{"id":12430,"date":"2018-08-03T17:18:37","date_gmt":"2018-08-04T00:18:37","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=12430"},"modified":"2018-08-03T17:18:37","modified_gmt":"2018-08-04T00:18:37","slug":"linear-approximation-ap-calculus-exam-review","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/","title":{"rendered":"Linear Approximation: AP Calculus Exam Review"},"content":{"rendered":"

What is linear approximation? Basically, it’s a method from calculus used to “straighten out” the graph of a function near a particular point. Scientists often use linear approximation to understand complicated relationships among variables.<\/p>\n

In this review article, we’ll explore the methods and applications of linear approximation. We’ll also take a look at plenty of examples along the way to better prepare you for the AP Calculus exams.<\/p>\n

Linear Approximation and Tangent Lines<\/h2>\n

By definition the linear approximation<\/strong> for a function f<\/em>(x<\/em>) at a point x<\/em> = a<\/em> is simply the equation of the tangent line<\/em> to the curve at that point. And that means that derivatives<\/em> are key! (Check out How to Find the Slope of a Line Tangent to a Curve<\/a> or Is the Derivative of a Function the Tangent Line?<\/a> for some background material.)<\/p>\n

\"three
Three different tangent lines for a curve y<\/em> = f<\/em>(x<\/em>).<\/figcaption><\/figure>\n

Formula for the Linear Approximation<\/h3>\n

Given a point x<\/em> = a<\/em> and a function f<\/em> that is differentiable at a<\/em>, the linear approximation<\/strong> L<\/em>(x<\/em>) for f<\/em> at x<\/em> = a<\/em> is:<\/p>\n

L<\/em>(x<\/em>) = f<\/em>(a<\/em>) + f<\/em> '(a<\/em>)(x<\/em> – a<\/em>)<\/p>\n

The main idea behind linearization is that the function L<\/em>(x<\/em>) does a pretty good job approximating values of f<\/em>(x<\/em>), at least when x<\/em> is near a<\/em>.<\/p>\n

In other words, L<\/em>(x<\/em>) ≈ f<\/em>(x<\/em>) whenever x<\/em> ≈ a<\/em>.<\/p>\n

Example 1 — Linearizing a Parabola<\/h3>\n

Find the linear approximation of the parabola f<\/em>(x<\/em>) = x<\/em>2<\/sup> at the point x<\/em> = 1.<\/p>\n

  A. x<\/em>2<\/sup> + 1<\/p>\n

  B. 2x<\/em> + 1<\/p>\n

  C. 2x<\/em> – 1<\/p>\n

  D. 2x<\/em> – 2<\/p>\n

Solution<\/h4>\n

C<\/strong>. <\/p>\n

Note that f<\/em> '(x<\/em>) = 2x<\/em> in this case. Using the formula above with a<\/em> = 1, we have:<\/p>\n

L<\/em>(x<\/em>) = f<\/em>(1) + f<\/em> '(1)(x<\/em> – 1)<\/p>\n

L<\/em>(x<\/em>) = 12<\/sup> + 2(1)(x<\/em> – 1) = 2x<\/em> – 1<\/p>\n

Follow-Up: Interpreting the Results<\/h4>\n

Clearly, the graph of the parabola f<\/em>(x<\/em>) = x<\/em>2<\/sup> is not a straight line. However, near any particular point, say x<\/em> = 1, the tangent line does a pretty good job following the direction of the curve. <\/p>\n

\"Graph
The parabola y<\/em> = x<\/em>2<\/sup> can be approximated by the line y<\/em> = 2x<\/em> – 1 when x<\/em> ≈ 1.<\/figcaption><\/figure>\n

How good is this approximation? Well, at x<\/em> = 1, it’s exact! L<\/em>(1) = 2(1) – 1 = 1, which is the same as f<\/em>(1) = 12<\/sup> = 1.<\/p>\n

But the further away you get from 1, the worse the approximation becomes.<\/p>\n\n\n\n\n\n\n\n\n
x<\/em><\/th>\nf<\/em>(x<\/em>) = x<\/em>2<\/sup><\/th>\nL<\/em>(x<\/em>) = 2x<\/em> – 1<\/th>\n<\/tr>\n<\/thead>\n
1.1<\/td>\n1.21<\/td>\n1.2<\/td>\n<\/tr>\n
1.2<\/td>\n1.44<\/td>\n1.4<\/td>\n<\/tr>\n
1.5<\/td>\n2.25<\/td>\n2<\/td>\n<\/tr>\n
2<\/td>\n4<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

<\/p>\n

Approximating Using Differentials<\/h2>\n

The formula for linear approximation can also be expressed in terms of differentials<\/strong>. Basically, a differential is a quantity that approximates a (small) change in one variable due to a (small) change in another. The differential of x<\/em> is dx<\/em>, and the differential of y<\/em> is dy<\/em>. <\/p>\n

