Class Summary Page

for

MATH 162

Calculus with Applications, II

September 2, 1997 -- September 22, 1997

  • Wednesday, September 3. We went over the syllabus and worked in groups on some problems that reviewed material from MATH 160. The review problems are keyed to certain specific sections of the course text. If you have difficulty doing these, please reread the appropriate section (soon!!!). Information on what students should remember about trigonometry was distributed, as was the first homework assignment (on mathematical writing).

  • Friday, September 5. We worked through a handout and conducted a brief review of functions and inverse functions. A key point covered was that a function which fails the "Horizontal Line Test" might be able to pass it if we suitably restrict its domain. This is an old story for the "squaring function" f(x)=x^2, but today we began to apply that idea to the trigonometric functions. We restricted the domain of the sine function to the interval [-Pi/2,Pi/2] so that it had an inverse, which we call the arcsine function. We evaluated a number of expressions involving arcsines of various quantities; we saw that although arcsin(sin(x)) = x for every x in [-Pi/2,Pi/2], arcsin(sin(Pi)) = arcsin(0) = 0 -- so we need to remember how the inverse sine function was constructed in doing similar problems. We ended by working an important example in which we saw that it could be useful it was to construct a right triangle diagram whenever we wanted to apply a trigonometric function to the inverse of some other trigonometric function -- we will use this trick several times this semester.

  • Monday, September 8. A handout was distributed which contained a chart for us to fill in regarding several trigonometric functions and their inverses (the domain of the original function, the domain of the restricted function, the range of these functions, the domain and range of the inverse function and the derivative of the inverse function. In the case of the sine function, we had already done most of this on Friday; all we had to do was compute the derivative of arcsin(x). There was a brief discussion of the origin of the terminology "arcsine." We then filled out the table for the tangent function; one unusual property that this function has is that it possesses different finite limits as its argument tends to +Infinity and -Infinity (namely, Pi/2 and -Pi/2). Students were asked to use their readings in section 6.6 in order to complete the table on their own, and warned that the choice of domain for the restricted secant function might appear stange at first, but that the reason for this choice would be made clear later on in the course. We then worked with the computers in the lab, first using Netscape to visit the web pages for this course (including this one!) and then using a Maple worksheet "Inverse Functions" to reinforce intuition regarding inverse functions (particularly inverse trigonometric functions). The example that we worked through in class was that of the cosine/arccosine pair. I strongly encourage students to work through the other examples in ths notebook on their own. Finally we ended the class by rewriting the differentiation formulas as integration formulas and computing a pair of similar-looking examples: Int(1/(1+4*x^2),x) and Int(x/(1+4*x^2),x). Although the integrals look similar, they ended up yielding quite different functions. The first was (1/2)*arctan(2*x) + c, while the second was (1/8)*ln(1+4*x^2) + c. Note that at this point in the course we only have two interation techniques: (i) recognizing the integrand as being the derivative of some known function, and (ii) making a u-substitution that brings us back to case (i).

  • Wednesday, September 10. In response to a question about the homework, I outlined the solution of Exercise 93 of section 6.6 (page 412); this problem is a special case of the more general problem of integrating inverse functions which we will have more to say about in section 7.1. We introduced a pair of functions: the first was the average of e^x and e^(-x), and the second was the average of e^x and -e^(-x), and used a handout as an aid in sketching their graphs. (The handout can also be used as a guide for how to instruct Maple to plot pairs of functions; to plot just a single function omit the pair of curly braces.) We saw that the first function was even (symmetric about the x-axis) and we called it the hyperbolic cosine of x or cosh(x). We saw that the second function was odd (symmetric about the origin) and we called it the hyperbolic sine of x or sinh(x). In analogy with the ordinary trigonometic functions, we also defined the hyperbolic tangent [tanh(x)], the hyperbolic secant [sech(x)], the hyperbolic cotangent [coth(x)], and the hyperbolic cosecant [csch(x)]. We computed the derivatives of sinh(x) and cosh(x), and found these to be reminiscent of the derivatives of cos(x) and sin(x): (sinh(x))' = cosh(x) and (cosh(x))' = sinh(x). We ended by searching for a fundamental hyperbolic identity which might be analogous to the fundamental trigonometric identity: (cos(x))^2 + (sin(x))^2 = 1. We found that (cosh(x))^2 + (sinh(x))^2 = cosh(2x) -- note that for the trigonometric functions we have that (cos(x))^2 - (sin(x))^2 = cos(2x) -- but (cos(x))^2 - (sin(x))^2 = 1. It seems that the hyperbolic functions satisfy many relations similar to those of the trigonometric functions except that sometimes there is a minus sign difference.

