{"id":628,"date":"2025-10-09T15:26:48","date_gmt":"2025-10-09T15:26:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=628"},"modified":"2025-10-29T14:28:34","modified_gmt":"2025-10-29T14:28:34","slug":"tangent-planes-and-linear-approximations-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/tangent-planes-and-linear-approximations-fresh-take\/","title":{"raw":"Tangent Planes and Linear Approximations: Fresh Take","rendered":"Tangent Planes and Linear Approximations: Fresh Take"},"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<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Think of a tangent plane as the 3D version of a tangent line. In single-variable calculus, you found the line that just barely touches a curve at one point. Now you're doing the same thing with surfaces\u2014finding the plane that grazes the surface at a specific point.<\/p>\r\n<p class=\"whitespace-normal break-words\">Here's the key insight: at any point on a smooth surface, there are infinitely many tangent lines. You get one in every direction you could move along the surface. If all these lines lie flat in the same plane, that's your tangent plane.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Formula\u00a0<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a surface [latex]z = f(x, y)[\/latex] at point [latex](x_0, y_0)[\/latex], the tangent plane equation is:<\/p>\r\n<p class=\"whitespace-normal break-words\" 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]<\/p>\r\nThe formula comes from finding two tangent lines. One goes in the [latex]x[\/latex]-direction, the other in the [latex]y[\/latex]-direction. Then you determine the plane that contains both lines. The partial derivatives [latex]f_x(x_0, y_0)[\/latex] and [latex]f_y(x_0, y_0)[\/latex] tell you how steeply the surface is tilting in each direction at that point.\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate the function<\/strong> at your point to find [latex]f(x_0, y_0)[\/latex]\u2014this gives you the z-coordinate<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find both partial derivatives<\/strong> [latex]f_x[\/latex] and [latex]f_y[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate the partials<\/strong> at [latex](x_0, y_0)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Plug everything into the formula<\/strong> and simplify<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Not every surface has a tangent plane at every point. A surface might have a sharp corner or crease. Approaching from different directions might give conflicting information about the slope. In these cases, a tangent plane won't exist. The function needs to be differentiable at the point\u2014something you'll explore more carefully later.<\/p>\r\n\r\n<\/div>\r\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)=x^{3}+x^{2}y+y^{2}-2x+3y-2[\/latex] at point [latex](-1,\\ 3)[\/latex].[reveal-answer q=\"fs-id1467793933124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793933124\"]\r\n<div style=\"text-align: center;\">[latex]z=7x+8y-3[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Linear Approximations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Linear approximation is about using something simple (a flat plane) to estimate something more complex (a curved surface). You've done this before with tangent lines in single-variable calculus. Now you're extending it to functions of two variables using tangent planes.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Core Idea<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When you need to approximate [latex]f(x, y)[\/latex] near a point [latex](x_0, y_0)[\/latex] where you know the exact value, use the tangent plane at that point. The tangent plane equation becomes your approximation formula:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[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]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is called the <strong>linear approximation<\/strong> of [latex]f[\/latex] at [latex](x_0, y_0)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Close to [latex](x_0, y_0)[\/latex], the surface and its tangent plane are nearly touching. The farther you move from [latex](x_0, y_0)[\/latex], the more the surface and plane separate. That's why this method works best for points near your base point.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate<\/strong> [latex]f(x_0, y_0)[\/latex] to get the exact function value at your base point<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find and evaluate<\/strong> both partial derivatives: [latex]f_x(x_0, y_0)[\/latex] and [latex]f_y(x_0, y_0)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Plug into the formula<\/strong> to build your linear approximation [latex]L(x, y)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Substitute<\/strong> the point you want to approximate into [latex]L(x, y)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Choose your base point [latex](x_0, y_0)[\/latex] strategically. Pick a point where calculations are easy and that's close to where you need the approximation. The closer your target point is to [latex](x_0, y_0)[\/latex], the better your approximation will be.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Given the function [latex]f\\,(x,\\ y)=e^{5-2x+3y}[\/latex] approximate [latex]f\\,(4.1,\\ 0.9)[\/latex] using point [latex](4,\\ 1)[\/latex] for [latex](x_0,\\ y_0)[\/latex]. What is the approximate value of [latex]f\\,(4.1,\\ 0.9)[\/latex] to four decimal places?[reveal-answer q=\"fs-id1467793733124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793733124\"]\r\n<div style=\"text-align: center;\">[latex]L\\,(x,\\ y)=6-2x+3y[\/latex], so [latex]L\\,(4.1,\\ 0.9)=6-2(4.1)+3(0.9)=0.5[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex]f\\,(4.1,\\ 0.9)=e^{5-2(4.1)+3(0.9)}=e^{-0.5}\\approx 0.6065.