{"id":12430,"date":"2018-08-03t17:18:37","date_gmt":"2018-08-04t00:18:37","guid":{"rendered":"\/\/www.catharsisit.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":"\/\/www.catharsisit.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
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 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 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 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 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 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 <\/p>\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 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 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 approximate 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 then, using the differential, 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 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 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 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>\nformula for the linear approximation<\/h3>\n
example 1 — linearizing a parabola<\/h3>\n
solution<\/h4>\n
follow-up: interpreting the results<\/h4>\n
\n\n
\n x<\/em><\/th>\n f<\/em>(x<\/em>) = x<\/em>2<\/sup><\/th>\n l<\/em>(x<\/em>) = 2x<\/em> – 1<\/th>\n<\/tr>\n<\/thead>\n\n \n 1.1<\/td>\n 1.21<\/td>\n 1.2<\/td>\n<\/tr>\n \n 1.2<\/td>\n 1.44<\/td>\n 1.4<\/td>\n<\/tr>\n \n 1.5<\/td>\n 2.25<\/td>\n 2<\/td>\n<\/tr>\n \n 2<\/td>\n 4<\/td>\n 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n approximating using differentials<\/h2>\n
example 2 — using differentials with limited information<\/h3>\n
solution<\/h4>\n
example 3 — using differentials to approximate a value<\/h3>\n
using differentials. express your answer as a decimal rounded to the nearest hundred-thousandth.<\/p>\n
solution<\/h4>\n
. find its derivative (we’ll need it for the approximation formula).<\/p>\n
<\/p>\n
, we can estimate the required quantity.<\/p>\n
<\/p>\n
exact change versus approximate change<\/h2>\n
<\/p>\n
example 4 — comparing exact and approximate values<\/h3>\n
solution<\/h4>\n
application — finding zeros<\/h2>\n