Lecture for Week 7 (Secs. 3.10–12)Derivative Miscellany II1Related rates“Related rates” means “applications of thechain rule and implicit differentiation.”“Applications” in this context means “prac-tical application s ” (i.e., word problems).2Exercise 3.10.7A lamp is at the top of a 15-ft-tall pole. A man6 ft tall walks away from the pole with a spe ed of5 ft/s. How fast is the tip of his shadow movingwhen he is 40 ft from the pole? How fast is theshadow lengthening at that point?3Every nontrivial word problem’s solutionstarts with some hard work specific to that prob-lem, having nothing to do with the mathematicalconcept being demonstrated. Often the necessaryreasoning is geometrical. For example, in Exer-cise 3.10.5 we need to know the formula for thesurface area of a sphere; I chose not to do thatproblem, because it is so similar to the sphereproblem that I gave so much attention to whilewe were studying the chain rule. In the presentproblem we need similar triangles.4.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................1560x yHere x and y are coordinates, not lengths. Thekey equation isy15=y − x6,5which simplifies to y =53x. Sodydt=53dxdt=253= 8.33 ft/s.That’s how fast the tip moves. The rate ofchange of the length isddt(y − x) =dydt−dxdt=103= 3.33 ft/s.6Notice that we never had to use the num-ber 40 ft , because the rates tu rned ou t to beconstants. That is an accid ent of this problem.However, the point of gene ral importance is thatit was necessary to consider a general x, not justx = 40, to derive the relation between the rates(derivatives). Do not plug in the instanta-neous values of changing quantities untilyou have finished differentiating!7Exercise 3.10.33A runner runs around a circular track of ra-dius 100 m at speed 7 m/s. Her friend stands adistance 200 m from the cente r. How fast is thedistance between them ch anging when the dis-tance is 200 m?Start by drawing a figure, of course!8...............................................................................................................................................................................................................................................................................................................................................................................................................................................................θ.................................................................................................................................................................................................................................sThis time we need the law of cosines, the gener-alization of the Pythagorean theorem to a non-right triangle. Recall that the known sides are100 and 200.s2= 1002+ 2002− 2 × 100 × 200 cos θ.9So s = 200 when 100 = 400 cos θ, or cos θ =14.Don’t reach for the calculator’s arccos key yet —we may not need i t.Differentiate the formula for s2with respectto t:2sdsdt= 40, 000 sin θdθdt.Two things to note: (1) Don’t plug in num-bers for s and θ first. (2) Taking the square rootfirst would make the calculus harder, not easier.10Nowsin θ =p1 − cos2θ =r1516,so after dividing by 2s = 400 we getdsdt= 100√154dθdt.And the angular velocity i sdθdt=7100, so finallydsdt=74√15 m/s.11The language of differentialsThe key idea: f′(a) tells us how to build alinear approximation to f(x) that is good for xnear a.f(x) ≈ fapprox(x) ≡ f(a) + f′(a)(x − a).(fapproxis the funct i on whose graph is the tan-gent line at a.)12Recall that f′(a) = lim∆x→0∆y∆x, where ∆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.You have been taught to interpretdydxasddxy,an operation on the function y = f (x). Now we seethat, at least intuitively,dydxcan be broken up likea fraction,(dy)(dx).13“dx” is what ∆x is called when you areannouncin g you r int ention to ign ore powers of∆x — or to ignore anything that vanishes fasterthan ∆x as ∆x → 0.(More precisely, one ignores things that vanisheven after being divided by ∆x; for example,(∆x)3∆x= (∆x)2→ 0.)In that approximation, ∆y equals dy.14Thusdydxis not just a notation for f′, butcan be i nterpreted as a ratio of two nu mbers:dx is any change in x, and dy is the correspond-ing change in y when you adopt the tangent-line approximation.∆x and ∆y are called increments. dx and dyare called differentials.15Leibniz and other early mathematiciansthought of dx and dy as infinitesimal numbers —so small that the approximate equation∆y ≈dydx∆xbecame exact:dy =dydxdx.Today we avoid that kind of talk in discussing fun-damentals; we use limits instead. But thinking off′=dydxas a ratio of small changes is still veryhelpful in applying calculus.16Example: Find a formula for the rate ofchange of the area of a circle with respect to itsradius. (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 goesfrom πr2to π(r + dr)2. (I leave it to you to drawthe obvious diagram.)dA = π(r + dr)2− πr2= π[r2+ 2r dr + (dr)2] − πr2.17The terms without any dr factors cancel. Weneglect the second-order term d r2because it isvery small. (Whereas dr is ju s t “small”.)dA = π[r2+ 2r dr + (dr)2] − πr2= 2πr dr.Conclusion:dAdr= 2πr.This kind of argument appears often in sci-ence and engineering textbooks. Note that the dis-tinction between dA and ∆A has been blurred.182. Careful restatement in the language ofdifference quotients and limits. When the radiuschanges from r to r + ∆r, the area changes fromπr2to π(r + ∆ r)2.∆A = π[r2+ 2r ∆r + (∆r)2] − πr2= π(2r + ∆r) ∆r.Thus∆A∆r= π(2r + ∆r), and thereforedAdr= lim∆r→0π(2r + ∆r) = 2πr.19The differential argument is just shorthand forthis.Better than the linear approximation is thequadratic
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