[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186157&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=gRYBbN5jrZM&amp;video_target=tpm-plugin-fvlqk2f9-gRYBbN5jrZM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.20_transcript.html\">transcript for \u201cCP 4.20\u201d here (opens in new window).<\/a><\/p>\r\n\r\n<\/section>\r\n<h2>Differentiability<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">In single-variable calculus, a function is differentiable at a point if its derivative exists there. For functions of two variables, differentiability means something more: the surface must be genuinely smooth at that point, with a well-defined tangent plane.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Key Issue<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Just having both partial derivatives [latex]f_x(x_0, y_0)[\/latex] and [latex]f_y(x_0, y_0)[\/latex] exist isn't enough. A function can have partial derivatives at a point but still have a corner or crease there. True differentiability requires the tangent plane to approximate the surface well in all directions near the point.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Formal Definition<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A function [latex]f(x, y)[\/latex] is differentiable at [latex](x_0, y_0)[\/latex] if you can write:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]f(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) + E(x, y)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where the error term [latex]E(x, y)[\/latex] shrinks faster than the distance to [latex](x_0, y_0)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x, y)\\to(x_0, y_0)}\\dfrac{E(x, y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}=0[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Practical Test<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Checking the formal definition is tedious. Here's an easier way: if [latex]f[\/latex], [latex]f_x[\/latex], and [latex]f_y[\/latex] are all continuous at [latex](x_0, y_0)[\/latex], then [latex]f[\/latex] is automatically differentiable there. Most functions you'll work with pass this test.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Important Connections<\/strong><\/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\"><strong>Differentiability implies continuity<\/strong>: If [latex]f[\/latex] is differentiable at a point, it's continuous there<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>The reverse isn't true<\/strong>: A function can be continuous at a point without being differentiable there<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Watch for trouble spots<\/strong>: Points where partial derivatives don't exist or aren't continuous are where differentiability typically fails<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Show that the function [latex]f\\,(x,\\ y)=3x-4y^{2}[\/latex] is differentiable at point [latex](-1,\\ 2)[\/latex].[reveal-answer q=\"fs-id1467793633124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1467793633124\"]\r\n<div style=\"text-align: center;\">[latex]f\\,(-1,\\ 2)=-19[\/latex], [latex]f_x\\,(-1,\\ 2)=3[\/latex], [latex]f_y\\,(-1,\\ 2)=-16[\/latex], [latex]E\\,(x,\\ y)=-4(y-2)^{2}[\/latex].<\/div>\r\n&nbsp;\r\n<div>[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to (x_0,\\ y_0)}\\frac{E\\,(x,\\ y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}}&amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to (-1,\\ 2)}\\frac{-4(y-2)^{2}}{\\sqrt{(x+1)^{2}+(y-2)^{2}}}}\\hfill \\\\ \\hfill &amp; \\leq\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to (-1,\\ 2)}\\frac{-4((x+1)^{2}+(y-2)^{2})}{\\sqrt{(x+1)^{2}+(y-2)^{2}}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}-4\\sqrt{(x+1)^{2}+(y-2)^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186158&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=sHwCY2cb57M&amp;video_target=tpm-plugin-vu4834g5-sHwCY2cb57M\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.21_transcript.html\">transcript for \u201cCP 4.21\u201d here (opens in new window).<\/a><\/p>\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Differentials<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The total differential gives you a way to estimate how much a function's output changes when you make small adjustments to its inputs. It's the multivariable version of the differential [latex]dy = f'(x)dx[\/latex] from single-variable calculus.<\/p>\r\n<p class=\"whitespace-normal break-words\">For a function [latex]z = f(x, y)[\/latex], the total differential at [latex](x_0, y_0)[\/latex] is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]dz = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]dx = \\Delta x[\/latex] and [latex]dy = \\Delta y[\/latex] are small changes in [latex]x[\/latex] and [latex]y[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This approximates the actual change [latex]\\Delta z = f(x + \\Delta x, y + \\Delta y) - f(x, y)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>What This Means<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The total differential tells you: if [latex]x[\/latex] changes by a small amount [latex]dx[\/latex] and [latex]y[\/latex] changes by [latex]dy[\/latex], then [latex]z[\/latex] changes by approximately [latex]dz[\/latex]. Each partial derivative shows how sensitive [latex]z[\/latex] is to changes in that particular variable.