  • Friday, September 12. We picked up today on the fact that if you let x=cos(theta) and y=sin(theta), then the fundamental trigonometric identity tells you that x^2+y^2=1, i.e., the point (x,y) lies on the unit circle, while if x=cosh(theta) and y=sinh(theta), then the fundamental hyperbolic identity tells you that x^2-y^2=1, i.e., the point (x,y) lies on the right branch of a "unit" hyperbola. We discussed a curious geometric relation between the circle and hyperbola in terms of areas swept out by line segments drawn from the origin to a point moving on the circle or hyperbola (see the last paragraph on page 414). The calculation of the area in the case of the hyperbola is a homework problem (Exercise 64 in section 6.7), but we computed the area in the case of the circle by observing that (i) the area must grow linearly with the total angle swept out, i.e. A(theta) = m*theta for some constant m, and (ii) when theta=2*Pi, the area swept out is just the area of the unit circle, i.e., A(2*Pi) = Pi*1^2 = Pi. From these observations it follows that A(theta) = theta/2, and you will see (by a different argument) that this is also true for the hyperbola. We next moved to the handout for the day where we completed the differentiation table for the hyperbolic functions. We worked out the differentiation formula for tanh(x) in class using the Quotient Rule, the differentiation formulas for sinh(x) and cosh(x), and the Fundamental Hyperbolic Identity; for homework you will repeat this kind of argumnt for sech(x), coth(x) and csch(x). We observed that rewriting the differentiation formula for tanh(x) gave us an integration formula for Int((sech(x))^2,x), and then we made a u-substitution to also evaluate Int(x*(sech(x^2))^2,x). Following the steps on the reverse side of the handout, we saw how to express the inverse hyperbolic sine in terms of familiar functions. It is not necessary that you memorize the formula that we obtained, but you should be able to reconstruct the procedure that we used to get this formula. Also, you should perform the last steps (skipped in class): differentiate the inverse hyperbolic sine and compare the answer obtained to the formula for the derivative of the inverse (trigonometric) sine. We concluded by working Exercise 62 from section 6.7; this problem was typed onto the handout. This problem which dealt with the shape of a hangiing cable not only presented a physical application of one of the hyperbolic functions, but also introduced the concept of a differential equation to which we shall return in section 8.1.

  • Monday, September 15. We began with a Maple demonstration of the growth properties as x gets large in magnitude (either positively or negatively) of sinh(x). We saw plots that suggested that sinh(x) grows faster "at infinity" than any (odd) power of x. As sinh(x) is approximately (1/2)*e^x for large positive values of x this raised the question of whether we could show that the limit of e^x/x^n is infinite as x tends to infinity. We recognized that this limit is an indeterminate form of type Infinity/Infinity, and then reviewed some examples from first semester calculus to see that limits of this type can converge to 0, diverge to Infinity, or converge to any value in between. (Although we didn't see an example, it's also true that such limits can diverge through oscillations.) Following the handout for the day, we next looked at all possible limit forms of quotients, products, differences, and powers involving functions tending to 0, 1 or Infinity, and classified these as determinate or indeterminate. We found seven types of indeterminate forms: Infinity/Infinity, 0/0, Infinity*0, Infinity-Infinity, Infinity^0, 1^Infinity, and 0^0. (Make certain that you understand why these are indeterminate, and why the other limits on that handout are not!) We then specialized to the case of indeterminate limits of type 0/0. We warmed up with a limit that we could evaluate by first semester techniques (canceling vanishing terms in the numerator and denominator). To catch our breaths, we paused for a historical interlude which focused on the Bernoulli brothers, Jakob and Johann, and their connection to the problem of the shape of a cable (discussed last class period) and the Marquis de L'Hospital (whose calculus text contains the first appearance of the rule that we were about to discuss) -- and the issue of who was the real discoverer of "L'Hospital's Rule." Next we looked at an example where first semester techniques could not be directly applied, the limit of cos(x)/(2x-Pi) as x tends to Pi/2. We replaced cos(x) by its linearization at Pi/2 (review the "Linear Approximations" part of section 2.9 if you have forgotten how to do this), and saw that the limit was -1/2. A closer (hand-waving) inspection of the guts of this method yielded L'Hospital's Rule for indeterminate forms of type 0/0 (for limits at a point). The class ended with a proof of this rule that was simplified by making the extra assumption that the functions in the numerator and denominator of the original limit had derivatives that were continuous at the point where the limit was being computed. (See either the text or the handout for the more general assumptions under which the result is still true.)