Based upon the formula dy<\/em>\/dx<\/em> = f<\/em> '(x<\/em>), we may identify:<\/p>\n

dy<\/em> = f<\/em> '(x<\/em>) dx<\/em><\/p>\n

The related formula allows one to approximate near a particular fixed point:<\/p>\n

f<\/em>(x<\/em> + dx<\/em>) ≈ y<\/em> + dy<\/em><\/p>\n

Example 2 — Using Differentials With Limited Information<\/h3>\n

Suppose g<\/em>(5) = 30 and g<\/em> '(5) = -3. Estimate the value of g<\/em>(7).<\/p>\n

  A. 24<\/p>\n

  B. 27<\/p>\n

  C. 28<\/p>\n

  D. 33<\/p>\n

Solution<\/h4>\n

A<\/strong>. <\/p>\n

In this example, we do not know the expression for the function g<\/em>. Fortunately, we don’t need to know!<\/p>\n

First, observe that the change in x<\/em> is dx<\/em> = 7 – 5 = 2.<\/p>\n

Next, estimate the change in y<\/em> using the differential formula.<\/p>\n

dy<\/em> = g<\/em> '(x<\/em>) dx<\/em> = g<\/em> '(5) · 2 = (-3)(2) = -6.<\/p>\n

Finally, put it all together:<\/p>\n

g<\/em>(5 + 2) ≈ y<\/em> + dy<\/em> = g<\/em>(5) + (-6) = 30 + (-6) = 24<\/p>\n

Example 3 — Using Differentials to Approximate a Value<\/h3>\n

Approximate \"cube_root_1.1\" using differentials. Express your answer as a decimal rounded to the nearest hundred-thousandth.<\/p>\n

Solution<\/h4>\n

1.03333<\/strong>. <\/p>\n

Here, we should realize that even though the cube root of 1.1 is not easy to compute without a calculator, the cube root of 1 is trivial. So let’s use a<\/em> = 1 as our basis for estimation.<\/p>\n

Consider the function \"cube_root_x\". Find its derivative (we’ll need it for the approximation formula).<\/p>\n

\"derivative_of_cube_root_x\"<\/p>\n

Then, using the differential, \"differential_of_cube_root_x\", we can estimate the required quantity.<\/p>\n

\"estimating_cube_root_1.1\"<\/p>\n

Exact Change Versus Approximate Change<\/h2>\n

Sometimes we are interested in the exact<\/em> change of a function’s values over some interval. Suppose x<\/em> changes from x<\/em>1<\/sub> to x<\/em>2<\/sub>. Then the exact change<\/strong> in f<\/em>(x<\/em>) on that interval is:<\/p>\n

Δy<\/em> = f<\/em>(x<\/em>2<\/sub>) – f<\/em>(x<\/em>1<\/sub>)<\/p>\n

We also use the “Delta” notation for change in x<\/em>. In fact, Δx<\/em> and dx<\/em> typically mean the same thing:<\/p>\n

Δx<\/em> = dx<\/em> = x<\/em>2<\/sub> – x<\/em>1<\/sub><\/p>\n

However, while Δy<\/em> measures the exact change in the function’s value, dy<\/em> only estimates the change based on a derivative value.<\/p>\n

\"Linear<\/p>\n

Example 4 — Comparing Exact and Approximate Values<\/h3>\n

Let f<\/em>(x<\/em>) = cos(3x), and let L<\/em>(x<\/em>) be the linear approximation to f<\/em> at x<\/em> = π\/6. Which expression represents the absolute error in using L<\/em> to approximate f<\/em> at x<\/em> = π\/12?<\/p>\n

  A. π\/6 – √2<\/span>\/2<\/p>\n

  B. π\/4 – √2<\/span>\/2<\/p>\n

  C. √2<\/span>\/2 – π\/6<\/p>\n

  D. √2<\/span>\/2 – π\/4<\/p>\n

Solution<\/h4>\n

B.<\/strong><\/p>\n

Absolute error<\/strong> is the absolute difference between the approximate and exact values, that is, E<\/em> = | f<\/em>(a<\/em>) – L<\/em>(a<\/em>) |.<\/p>\n

Equivalently, E<\/em> = | Δy<\/em> – dy<\/em> |.<\/p>\n

Let’s compute dy<\/em> ( = f<\/em> '(x<\/em>) dx<\/em> ). Here, the change in x<\/em> is negative. dx<\/em> = π\/12 – π\/6 = -π\/12. Note that by the Chain Rule, we obtain: f<\/em> '(x<\/em>) = -3 sin(3x<\/em>). Putting it all together,<\/p>\n

dy<\/em> = -3 sin(3 π\/6 ) (-π\/12) = -3 sin(π\/2) (-π\/12) = 3π\/12 = π\/4<\/p>\n