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find the partial derivatives<\/strong> [latex]f_x[\/latex] and [latex]f_y[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate them<\/strong> at your base point [latex](x_0, y_0)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Identify the changes<\/strong> [latex]dx[\/latex] and [latex]dy[\/latex] in your input variables<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Calculate<\/strong> [latex]dz = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Use this to approximate<\/strong> the new function value: [latex]f(x + dx, y + dy) \\approx f(x, y) + dz[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Practical Application<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Total differentials are especially useful for error propagation. If your measurements have small errors [latex]dx[\/latex] and [latex]dy[\/latex], the differential tells you approximately how much error to expect in your calculated result.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>For Three Variables<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The same idea extends naturally: [latex]dw = f_x dx + f_y dy + f_z dz[\/latex] for functions [latex]w = f(x, y, z)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the differential [latex]dz[\/latex] of the function [latex]f(x, y)=4y^{2}+x^{2}y-2xy[\/latex] and use it to approximate [latex]\\Delta z[\/latex] at point [latex](1, -1)[\/latex]. Use [latex] \\Delta x=0.03[\/latex] and [latex]\\Delta y=-0.02[\/latex]. What is the exact value of [latex]\\Delta z[\/latex]<span class=\"os-math-in-para\"><span id=\"MathJax-Element-249-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;\/p&gt; &lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;inline&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mrow&gt;&lt;mtext&gt;&amp;#x394;&lt;\/mtext&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mo&gt;?&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mrow&gt;&lt;mtext&gt;\u0394&lt;\/mtext&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mo&gt;?&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt; &lt;p&gt;\"><span id=\"MathJax-Span-7297\" class=\"math\"><span id=\"MathJax-Span-7298\" class=\"mrow\"><span id=\"MathJax-Span-7299\" class=\"semantics\"><span id=\"MathJax-Span-7300\" class=\"mrow\"><span id=\"MathJax-Span-7301\" class=\"mrow\"><span id=\"MathJax-Span-7304\" class=\"mo\">?<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>[reveal-answer q=\"723995431\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"723995431\"]\r\n<p style=\"padding-left: 300px; text-align: left;\">[latex]dz=0.18[\/latex]<\/p>\r\n<p style=\"padding-left: 300px; text-align: left;\">[latex]\\Delta z =f(1.03,-1.02)-f(1,-1)=0.180682[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186159&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=iYp0IP1wN0w&amp;video_target=tpm-plugin-l2ka20sb-iYp0IP1wN0w\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.22_transcript.html\">transcript for \u201cCP 4.22\u201d here (opens in new window).<\/a><\/center><\/section>","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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of a tangent plane as the 3D version of a tangent line. In single-variable calculus, you found the line that just barely touches a curve at one point. Now you&#8217;re doing the same thing with surfaces\u2014finding the plane that grazes the surface at a specific point.<\/p>\n<p class=\"whitespace-normal break-words\">Here&#8217;s the key insight: at any point on a smooth surface, there are infinitely many tangent lines. You get one in every direction you could move along the surface. If all these lines lie flat in the same plane, that&#8217;s your tangent plane.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Formula\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a surface [latex]z = f(x, y)[\/latex] at point [latex](x_0, y_0)[\/latex], the tangent plane equation is:<\/p>\n<p class=\"whitespace-normal break-words\" 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]<\/p>\n<p>The formula comes from finding two tangent lines. One goes in the [latex]x[\/latex]-direction, the other in the [latex]y[\/latex]-direction. Then you determine the plane that contains both lines. The partial derivatives [latex]f_x(x_0, y_0)[\/latex] and [latex]f_y(x_0, y_0)[\/latex] tell you how steeply the surface is tilting in each direction at that point.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Evaluate the function<\/strong> at your point to find [latex]f(x_0, y_0)[\/latex]\u2014this gives you the z-coordinate<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find both partial derivatives<\/strong> [latex]f_x[\/latex] and [latex]f_y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate the partials<\/strong> at [latex](x_0, y_0)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Plug everything into the formula<\/strong> and simplify<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Not every surface has a tangent plane at every point. A surface might have a sharp corner or crease. Approaching from different directions might give conflicting information about the slope. In these cases, a tangent plane won&#8217;t exist. The function needs to be differentiable at the point\u2014something you&#8217;ll explore more carefully later.<\/p>\n<\/div>\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)=x^{3}+x^{2}y+y^{2}-2x+3y-2[\/latex] at point [latex](-1,\\ 3)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1467793933124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1467793933124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]z=7x+8y-3[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2>Linear Approximations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Linear approximation is about using something simple (a flat plane) to estimate something more complex (a curved surface). You&#8217;ve done this before with tangent lines in single-variable calculus. Now you&#8217;re extending it to functions of two variables using tangent planes.