  • Wednesday, September 17. Today is Riemann's birthday! A list of all the types of indeterminate forms and the corresponding examples in the course text was given; recall that there were only two examples in L'Hospital's original calculus book. We then spent the entire class period working several examples that demonstrated how to use L'Hospital's Rule (and also how not to use this rule). We began with a pair of indeterminate forms of type 0/0: the limits of sin(x)/sinh(x) and (cos(x)-1)/(cosh(x)-1) as x tends to 0. Both of these limits were equal to 1, which is yet another point of similarity between the trigonometric and hyperbolic functions. Also, in computing the second of these we had to make a double use of L'Hospital's rule (... well actually, we just used the first limit, but if we hadn't already computed that we would have needed to use L'Hospital's Rule). In our next example, the limit of (1-sin(x))/x as x tends to 0 we saw the importance of making certain that we only apply L'Hospital's Rule to indeterminate forms. This limit turns out to not exist -- as we let x tend to 0 from above we find that (1-sin(x))/x tends to Infinity, while it tends to -Infinity as x tends to 0 from below -- but if we mistakenly try to replace the limit of the ratio of the functions (1-sin(x)) and x with the limit of the ratio of their derivatives we get an answer of 1 (which is wrong!!). The next example, the limit of sec(x)/tan(x) demonstrated the importance of remaining open to simpler ways of evaluating limits. Even though this limit is indeterminate (of type Infinity/Infinity), the most that we can conclude about it from using L'Hospital's Rule is that if it converges, then it must converge to 1. However, if we re-express the secant and tangent in terms of sines and cosines, and cancel the cosines, then the convergence of this limit to 1 follows in a straightforward way. We concluded by looking at the example of the limit of x*e^(-x) as x tends to Infinity. This is an indeterminate form of type 0*Infinity. By thinking of multiplication by one of the factors as division by its reciprocal, we saw that we could turn this into an indeterminate form of either type 0/0 or Infinity/Infinity. It turned out that for this particular limit, it was much easier to evaluate the limit of x/e^x as x went to Infinity (the form of type Infinity/Infinity) than it was to evaluate the other possibility. The limit was 0, which means that e^(-x) goes to 0 more strongly than x goes to Infinity (as x goes to Infinity), or alternatively, e^x diverges much faster than x itself as x tends to Infinity. Being able to rapidly determine the best way to rewrite a limit of type 0*Infinity is a matter of experience that is gained by doing several examples.

