Lecture for Week 7 Secs 3 10 12 Derivative Miscellany II 1 Related rates Related rates means applications of the chain rule and implicit differentiation Applications in this context means practical applications i e word problems 2 Exercise 3 10 7 A lamp is at the top of a 15 ft tall pole A man 6 ft tall walks away from the pole with a speed of 5 ft s How fast is the tip of his shadow moving when he is 40 ft from the pole How fast is the shadow lengthening at that point 3 Every nontrivial word problem s solution starts with some hard work specific to that problem having nothing to do with the mathematical concept being demonstrated Often the necessary reasoning is geometrical For example in Exercise 3 10 5 we need to know the formula for the surface area of a sphere I chose not to do that problem because it is so similar to the sphere problem that I gave so much attention to while we were studying the chain rule In the present problem we need similar triangles 4 15 6 x 0 Here x and y are coordinates not lengths The key equation is y x y 15 6 5 y which simplifies to y 53 x So 5 dx 25 dy 8 33 ft s dt 3 dt 3 That s how fast the tip moves The rate of change of the length is d dy dx 10 y x 3 33 ft s dt dt dt 3 6 Notice that we never had to use the number 40 ft because the rates turned out to be constants That is an accident of this problem However the point of general importance is that it was necessary to consider a general x not just x 40 to derive the relation between the rates derivatives Do not plug in the instantaneous values of changing quantities until you have finished differentiating 7 Exercise 3 10 33 A runner runs around a circular track of radius 100 m at speed 7 m s Her friend stands a distance 200 m from the center How fast is the distance between them changing when the distance is 200 m Start by drawing a figure of course 8 s This time we need the law of cosines the generalization of the Pythagorean theorem to a nonright triangle Recall that the known sides are 100 and 200 s2 1002 2002 2 100 200 cos 9 So s 200 when 100 400 cos or cos 14 Don t reach for the calculator s arccos key yet we may not need it Differentiate the formula for s2 with respect to t d ds 40 000 sin 2s dt dt Two things to note 1 Don t plug in numbers for s and first 2 Taking the square root first would make the calculus harder not easier 10 Now r 15 sin 1 16 so after dividing by 2s 400 we get 15 d ds 100 dt 4 dt d 7 And the angular velocity is so finally dt 100 7 ds 15 m s dt 4 p cos2 11 The language of differentials The key idea f a tells us how to build a linear approximation to f x that is good for x near a f x fapprox x f a f a x a fapprox is the function whose graph is the tangent line at a 12 y Recall that f a lim where x x 0 x often called h is a change in x and y f a h f a is the resulting change in y Now let s turn that equation around y f a x if x is small dy d y You have been taught to interpret dx as dx an operation on the function y f x Now we see dy that at least intuitively dx can be broken up like dy a fraction dx 13 dx is what x is called when you are announcing your intention to ignore powers of x or to ignore anything that vanishes faster than x as x 0 More precisely one ignores things that vanish even after being divided by x for example x 3 x 2 0 x In that approximation y equals dy 14 dy Thus dx is not just a notation for f but can be interpreted as a ratio of two numbers dx is any change in x and dy is the corresponding change in y when you adopt the tangentline approximation x and y are called increments dx and dy are called differentials 15 Leibniz and other early mathematicians thought of dx and dy as infinitesimal numbers so small that the approximate equation y dy x dx became exact dy dy dx dx Today we avoid that kind of talk in discussing fundamentals we use limits instead But thinking of dy as a ratio of small changes is still very f dx helpful in applying calculus 16 Example Find a formula for the rate of change of the area of a circle with respect to its radius A plays the role of y r the role of x 1 Derivation in the language of differentials As the radius goes from r to r dr the area goes from r 2 to r dr 2 I leave it to you to draw the obvious diagram dA r dr 2 r 2 r 2 2r dr dr 2 r 2 17 The terms without any dr factors cancel We neglect the second order term dr 2 because it is very small Whereas dr is just small dA r 2 2r dr dr 2 r 2 2 r dr Conclusion dA 2 r dr This kind of argument appears often in science and engineering textbooks Note that the distinction between dA and A has been blurred 18 2 Careful restatement in the language of difference quotients and limits When the radius changes from r to r r the area changes from r 2 to r r 2 A r 2 2r r r 2 r 2 2r r r Thus A r 2r r and therefore dA lim 2r r 2 r r 0 dr 19 The differential argument is just shorthand for this Better than the linear approximation is the quadratic approximation which fits the graph not with the line that matches it best but with the parabola that matches it best f a 2 x a f x f a f a x a 2 You can go on to even higher degree polynomials called Taylor approximations These are covered in Chapter 10 20 Exercise 19 p 228 extended Use differentials to approximate 40 21 36 1 and We know that 36 6 so that is a good starting point a 1 f x f x x 2 x x f a f a x a 1 6 x 36 12 1 36 1 6 6 008333 120 Check 6 008332 36 10007 Pretty good 22 40 6 4 6 333333 12 Check 6 3333332 40 11111 Not so good The approximation is better when x a is smaller This …
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