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Core Idea<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When you need to approximate [latex]f(x, y)[\/latex] near a point [latex](x_0, y_0)[\/latex] where you know the exact value, use the tangent plane at that point. The tangent plane equation becomes your approximation formula:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[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]<\/p>\n<p class=\"whitespace-normal break-words\">This is called the <strong>linear approximation<\/strong> of [latex]f[\/latex] at [latex](x_0, y_0)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Close to [latex](x_0, y_0)[\/latex], the surface and its tangent plane are nearly touching. The farther you move from [latex](x_0, y_0)[\/latex], the more the surface and plane separate. That&#8217;s why this method works best for points near your base point.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Evaluate<\/strong> [latex]f(x_0, y_0)[\/latex] to get the exact function value at your base point<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find and evaluate<\/strong> both partial derivatives: [latex]f_x(x_0, y_0)[\/latex] and [latex]f_y(x_0, y_0)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Plug into the formula<\/strong> to build your linear approximation [latex]L(x, y)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Substitute<\/strong> the point you want to approximate into [latex]L(x, y)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Choose your base point [latex](x_0, y_0)[\/latex] strategically. Pick a point where calculations are easy and that&#8217;s close to where you need the approximation. The closer your target point is to [latex](x_0, y_0)[\/latex], the better your approximation will be.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Given the function [latex]f\\,(x,\\ y)=e^{5-2x+3y}[\/latex] approximate [latex]f\\,(4.1,\\ 0.9)[\/latex] using point [latex](4,\\ 1)[\/latex] for [latex](x_0,\\ y_0)[\/latex]. What is the approximate value of [latex]f\\,(4.1,\\ 0.9)[\/latex] to four decimal places?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1467793733124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1467793733124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]L\\,(x,\\ y)=6-2x+3y[\/latex], so [latex]L\\,(4.1,\\ 0.9)=6-2(4.1)+3(0.9)=0.5[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]f\\,(4.1,\\ 0.9)=e^{5-2(4.1)+3(0.9)}=e^{-0.5}\\approx 0.6065.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186157&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=gRYBbN5jrZM&amp;video_target=tpm-plugin-fvlqk2f9-gRYBbN5jrZM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.20_transcript.html\">transcript for \u201cCP 4.20\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<h2>Differentiability<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">In single-variable calculus, a function is differentiable at a point if its derivative exists there. For functions of two variables, differentiability means something more: the surface must be genuinely smooth at that point, with a well-defined tangent plane.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Key Issue<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Just having both partial derivatives [latex]f_x(x_0, y_0)[\/latex] and [latex]f_y(x_0, y_0)[\/latex] exist isn&#8217;t enough. A function can have partial derivatives at a point but still have a corner or crease there. True differentiability requires the tangent plane to approximate the surface well in all directions near the point.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Formal Definition<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A function [latex]f(x, y)[\/latex] is differentiable at [latex](x_0, y_0)[\/latex] if you can write:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]f(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) + E(x, y)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where the error term [latex]E(x, y)[\/latex] shrinks faster than the distance to [latex](x_0, y_0)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x, y)\\to(x_0, y_0)}\\dfrac{E(x, y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}=0[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Practical Test<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Checking the formal definition is tedious. Here&#8217;s an easier way: if [latex]f[\/latex], [latex]f_x[\/latex], and [latex]f_y[\/latex] are all continuous at [latex](x_0, y_0)[\/latex], then [latex]f[\/latex] is automatically differentiable there. Most functions you&#8217;ll work with pass this test.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Important Connections<\/strong><\/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\"><strong>Differentiability implies continuity<\/strong>: If [latex]f[\/latex] is differentiable at a point, it&#8217;s continuous there<\/li>\n<li class=\"whitespace-normal break-words\"><strong>The reverse isn&#8217;t true<\/strong>: A function can be continuous at a point without being differentiable there<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Watch for trouble spots<\/strong>: Points where partial derivatives don&#8217;t exist or aren&#8217;t continuous are where differentiability typically fails<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Show that the function [latex]f\\,(x,\\ y)=3x-4y^{2}[\/latex] is differentiable at point [latex](-1,\\ 2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1467793633124\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1467793633124\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]f\\,(-1,\\ 2)=-19[\/latex], [latex]f_x\\,(-1,\\ 2)=3[\/latex], [latex]f_y\\,(-1,\\ 2)=-16[\/latex], [latex]E\\,(x,\\ y)=-4(y-2)^{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div>[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to (x_0,\\ y_0)}\\frac{E\\,(x,\\ y)}{\\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}}& =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (-1,\\ 2)}\\frac{-4(y-2)^{2}}{\\sqrt{(x+1)^{2}+(y-2)^{2}}}}\\hfill \\\\ \\hfill & \\leq\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (-1,\\ 2)}\\frac{-4((x+1)^{2}+(y-2)^{2})}{\\sqrt{(x+1)^{2}+(y-2)^{2}}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to (2,-3)}-4\\sqrt{(x+1)^{2}+(y-2)^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186158&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=sHwCY2cb57M&amp;video_target=tpm-plugin-vu4834g5-sHwCY2cb57M\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.21_transcript.html\">transcript for \u201cCP 4.21\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<h2 data-type=\"title\">Differentials<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The total differential gives you a way to estimate how much a function&#8217;s output changes when you make small adjustments to its inputs. It&#8217;s the multivariable version of the differential [latex]dy = f'(x)dx[\/latex] from single-variable calculus.<\/p>\n<p class=\"whitespace-normal break-words\">For a function [latex]z = f(x, y)[\/latex], the total differential at [latex](x_0, y_0)[\/latex] is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]dz = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]dx = \\Delta x[\/latex] and [latex]dy = \\Delta y[\/latex] are small changes in [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This approximates the actual change [latex]\\Delta z = f(x + \\Delta x, y + \\Delta y) - f(x, y)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>What This Means<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The total differential tells you: if [latex]x[\/latex] changes by a small amount [latex]dx[\/latex] and [latex]y[\/latex] changes by [latex]dy[\/latex], then [latex]z[\/latex] changes by approximately [latex]dz[\/latex]. Each partial derivative shows how sensitive [latex]z[\/latex] is to changes in that particular variable.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Find the partial derivatives<\/strong> [latex]f_x[\/latex] and [latex]f_y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate them<\/strong> at your base point [latex](x_0, y_0)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Identify the changes<\/strong> [latex]dx[\/latex] and [latex]dy[\/latex] in your input variables<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Calculate<\/strong> [latex]dz = f_x(x_0, y_0)dx + f_y(x_0, y_0)dy[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Use this to approximate<\/strong> the new function value: [latex]f(x + dx, y + dy) \\approx f(x, y) + dz[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Practical Application<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Total differentials are especially useful for error propagation. If your measurements have small errors [latex]dx[\/latex] and [latex]dy[\/latex], the differential tells you approximately how much error to expect in your calculated result.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>For Three Variables<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The same idea extends naturally: [latex]dw = f_x dx + f_y dy + f_z dz[\/latex] for functions [latex]w = f(x, y, z)[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the differential [latex]dz[\/latex] of the function [latex]f(x, y)=4y^{2}+x^{2}y-2xy[\/latex] and use it to approximate [latex]\\Delta z[\/latex] at point [latex](1, -1)[\/latex]. Use [latex]\\Delta x=0.03[\/latex] and [latex]\\Delta y=-0.02[\/latex]. What is the exact value of [latex]\\Delta z[\/latex]<span class=\"os-math-in-para\"><span id=\"MathJax-Element-249-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;\/p&gt; &lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;inline&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mrow&gt;&lt;mtext&gt;&amp;#x394;&lt;\/mtext&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mo&gt;?&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mrow&gt;&lt;mtext&gt;\u0394&lt;\/mtext&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mo&gt;?&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt; &lt;p&gt;\"><span id=\"MathJax-Span-7297\" class=\"math\"><span id=\"MathJax-Span-7298\" class=\"mrow\"><span id=\"MathJax-Span-7299\" class=\"semantics\"><span id=\"MathJax-Span-7300\" class=\"mrow\"><span id=\"MathJax-Span-7301\" class=\"mrow\"><span id=\"MathJax-Span-7304\" class=\"mo\">?<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q723995431\">Show Solution<\/button><\/p>\n<div id=\"q723995431\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 300px; text-align: left;\">[latex]dz=0.18[\/latex]<\/p>\n<p style=\"padding-left: 300px; text-align: left;\">[latex]\\Delta z =f(1.03,-1.02)-f(1,-1)=0.180682[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186159&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=iYp0IP1wN0w&amp;video_target=tpm-plugin-l2ka20sb-iYp0IP1wN0w\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.22_transcript.html\">transcript for \u201cCP 4.22\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":15,"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-628","chapter","type-chapter","status-publish","hentry"],"part":428,"module-header":"fresh_take","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\/628","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\/628\/revisions"}],"predecessor-version":[{"id":1837,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/628\/revisions\/1837"}],"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\/628\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=628"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=628"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=628"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=628"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}