{"id":198,"date":"2019-08-27T20:59:09","date_gmt":"2019-08-27T20:59:09","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=198"},"modified":"2019-12-02T21:43:02","modified_gmt":"2019-12-02T21:43:02","slug":"tangent-line-approximation","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/tangent-line-approximation\/","title":{"rendered":"Tangent Line Approximation"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>The Tangent Line Approximation &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nSuppose we want to find the tangent to a curve. Just how can we go about finding one?\n\n<\/p>\n\n\n\n<p>\nHere is one way:\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Pick a point $Q$ by clicking on the curve on the applet, which is the line that appears is the secant line between $P$ and $Q$.<br> <br><\/li><li> Now drag point $Q$ towards point $P$. <\/li><\/ul>\n\n\n\n<p>\nAs $Q$ approaches $P$, the secant line approximates the tangent line better \nand better.  The limiting position of the secant line as $Q$ approaches $P$ is \nthe tangent to the curve at $P$.\n\n<\/p>\n\n\n\n<p>\nIf the curve is given by $y=f(x)$ and $P$ has the coordinates $(x_0,y_0)$, \nthen the slope of the tangent line at $P$ is $f'(x_0)$, the derivative of f \nevaluated at $x_0$.\n\n<\/p>\n\n\n\n<p>\n<b>Let&#8217;s find the equation of the line tangent to the parabola at $(2,3)$.<\/b>\n\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Drag point $P$ to $(2,3)$.\n\t<br><br>\n\t<\/li><li> Now pick another point $Q$ on the parabola and drag $Q$ towards $P$ to \n\t\tfind the tangent to the curve at $P$.\n<\/li><\/ul>\n\n\n\n<p> The slope of the tangent is just $f'(x)$ evaluated at x. \\begin{eqnarray*} f(x) &amp;=&amp; x^2-1 \\\\ f'(x) &amp;=&amp; 2x \\\\ f'(2) &amp;=&amp; 4. \\end{eqnarray*} Now, the equation of the line can be written in  <b>point-slope form<\/b>.  The <strong>point-slope form<\/strong> of the equation of the line through the point $(x_0,y_0)$  with slope $m$ is given by  \\[y-y_0 = m(x-x_0).\\]   <\/p>\n\n\n\n<p>So we can write like this: \\begin{eqnarray*} y-y_0 &amp;=&amp; m(x-x_0)\\\\ y-y_0 &amp;=&amp; f'(x_0)(x-x_0)\\\\ y-3 &amp;=&amp; 4(x-2) \\end{eqnarray*} since the line passes through the point $(2,3)$ and has slope $4$. <\/p>\n\n\n\n<p> In <b>slope-intercept  form<\/b>,  the line with slope $m$  and $y$-intercept $b$ is given by  \\[y = mx+b.\\], so the equation of the tangent line becomes $$ y=4x-5. $$<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Drag $P$ along the parabola or enter the x-coordinate for point $P$.\n\t\t<p>\n\t\t  <\/p><center>\n\t\t    <applet codebase=\"..\/..\/java\/\" code=\"tla2.class\" width=\"320\" height=\"420\">\n\t\t    <\/applet>\n\t\t  <\/center>\n\t\t<p><\/p>\n\n\t<br><br>\n\t<\/li><li> Notice how the equation of the tangent line changes as you move \n\t\tpoint $P$.\n<\/li><\/ul>\n\n\n\n<p>\n<b>What happens when $x=0$ for this function?  What about as $|x|$ gets large?<\/b>\n\n<\/p>\n\n\n\n<p>\n<b>Now that we can find the tangent to a curve at a point, of what use is this?<\/b>\n\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> &#8220;Magnify&#8221; the parabola by zooming in on point $P$.\n<\/li><\/ul>\n\n\n\n<p>\n<b>Do you notice that as you zoom in on $P$ the curve looks more \nand more linear and is approximated better and better by the tangent line?