<\/li>\n<li>Use tangent planes for approximation<\/li>\n<li>Explain when a function of two variables is differentiable<\/li>\n<li>Use the total differential to approximate the change in a function of two variables<\/li>\n<\/ul>\n<\/section>\n<h2>Tangent Planes<\/h2>\n<p class=\"whitespace-normal break-words\">You&#8217;ve already worked with tangent lines to curves when studying functions of one variable. For [latex]y=f(x)[\/latex], the slope of the tangent line at [latex]x=a[\/latex] is [latex]m=f^{\\prime}(a)[\/latex]. Now we&#8217;re extending this idea to three dimensions\u2014finding tangent planes to surfaces.<\/p>\n<p class=\"whitespace-normal break-words\">In two dimensions, only one line can be tangent to a curve at a point. In three dimensions, things get more interesting\u2014infinitely many lines can be tangent to a surface at a single point. If all these tangent lines lie in the same plane, that plane is the tangent plane.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>tangent plane<\/h3>\n<p>Let [latex]P_{0}=(x_0, y_0, z_0)[\/latex] be a point on a surface [latex]S[\/latex], and let [latex]C[\/latex] be any curve passing through [latex]P_0[\/latex] and lying entirely in [latex]S[\/latex]. If the tangent lines to all such curves [latex]C[\/latex] at [latex]P_0[\/latex] lie in the same plane, then this plane is called the <strong>tangent plane<\/strong> to [latex]S[\/latex] at [latex]P_0[\/latex].<\/p>\n<\/section>\n<figure id=\"attachment_1230\" aria-describedby=\"caption-attachment-1230\" style=\"width: 697px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1230 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22215930\/4-4-1.jpeg\" alt=\"A surface S is shown with a point P0 = (x0, y0, z0). There are two intersecting curves shown on S that pass through P0. There are tangents drawn for each of these curves at P0, and these tangent lines create a plane, namely, the tangent plane at P0.\" width=\"697\" height=\"684\" \/><figcaption id=\"caption-attachment-1230\" class=\"wp-caption-text\">Figure 1. The tangent plane to a surface [latex]\\small{S}[\/latex] at a point [latex]\\small{P_{0}}[\/latex] contains all the tangent lines to curves in [latex]\\small{S}[\/latex] that pass through [latex]\\small{P_{0}}[\/latex].<\/figcaption><\/figure>\n<p>For a tangent plane to exist at a point, the function defining the surface needs to be differentiable at that point. We&#8217;ll define differentiability more carefully later in this section. For now, here&#8217;s the formula for finding tangent planes.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>equation of a tangent plane<\/h3>\n<p>Let [latex]S[\/latex] be a surface defined by a differentiable function [latex]z=f\\,(x,\\ y)[\/latex], and let [latex]P_0=(x_0,\\ y_0)[\/latex] be a point in the domain of [latex]f[\/latex]. Then, the equation of the tangent plane to [latex]S[\/latex] at [latex]P_0[\/latex] is given by<\/p>\n<div style=\"text-align: center;\">[latex]z=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<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Where Does This Formula Come From?<\/h3>\n<p class=\"whitespace-normal break-words\">To understand why this works, let&#8217;s build it from the ground up using two tangent lines to the surface [latex]S[\/latex]. First, recall what you learned about tangent lines in single-variable calculus.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Definition of a Tangent Line<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]y = f(x)[\/latex] is differentiable at [latex]x = x_0[\/latex], then the equation of the tangent line to the curve at [latex]x=x_0[\/latex] is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]y = f(x_0)+f'(x_0)(x-x_0)[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Now, consider two curves on our surface [latex]S[\/latex]. The first curve is where [latex]S[\/latex] intersects the vertical plane [latex]x=x_0[\/latex]. The tangent line to this curve at our point is [latex]z=f(x_0, y_0)+f_y(x_0, y_0)(y-y_0)[\/latex]. The second curve is where [latex]S[\/latex] intersects the vertical plane [latex]y=y_0[\/latex]. Its tangent line is [latex]z=f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">These two tangent lines determine our tangent plane. To find the equation of the plane, we need a normal vector. Let&#8217;s find vectors parallel to each tangent line:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Parallel to the