  • Friday, September 19. We began with one last example of L'Hospital's Rule: computing the limit of x^(1/x) as x tend to Infinity. This is a limit is an indeterminate form of type Infinity^0. The idea for handling any indeterminate form involving a power f(x)^g(x) is to express such a power in terms of the natural exponential. An example, the differentiation of 2^x, was worked at this point as a reminder that the idea of converting powers into exponentials was something that had already been introduced in first semester calculus. We write f(x)^g(x) = e^(ln(f(x)^g(x)), and then use a property of logarithms to write g(x)*ln(f(x)), so f(x)^g(x) = e^( g(x)*ln(f(x))). Finally, since exponentiation is a continuous operation [note that if h is continuous at a, then lim_{x->a} h(x) = h(a) = h(lim_{x->a}), so one can take the limit inside a continuous function], it follows that lim_{x->Infinity} x^(1/x) = lim_{x->Infinity} e^((1/x)*ln(x)) = e^(lim_{x->Infinity} (1/x)*ln(x)). When one interchanges the limit and the exponentiation, there is the understanding that if the inside limit diverges to infinity, then we say the original limit is Infinity (because e^x goes to Infinity as x goes to Infinity) and if the inside limit goes to -Infinity, then we say that the original limit goes to 0 (since e^x goes to 0 as x goes to -Infinity). The computation of this limit was not difficult, indeed it was just a special case of HW point out that to handle indeterminate forms that arise from powers, one just needs to re-express the power as an exponential base e and then interchange the limit and the exponentiation. The resulting limit will be of the type 0*Infinity, which can then be transformed into either type 0/0 or Infinity/Infinity in preparation for L'Hospital's Rule. A handout, a preprint of the paper "A Geometric Approach to the Integration of Inverse Functions" by the instructor, was handed out. This method is a generalization of the procedure demonstrated last Wednesday in the outline the solution of Exercise 93 of section 6.6 (page 412). Today we worked through the method to see what it says about the antiderivative of ln(x) -- Example 1 in the handout. The pictures that aren't in the preprint were illustrated on a (reversible) transparency overhead -- the steps we followed are described in the proof at the bottom of page 1 and the top of page 2. We double-checked our answer by differentiating it to see that its derivative really was ln(x). Of special interest was the observation that we need to use the Product Rule for Differentiation. This suggests that we should look for an Integration version of the Product Rule -- just as the Sum Rule for Integration, Int(f(x)+g(x),x) = Int(f(x),x) + Int(g(x),x), is just the Integration version of the Sum Rule for Differentiation, D(f(x)+g(x),x) = D(f(x),x) + D(g(x),x). This is what section 7.1 is all about, and we'll begin studying it on Monday.

  • Monday, September 22. After a reminder that it was always possible to re-express differentiation rules as integration rules (some familiar and important examples are the Sum, Difference and Constant Multiple Rules of Differentiation/Integration, and the Chain Rule of Differentiation which becomes the Substitution Rule of Integration), we set out to discover what the integration rule corresponding to the Product Rule of Differentiation was. We found the "Integration by Part Formula," most succinctly expressed using differential notation: int(u dv) = u*v - int(v du). We also derived a definite integral version of this formula. There are three basic classes of integrals to which integration by parts is well-suited: (1) Integrals where you know how to differentiate the integrand [here choose u to be the function in the inegrand, and take dv=dx], (2) Integrals of the form Int(x^n*f(x),x) where f is one of the functions e^(a*x), cos(a*x), sin(a*x), cosh(a*x) and sinh(a*x) [here set u= x^n and take dv=f(x)dx --- repeatedly apply integration by parts until the monomial disappers (with each integration by parts the power in the monomial is knocked down by one until it disappears)], and (3) Integrals of the form Int(f(x)*g(x),x) where both f and g are from the class of functions functions e^(a*x), cos(a*x), sin(a*x), cosh(a*x) and sinh(a*x) [here after two integrations by parts we end up with an algrebraic equation for the integral that we can solve without having to do any more calculus]. Sometimes when we integrate by parts we can relate one integral to another which is similar but maybe a little simpler in the sense that it involves some common term now raised to a lower power. When this happens, we call this relation a reduction formula. We worked several examples which demonstrated all of these different uses of integration by parts. We started with a pair of "I can differentiate that integrals": Int(ln(x),x) [which we could compare to the formula that we obtained geometrically last class] and Int((ln(x))^2,x=1..e) [our token definite integral example]. We then worked some "kill the x" integrals: Int(x*cosh(x),x), Int(x^n*cosh(x),x), and Int(x^n*sinh(x),x). Combining the last two of these, we obtained a reduction formula that relates Int(x^n*cosh(x),x) to Int(x^(n-2)*cosh(x),x), and then we used that formula twice to rather quickly evaluate Int(x^4*cosh(x),x). Next, we looked at the "cyclic" integral Int(e^(3*x)*sin(7*x),x). Finally, we observed that we could use a similar procedure to evaluate Int(e^(3*x)*sinh(7*x),x), but that it was much simpler to instead use the definition of the hyperbolic sine to convert this to an integral of exponentials. (Remember: There's no need to use newer, more complicated integration techniques on problems where older, simpler techniques work well.)

