{"id":11920,"date":"2018-04-26t11:08:00","date_gmt":"2018-04-26t18:08:00","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=11920"},"modified":"2018-04-26t18:08:10","modified_gmt":"2018-04-27t01:08:10","slug":"ap-calculus-review-applications-derivatives","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-review-applications-derivatives\/","title":{"rendered":"ap calculus review: applications of derivatives"},"content":{"rendered":"
calculus is primarily the study of rates of change<\/em>. however, there are numerous applications of derivatives beyond just finding rates and velocities.<\/p>\n in this review article, we will highlight the most important applications of derivatives for the ap calculus ab\/bc exams.<\/p>\n here is a list of some of the essential derivative applications.<\/p>\n perhaps the most important application of the derivative is in finding rates of change. in fact, this is why newton (and leibniz independently) invented calculus in the first place!<\/p>\n suppose y<\/em> = f<\/em>(x<\/em>) is a function that models some real-world behavior. here, the input is x<\/em> and output is y<\/em>. a natural question in the sciences is: how does the output change as one changes the input?<\/em> in other words, what is the change in y<\/em> resulting from a change in x<\/em>?<\/p>\n for most purposes, it makes sense to consider the ratio of the changes. for example, if y<\/em> changes twice as quickly as x<\/em>, then that would be a rate of change of 2.<\/p>\n so this leads to the concept of average rate of change, \u0394y<\/em>\/\u0394x<\/em>. (see ap calculus review: average rate of change<\/a> for more about this.)<\/p>\n finally, by allowing the change in input to approach 0 (by taking a limit<\/strong>), we get the definition of derivative:<\/p>\n of course after you learn the definition, then you get to learn all of the shortcut rules for derivatives that let you avoid taking limits at all!<\/p>\n suppose the population of a small city can be modeled by p<\/em> = 40532 e<\/em>0.03t<\/sup>, where p<\/em> is total population and t<\/em> is time in years after the year 2000. how quickly was the population increasing in 2002?<\/p>\n the phrase “how quickly” means “at what rate”? so we use a derivative to find the rate of change.<\/p>\n p<\/em>\u00a0‘ = 40532 e<\/em>0.03t<\/sup> (0.03) = 1215.96 e<\/em>0.03t<\/sup><\/p>\n the year 2002 corresponds to t<\/em> = 2.<\/p>\n p<\/em>\u00a0‘(2) = 1215.96 e<\/em>0.03(2)<\/sup> = 1291.15<\/p>\n thus the city was growing at a rate of about 1291 people per year.<\/p>\n by definition, velocity<\/strong> is the rate of change of position<\/strong>. so, if you have a position function s<\/em>(t<\/em>), then the velocity function is nothing more than the derivative of s<\/em>(t<\/em>).<\/p>\n analogously, acceleration<\/strong> is the rate of change of velocity – hence the second derivative<\/em> of position.<\/p>\n the derivative value f<\/em>\u00a0‘(a<\/em>) measures the slope<\/strong> of the curve y<\/em> = f<\/em>(x<\/em>) at the point x<\/em> = a<\/em>. by extension, then the derivative also gives you the slope of the tangent line<\/strong> there.<\/p>\n here is another reference for tangent lines and slopes of curves: how to find the slope of a line tangent to a curve<\/a>.<\/p>\n find the equation of the tangent line to the curve y<\/em> = (1\/3)x<\/em>3<\/sup> – (1\/2)x<\/em>2<\/sup> + 3x<\/em> – 5 at x<\/em> = 6.<\/p>\n slope = derivative. here, dy<\/em>\/dx<\/em> = (1\/3)3x<\/em>2<\/sup> – (1\/2)2x<\/em> + 3 = x<\/em>2<\/sup> – x<\/em> + 3.<\/p>\n therefore, the slope at x<\/em> = 6 is:<\/p>\n m<\/em> = (6)2<\/sup> – (6) + 3 = 33.<\/p>\n we also need the y<\/em>-coordinate of the point. so plug x<\/em> = 6 into the original function as well.<\/p>\n (1\/3)(6)3<\/sup> – (1\/2)(6)2<\/sup> + 3(6) – 5 = 67.<\/p>\n finally, using point-slope form, we construct the equation of the tangent line.<\/p>\n y<\/em> – 67 = 33(x<\/em> – 6) = 33x<\/em> – 198, or<\/p>\n y<\/em> = 33x<\/em> – 131<\/p>\n there’s quite a lot that calculus can tell us about the shape of a graph. we’ll focus on what the derivative and second derivative can say about a graph.<\/p>\n for a detailed review of these topics, you can also check out: ap calculus exam review: analysis of graphs<\/a><\/p>\n the first derivative measures the slope<\/strong> of the curve. so just as slopes of lines can tell you whether the line increases, decreases, or is horizontal, the derivative can be used to locate intervals of increase and decrease, as well as any turnaround points.<\/p>\n <\/p>\n at a critical point<\/strong>, the graph may<\/em> turn around (but it’s not a guarantee). then you can use the first derivative test<\/strong> to decide whether each critical point is a relative maximum, minimum, or neither.<\/p>\n first derivative test<\/strong>: suppose x<\/em> = a<\/em> is a critical point for a function f<\/em>.<\/p>\n second derivatives give you information about concavity<\/strong> (direction of curving), and inflection points<\/strong> (transitions between different concavities).<\/p>\napplications of derivatives<\/h2>\n
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rates of change<\/h2>\n
<\/p>\n
example: rates of change<\/h3>\n
solution<\/h4>\n
velocity and acceleration<\/h3>\n
<\/p>\n
slope and tangent lines<\/h2>\n
example: tangent lines<\/h3>\n
solution<\/h4>\n
analysis of graphs<\/h2>\n
first derivatives — increase, decrease, and relative extrema<\/h3>\n
\n\n
\n if f<\/em> ' > 0<\/td>\n f<\/em> is increasing (on an interval)<\/td>\n<\/tr>\n \n if f<\/em> ' < 0<\/td>\n f<\/em> is decreasing (on an interval)<\/td>\n<\/tr>\n \n if f<\/em> '(a<\/em>) = 0<\/td>\n f<\/em> has a critical point at x<\/em> = a<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \n
<\/p>\n
second derivatives — concavity and inflection points<\/h3>\n