first tangent line: [latex]{\\bf{a}}={\\bf{j}}+f_y(x_0, y_0){\\bf{k}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Parallel to the second tangent line: [latex]{\\bf{b}}={\\bf{i}}+f_x(x_0, y_0){\\bf{k}}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The cross product of these vectors gives us a normal vector to the plane:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{a}}\\times{\\bf{b}}} & =\\hfill & {({\\bf{j}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{k}})\\times({\\bf{i}}+f_x\\,(x_0,\\ y_0)\\,{\\bf{k}})} \\hfill \\\\ \\hfill & =\\hfill & {\\begin{vmatrix}{\\bf{i}}&{\\bf{j}}&{\\bf{k}}\\\\0&1&f_y\\,(x_0,\\ y_0)\\\\1&0&f_x(x_0,\\ y_0)\\end{vmatrix}} \\hfill \\\\ \\hfill & =\\hfill & {f_x\\,(x_0,\\ y_0)\\,{\\bf{i}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{j}}-{\\bf{k}}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>This vector is perpendicular to both tangent lines, so it&#8217;s perpendicular to the tangent plane. We can now use this normal vector [latex]{\\bf{n}}[\/latex] and the point [latex]P_0=(x_0, y_0, f(x_0, y_0))[\/latex] to write the equation of the plane:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {{\\bf{n}}\\cdot ((x-x_0)\\,{\\bf{i}}+(y-y_0)\\,{\\bf{j}}+(z-f\\,(x_0,\\ y_0))\\,{\\bf{k}})} & =\\hfill & {0} \\hfill \\\\ \\hfill {(f_x\\,(x_0,\\ y_0)\\,{\\bf{i}}+f_y\\,(x_0,\\ y_0)\\,{\\bf{j}}-{\\bf{k}})\\cdot ((x-x_0)\\,{\\bf{i}}+(y-y_0)\\,{\\bf{j}}+(z-f\\,(x_0,\\ y_0))\\,{\\bf{k}})} & =\\hfill & {0} \\hfill \\\\ \\hfill {f_x\\,(x_0,\\ y_0)(x-x_0)+f_y\\,(x_0,\\ y_0)(y-y_0)-(z-f\\,(x_0,\\ y_0))} & =\\hfill & {0} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p class=\"whitespace-normal break-words\">Solving for [latex]z[\/latex] gives us the tangent plane equation.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the equation of the tangent plane to the surface defined by the function\u00a0[latex]f\\,(x,\\ y)=2x^{2}-3xy+8y^{2}+2x-4y+4[\/latex] at point [latex](2,-1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1367793933124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1367793933124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">First, we must calculate [latex]f_x\\,(x,\\ y)[\/latex] and [latex]f_y\\,(x,\\ y)[\/latex], then use our definition with [latex]x_0=2[\/latex] and [latex]y_0=-1[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f_x\\,(x,\\ y)} & =\\hfill & {4x-3y+2}\\hfill \\\\ \\hfill {f_y\\,(x,\\ y)} & =\\hfill & {-3x+16y-4} \\hfill \\\\ \\hfill {f\\,(2,\\ -1)} & =\\hfill & {2(2)^{2}-3(2)(-1)+8(-1)^{2}+2(2)-4(-1)+4=34.}\\hfill \\\\ \\hfill {f_x\\,(2,\\ -1)} & =\\hfill & {4(2)-3(-1)+2=13} \\hfill \\\\ \\hfill {f_y\\,(2,\\ -1)} & =\\hfill & {-3(2)+16(-1)-4=-26.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Then our equation becomes<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {z} & =\\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 {z} & =\\hfill & {34+13(x-2)-26(y-(-1))}\\hfill \\\\ \\hfill {z} & =\\hfill & {34+13x-26-26y-26}\\hfill \\\\ \\hfill {z} & =\\hfill & {13x-26y-18.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>(See the following figure).<\/p>\n<div>\n<figure id=\"attachment_1232\" aria-describedby=\"caption-attachment-1232\" style=\"width: 764px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1232\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22220223\/4-4-2.jpeg\" alt=\"A curved surface f(x, y) = 2x2 \u2013 3xy + 8y2 + 2x + 4y + 4 with tangent plane z = 13x \u2013 26y \u2013 18 at point (2, \u20131, 34).\" width=\"764\" height=\"474\" \/><figcaption id=\"caption-attachment-1232\" class=\"wp-caption-text\">Figure 2.\u00a0Calculating the equation of a tangent plane to a given surface at a given point.<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the equation of the tangent plane to the surface defined by the function [latex]f\\,(x,y)=\\sin{(2x)}\\cos{(3y)}[\/latex] at point [latex](\\pi \/3,\\pi \/4)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1377793933124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1377793933124\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">First, calculate [latex]f_x\\,(x,\\ y)[\/latex] and [latex]f_y\\,(x,\\ y)[\/latex], then use our definition with [latex]x_0=\\pi \/3[\/latex] and [latex]y_0=\\pi \/4[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {f_x\\,(x,\\ y)} & =\\hfill & {2\\cos{(2x)}\\cos{(3y)}}\\hfill \\\\ \\hfill {f_y\\,(x,y)} & =\\hfill & {-3\\sin{(2x)}\\sin{(3y)}}\\hfill \\\\ \\hfill {f,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} & =\\hfill & {\\sin{(2(\\frac{\\pi}{3}))}\\cos{(3\\,(\\frac{\\pi}{4}))}=(\\frac{\\sqrt{3}}{2})(-\\frac{\\sqrt{2}}{2})=-\\frac{\\sqrt{6}}{4}}\\hfill \\\\ \\hfill {f_x\\,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} & =\\hfill & {2\\cos{(2(\\frac{\\pi}{3}))}\\cos{(3,(\\frac{\\pi}{4}))}=2\\,(-\\frac{1}{2})(-\\frac{\\sqrt{2}}{2})=\\frac{\\sqrt{2}}{2}}\\hfill \\\\ \\hfill {f_y\\,(\\frac{\\pi}{3},\\ \\frac{\\pi}{4})} & =\\hfill & {-3\\sin{(2\\,(\\frac{\\pi}{3}))}\\sin{(3\\,(\\frac{\\pi}{4}))}=-3\\,(\\frac{\\sqrt{3}}{2})(\\frac{\\sqrt{2}}{2})=-\\frac{3\\sqrt{6}}{4}.} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<div>\n<p>Then our definition becomes<\/p>\n<p>[latex]\\begin{alignat}{2} \\hspace{6cm}z&=f(x_0,y_0)+f_x(x_0,y_0) (x-x_0)+f_y(x_0,y_0) (y-y_0)\\\\    z&=-\\frac{\\sqrt6}{4}+\\frac{\\sqrt2}{2}\\left(x-\\frac{\\pi}3\\right)-\\frac{3\\sqrt6}4\\left(y-\\frac{\\pi}4\\right)&\\quad\\\\    z&=\\frac{\\sqrt2}{2}x-\\frac{3\\sqrt6}{4}y-\\frac{\\sqrt6}4-\\frac{\\pi\\sqrt2}6+\\frac{3\\pi\\sqrt6}{16}.\\\\    \\end{alignat}[\/latex]<\/p>\n<\/div>\n<div><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"><\/div>\n<\/div>\n<p><\/span><\/div>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">When Tangent Planes Don&#8217;t Exist<\/h3>\n<p class=\"whitespace-normal break-words\">Not every surface has a tangent plane at every point. Consider this function:<\/p>\n<div style=\"text-align: center;\">[latex]f\\,(x,\\ y)=\\begin{cases}\\frac{xy}{\\sqrt{x^{2}+y^{2}}}\\ \\ (x,\\ y)\\neq(0,\\ 0) \\\\ 0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (x,\\ y)=(0,\\ 0).\\end{cases}[\/latex]<\/div>\n<p>Here&#8217;s what its graph looks like:<\/p>\n<figure id=\"attachment_1234\" aria-describedby=\"caption-attachment-1234\" style=\"width: 581px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1234\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/22220328\/4-4-3.jpeg\" alt=\"A curved surface that passes through (0, 0, 0) and that folds up on either side of the z axis.\" width=\"581\" height=\"418\" \/><figcaption id=\"caption-attachment-1234\" class=\"wp-caption-text\">Figure 3.\u00a0Graph of a function that does not have a tangent plane at the origin.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">When either [latex]x=0[\/latex] or [latex]y=0[\/latex], we have [latex]f(x, y)=0[\/latex]. The function doesn&#8217;t change along either axis, so [latex]f_x(x, 0)=f_y(0, y)=0[\/latex]. As [latex]x[\/latex] or [latex]y[\/latex] approach zero along the axes, these partial derivatives stay zero. Using our formula, we&#8217;d get [latex]z=0[\/latex] as the equation of the tangent plane.<\/p>\n<p class=\"whitespace-normal break-words\">But watch what happens when we approach the origin from a different direction. Let&#8217;s approach along the line [latex]y=x[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]f\\,(x,\\ x)=\\dfrac{x(x)}{\\sqrt{x^{2}+(x)^{2}}}=\\dfrac{x^{2}}{\\sqrt{2x^{2}}}=\\dfrac{|x|}{\\sqrt{2}}.[\/latex]<\/div>\n<p class=\"whitespace-normal break-words\">When [latex]x>0[\/latex], this curve has slope [latex]\\sqrt{2}\/2[\/latex]. When [latex]x<0[\/latex], the slope is [latex]-\\sqrt{2}\/2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This creates a problem. Our definition of tangent plane assumes all tangent lines through a point lie in the same plane. That&#8217;s clearly not happening here\u2014we&#8217;re getting different slopes depending on which direction we approach from. When we discuss differentiability later, we&#8217;ll see that this function isn&#8217;t differentiable at the origin, which is why a tangent plane doesn&#8217;t exist there.<\/p>\n","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-626","chapter","type-chapter","status-publish","hentry"],"part":428,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/626","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/626\/revisions"}],"predecessor-version":[{"id":1793,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/626\/revisions\/1793"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/428"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/626\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=626"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=626"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=626"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=626"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}