\n2<\/td>\n | 4<\/td>\n | 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n <\/p>\n approximating using differentials<\/h2>\nthe 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>\nbased upon the formula dy<\/em>\/dx<\/em> = f<\/em> '(x<\/em>), we may identify:<\/p>\ndy<\/em> = f<\/em> '(x<\/em>) dx<\/em><\/p>\nthe related formula allows one to approximate near a particular fixed point:<\/p>\n f<\/em>(x<\/em> + dx<\/em>) ≈ y<\/em> + dy<\/em><\/p>\nexample 2 — using differentials with limited information<\/h3>\nsuppose 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>\na<\/strong>. <\/p>\nin this example, we do not know the expression for the function g<\/em>. fortunately, we don’t need to know!<\/p>\nfirst, observe that the change in x<\/em> is dx<\/em> = 7 – 5 = 2.<\/p>\nnext, estimate the change in y<\/em> using the differential formula.<\/p>\ndy<\/em> = g<\/em> '(x<\/em>) dx<\/em> = g<\/em> '(5) · 2 = (-3)(2) = -6.<\/p>\nfinally, put it all together:<\/p>\n g<\/em>(5 + 2) ≈ y<\/em> + dy<\/em> = g<\/em>(5) + (-6) = 30 + (-6) = 24<\/p>\nexample 3 — using differentials to approximate a value<\/h3>\napproximate using differentials. express your answer as a decimal rounded to the nearest hundred-thousandth.<\/p>\n solution<\/h4>\n1.03333<\/strong>. <\/p>\nhere, 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>\nconsider the function . find its derivative (we’ll need it for the approximation formula).<\/p>\n <\/p>\n then, using the differential, , we can estimate the required quantity.<\/p>\n <\/p>\n exact change versus approximate change<\/h2>\nsometimes 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>\nδy<\/em> = f<\/em>(x<\/em>2<\/sub>) – f<\/em>(x<\/em>1<\/sub>)<\/p>\nwe also use the “delta” notation for change in x<\/em>. in fact, δx<\/em> and dx<\/em> typically mean the same thing:<\/p>\nδx<\/em> = dx<\/em> = x<\/em>2<\/sub> – x<\/em>1<\/sub><\/p>\nhowever, 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<\/p>\n example 4 — comparing exact and approximate values<\/h3>\nlet 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>\nsolution<\/h4>\nb.<\/strong><\/p>\nabsolute 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>\nequivalently, e<\/em> = | δy<\/em> – dy<\/em> |.<\/p>\nlet’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>\ndy<\/em> = -3 sin(3 π\/6 ) (-π\/12) = -3 sin(π\/2) (-π\/12) = 3π\/12 = π\/4<\/p>\nok, 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>\nlastly, we take the absolute difference to compute the error, <\/p>\n e<\/em> = | √2<\/span>\/2 – π\/4 | = π\/4 – √2<\/span>\/2.<\/p>\napplication — finding zeros<\/h2>\nlinear 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 |