  • Wednesday, September 24. Today's lecture laid out a basic theme that we will be repeating off and on for the next several weeks. Most of the applications we will see have the following structure: we need to calculate some quantity which at first seems to be quite hard to do because it involves involves some sort of complicated quantity (think, for example, of the area under a curve). On the other hand, we do know how to do the computation for some simpler looking objects (in the context of area, we have no difficulty calculating the area of rectangles). Our basic trick is to divide the thing that we want to calculate into many small objects, and then approximate each of these by a similar but simpler object -- one that we know how to compute. When we take the limit as the number of little pieces goes to infinity, and their individual sizes all go to zero, two things happen: (1) the approximation becomes an equality (indeed, often it is the way we define the quantity we are trying to compute, and (2) the sum (instead of becoming infinitely hard to do, as one might well fear -- afterall we are adding up more and more terms) turns into an integral, which we can evaluate if we are sufficiently skilled in integration techniques. We began to apply this basic strategy to the problem of calculating the volume of a pyramid. (Remark: The answer to this problem has been known for at least 4000 years; it appears in the Rhind papyrus which was written by the Egyptian scribe Ahmes circa 1650 BC.) We divided the pyramid into a number of layers, and then approximated each of these by an (inscribed) prism. In essence, the approximating sums that we computed are the volume of ziggurats inscribed inside the pyramid. A Play-Doh demonstration showed that as we increase the number of levels in the ziggurat, we decrease the amount of error in the approximation. (Of course, in the limit, the error disappears entirely.) We set up the integral. A key step in doing this was determining the cross-sectional area of a slice of the pyramid; if we call this area A(x) where x is a variable that indicates the depth (measered from the top vertex of the pyramid) of the slice, then we found that V = int(A(x),x=0..h), where h is the height of the pyramid. We determined A(x) to be (b*x/h)^2 [b is the length of one of the sides of the base] by drawing the intersection of the pyramid with a plane (parallel to a pair of sides of the base) through the pyramid's center axis, and then determining the equations of the resulting lines -- not too hard since they pass through the origin, so they must just be of the form y=m*x for some constant m. (An equivalent method involves drawing the same diagram, but then using similar triangles insead of the equations of lines.) We then completed the integration to get V = (1/3)*h*b^2. At the very end of class, another Play-Doh demonstration showed how (in the special case where h=b) it was possible to divide a cube of side-length b into three equal pyramids, each having b for both the height and the base-side-length.

  • Thursday, September 25. We held a review session from 7:00-9:00 in ACD 420.

  • Friday, September 26. The first exam was given today.

  • Monday, September 29. The graded exams (together with a solution key) were returned today. We continued the discussion of how to use integrals to compute volumes by studying a special situation: volumes of solids of revolution. These solids are generated by rotating a planar region about a line -- think if using a lathe to spin the region so rapidly that it appears to be a solid. (For simplicity, we only consider horizontal and vertical lines.) If there is no cavity in the resulting solid, then when we slice the solid perpendicular to the axis of rotation, the slices are all nearly disks. This point was demonstrated with a set of plastic disks. Thus the cross-sectional area is just the area of a circle, which is Pi*(radius)^2. We worked some examples. We saw that if you rotate the region bounded by the x-axis and the lines x=4 and y=sqrt(x) about the x-axis, then the volume is given by Int(Pi*(sqrt(x))^2,x=0..4) = ... = 8*Pi. Our next example was computing the volume of a cone of base radius R and height h. (In our calculation, we we used the formual for a line to find an equation for the intersection of the cone with a plane containing its central axis; a les sophisticated method would be to just use similar triangle to determine the radius of any cross-section parallel to the base.) We found that the volume was (1/3)*Pi*R^2*h, which is one-third of the volume of the cylinder that circumscribes the cone. We noted that this was reminiscent of the relation we had found last Wednesday between the volumes of a pyramid and its circumscribing prism. We then took a little detour and went back to the general method of computing volumes via cross-sections to show that if you took "any" base with an area A, any point P a height h above this base, and formed a solid by filling in all of the line segments between the vertex P and ponts in the base, then the area will be (1/3)*A*h. The argument was based on a(n area) similarity argument, and then an integration -- the volume formulas for the pyramid and the cone are just special cases of this formula. We then went back to studying solids of revolution, and ended the class by finding the volume of the solid produced by rotating the region between the curves y=x and y=x^4 about the y-axis. One interesting aspect of this calculation -- besides the fact that we were now doing the slicing along the y-axis, hence integrating with respect to y instead of x -- was that there was now a cavity in the solid of revolution. We computed the volume in two ways. One method was to compute the volume of the solid where the cavity had been filled in, then compute the volume of the cavity (which I also a solid of revolution), and then subtract the seconfd volume from the first. Alternatively, we could combine the calculations by taking slices of the original solid. Each such slice would approximately be a "washer:" a large disk with a smaller disk punched out of its center. This point was demonstrated with a set of set of plastic rings. Using the second of these methods, we got that the volume was Int(Pi*((y^(1/4))^2-y^2),y=0..1) = ... = Pi/3.