Ok, next we compute the exact change.<\/p>\n

Δy<\/em> = f<\/em>(π\/12) – f<\/em>(π\/6) = cos(π\/4) – cos(π\/2) = √2<\/span>\/2<\/p>\n

Lastly, we take the absolute difference to compute the error, <\/p>\n

E<\/em> = | √2<\/span>\/2 – π\/4 | = π\/4 – √2<\/span>\/2.<\/p>\n

Application — Finding Zeros<\/h2>\n

Linear approximations also serve to find zeros of functions. In fact Newton’s Method<\/em> (see AP Calculus Review: Newton’s Method<\/a> for details) is nothing more than repeated linear approximations to target on to the nearest root of the function.<\/p>\n

The method is simple. Given a function f<\/em>, suppose that a zero for f<\/em> is located near x<\/em> = a<\/em>. Just linearize f<\/em> at x<\/em> = a<\/em>, producing a linear function L<\/em>(x<\/em>). Then the solution to L<\/em>(x<\/em>) = 0 should be fairly close to the true zero of the original function f<\/em>.<\/p>\n

Example 5 — Estimating Zeros<\/h3>\n

Estimate the zero of the function \"Polynomial_function\" using a tangent line approximation at x<\/em> = -1.<\/p>\n

  A. -1.48<\/p>\n

  B. -1.53<\/p>\n

  C. -1.62<\/p>\n

  D. -1.71<\/p>\n

Solution<\/h4>\n

D<\/strong>.<\/p>\n

Remember, The purpose of this question is to estimate<\/em> the zero. First of all, the tangent line approximation is nothing more than a linearization. We’ll need to know the derivative:<\/p>\n

\"Polynomial_function_derivative\"<\/p>\n

Then find the expression for L<\/em>(x<\/em>). Note that g<\/em>(-1) ≈ 2.13 and g<\/em> '(-1) = 3.<\/p>\n

L<\/em>(x<\/em>) = 2.13 + 3(x<\/em> – (-1)) = 5.13 + 3x<\/em>.<\/p>\n

Finally to find the zero, set L<\/em>(x<\/em>) = 0 and solve for x<\/em><\/p>\n

0 = 5.13 + 3x<\/em>  →  x<\/em> = -5.13\/3 = -1.71<\/p>\n","protected":false},"excerpt":{"rendered":"

What is linear approximation? Basically, it’s a method from calculus used to “straighten out” the graph of a function near a particular point. Scientists often use linear approximation to understand complicated relationships among variables. In this review article, we’ll explore the methods and applications of linear approximation. We’ll also take a look at plenty of […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-12430","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"\nLinear Approximation: AP Calculus Exam Review - Magoosh Blog | High School<\/title>\n<meta name=\"description\" content=\"In this review article, we'll take a look at plenty of examples of linear approximation to better prepare you for the AP Calculus exams.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Linear Approximation: AP Calculus Exam Review\" \/>\n<meta property=\"og:description\" content=\"In this review article, we'll take a look at plenty of examples of linear approximation to better prepare you for the AP Calculus exams.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/\" \/>\n<meta property=\"og:site_name\" content=\"Magoosh Blog | High School\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/MagooshSat\/\" \/>\n<meta property=\"article:published_time\" content=\"2018-08-04T00:18:37+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/magoosh.com\/hs\/files\/2017\/01\/graph_with_tangent_lines.png\" \/>\n<meta name=\"author\" content=\"Shaun Ault\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ShaunAultMath\" \/>\n<meta name=\"twitter:site\" content=\"@MagooshSAT_ACT\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Shaun Ault\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/\"},\"author\":{\"name\":\"Shaun Ault\",\"@id\":\"https:\/\/magoosh.com\/hs\/#\/schema\/person\/f01e70874cef77d6f6392c12c43f6b6f\"},\"headline\":\"Linear Approximation: AP Calculus Exam Review\",\"datePublished\":\"2018-08-04T00:18:37+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/\"},\"wordCount\":1147,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/magoosh.com\/hs\/#organization\"},\"keywords\":[\"AP Calculus\"],\"articleSection\":[\"AP\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/\",\"url\":\"https:\/\/magoosh.com\/hs\/ap\/linear-approximation-ap-calculus-exam-review\/\",\"name\":\"Linear Approximation: AP Calculus Exam Review - 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Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!"}],"_links":{"self":[{"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/posts\/12430","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/users\/223"}],"replies":[{"embeddable":true,"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/comments?post=12430"}],"version-history":[{"count":0,"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/posts\/12430\/revisions"}],"wp:attachment":[{"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/media?parent=12430"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/categories?post=12430"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/tags?post=12430"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/magoosh.com\/hs\/wp-json\/wp\/v2\/ppma_author?post=12430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}