<\/b>\n\n<\/p>\n\n\n\n<p>\n<b>Let&#8217;s get more specific:<\/b>\n\n<\/p>\n\n\n\n<p>\nNear $x_0$, we saw that $y=f(x)$ can be approximated by the tangent line \n$y-y_0=f'(x_0)(x-x_0)$.  Writing this as $y=y_0+f'(x_0)(x-x_0)$ and noting \nthat $y=f(x_0)$, we find that\n\n<\/p>\n\n\n\n<center>\n<p align=\"center\">\n\t\t$f(x)\\approx f(x_0) + f'(x_0)(x-x_0).$\n<\/p>\n<\/center>\n\n\n\n<p> Suppose $f$ can be differentiated $n+1$ times at each point in some interval containing $x_0$. Then for $x$ in this interval about $x_0$,  $$ f(x) \\approx f(x_0) + f'(x_0)(x-x_0) + \\frac{f&#8221;(x_0)}{2!}(x-x_0)^2 + \\dots + \\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n. $$. Notice that the right-hand side is just the 2-term <strong>Taylor  Expansion<\/strong> of $f(x)$.  <\/p>\n\n\n\n<p>\nIf we know that value of $f$ at $x_0$, this gives us a way to approximate \nthe value of $f$ at $x$ near $x_0$.  We do this by starting at $(x_0,f(x_0))$ \nand moving along the tangent line to approximate the value of the function \nat $x$.\n\n<\/p>\n\n\n\n<p>\n<b>Look at $f(x) = \\arctan{x}$.<\/b>\n\n<\/p>\n\n\n\n<center>\n    <applet codebase=\"..\/..\/java\/\" code=\"tla3.class\" width=\"320\" height=\"400\">\n    <\/applet>\n  <\/center>\n\n\n\n<p>\nLet&#8217;s use the tangent approximation $f(x) \\approx f(x_0)+f'(x_0)(x-x_0)$ \nto approximate $f(1.04)$:\n\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Now $f'(x) = \\left[\\frac{1}{1+x^2}\\right]$ so \n\t\t$f'(1)=\\left[\\frac{1}{1+1^2}\\right]=\\frac{1}{2}$.\n\t<br><br>\n\t<\/li><li> Let $x_0 = 1$ and $x = 1.04$.\n\t<br><br>\n\t<\/li><li> Then $f(1.04) \\approx f(1) + f'(1)(1.04 &#8211; 1) \\approx \\frac{\\pi}{4} + \n\t\t\\frac{1}{2}(0.04) \\approx 0.81$.\n<\/li><\/ul>\n\n\n\n<p>\n<b>How well does this approximate $\\arctan(1.04)$?<\/b>\n\n<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li> Display the tangent through $\\left( 1, \\frac{\\pi}{4}\\right)$.\n\t<br><br>\n\t<\/li><li> Zoom in on the point to see geometrically how close together the \n\t\tcurve and the tangent line are at $x = 1.04$.\n<\/li><\/ul>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<p>\n<\/p><h3>Key Concepts<\/h3>\n<p>\n<\/p><\/center>\n\n\n\n<ul class=\"wp-block-list\"><li> For the curve $y = f(x)$, the slope of the tangent line at a point \n\t\t$(x_0,y_0)$ on the curve is $f'(x_0)$.  The equation of the tangent line \n\t\tis given by\n\t\t$$ y-y_0 = f'(x_0)(x-x_0). $$\n\t<br><br>\n\t<\/li><li> For $x$ close to $x_0$, the value of $f(x)$ may be approximated by\n\t\t$$ f(x) \\approx f(x_0)+f'(x_0)(x-x_0). $$\n<\/li><\/ul>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ0910\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Tangent Line Approximation &#8211; HMC Calculus Tutorial Suppose we want to find the tangent to a curve. Just how can we go about finding one? Here is one way: Pick a point $Q$ by clicking on the curve on the applet, which is the line that appears is the secant line between $P$ and&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-198","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/198","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=198"}],"version-history":[{"count":5,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/198\/revisions"}],"predecessor-version":[{"id":1113,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/198\/revisions\/1113"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=198"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}