  • Wednesday, October 1. We made a segue from computing volumes of revolution to integration of rational functions (i.e., ratios of polynomials) by doing a volume problem that led to an integral that we didn't (yet) know how to integrate. We rotated the region between the curve y=sqrt((4+x-x^2)/(3+2*x-x^2))+1 and the lines y=1, x=0 and x=1/2 about the line y=1. (Note that one feature of this problem that was different from the examples we did on Monday was that the axis of roation was not one of the coordinate axes.) We began by trying to determine what the curve y=sqrt((4+x-x^2)/(3+2*x-x^2))+1 looked like. It's not terribly important to have an exact sketch; in fact, we were content with noting that since we were only looking at x's near 0, this function would be approximately its value at x=0, namely sqrt(4/3)+1, which is a little larger than 2. We were then able to set up the integral for the volume, Int(Pi*((sqrt((4+x-x^2)/(3+2*x-x^2))+1)-1)^2,x-0..1/2), which simplifies to Pi*Int((4+x-x^2)/(3+2*x-x^2),x=0..1/2). At this point we stopped worrying about the definite integral and began concentrating on the indefinite integral, Int((x^2-x-4)/(x^2-2*x-3),x). We didn't see a good u-substitution to make. Hoping that we might be able to simplify the integrand by canceling a common factor in the integrand, we tried to factor each of these. The denominator is x^2-2*x-3=(x-3)(x+1). It wasn't immediately clear how to factor the numerator: x^2-x-4. We ended up using the quadratic formula to solve the equation x^2-x-4=0. The solutions to this equation are x=(1+sqrt(17))/2 and x=(1-sqrt(17))/2. Therefore, x^2-x-4 = (x-(1+sqrt(17))/2)*( x-(1-sqrt(17))/2). [Here's an important observation: if we had found that the solutions of the quadratic equation involved imaginary numbers (because there was a negative number in the square root of the quadratic formula), then we would say that this qradratic polynomial is irreducible -- because it cannot be "reduced" to being written as a product of two linear terms. For the time being, we will not worry about irreducible quadratic factors, but we'll eventually do so when we look at the second half of section 7.4.] So our plan to simplify the integrand by factoring didn't work. We next tried to get as much mileage as we could out of a certain analogy between numbers and function:
    Integers (e.g., 4, 3) <---> Polynomials (e.g., x+1, x^2+5)
    Fractions (e.g., 4/3, 3/4) <---> Rational Functions (e.g., (x+1)/(x^2+5), (x^2+5)/(x+1))
    Proper Fractions (e.g., 3/4) <---> Proper Rational Functions (e.g., (x+1)/(x^2+5))
    Improper Fractions (e.g., 4/3) <---> Improper Rational Functions (e.g., (x^2+5)/(x+1))
    In trying to distinguish between proper and improper rational functions, we had to decide what it meant to say that a certain rational function was larger than another; instead of using their numerical size (which could be tricky -- what would you do when the graph of one of the functions crosses the graph of the other?), we decided to use the order of the polynomials: the rational function p(x)/q(x) is proper if the oredr of the polynomial p is less than that of q. Our analogy suggests that just as we can rewrite improper fractions as the sum of an integer and a proper fraction, we should try to rewrite an improper rational function as the sum of a polynomial and a proper rational function. We saw two ways to do this, using long division and adding and subtracting in the numerator. The first of these is guaranteed to eventually give you the decompostion that you want, while the second is faster if you see how to do it. For example, in this situation, the second method consists of writing (x^2-x-4)/(x^2-2*x-3) = (x^2-2*x-3+x-1)/(x^2-2*x-3) = (x^2-2*x-3)/(x^2-2*x-3) + (x-1)/(x^2-2*x-3) = 1 + (x-1)/(x^2-2*x-3). At this point we ran out of time. [To be continued ...]

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