Math 1105: Calculus I (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 3

Size: px

Start display at page:

Download "Math 1105: Calculus I (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 3"

Isabella Burns
6 years ago
Views:

1 Math : alculus I Math/Sci majos MWF am / pm, ampion Witten homewok Review: p 94, p 977,8,9,6, 6: p 46, 6: p 4964b,c,69, 6: p 47,6 p 94, Evaluate the following it by identifying the integal that it epesents: n 4k + 4 n n n k= Solution The Riemann sum with n subintevals fo the function fx = x + on the inteval [, 4] is given by n fx k x k If we choose patitions with subintevals of equal length, then we have x k = 4 = 4 n n fo each k =,, n If, moeove, we choose ight endpoints then we have x k = + k 4 n = 4k n fo each k =,, n The Riemann sum now equals n 4k f 4 n 4k n n = + n k= k= Thus the it in question is the definite integal of f on [, 4]: n 4k n n n = x + dx = x6 6 + x p 97, 7 Evaluate x x et dt x x k= k= 4 4 n = = 6 Solution We attempt to use l Hôpital s Rule, and invoke the Fundamental Theoem of alculus pat : d x dx et dt Thus x x et dt x = e4 p 97, 8 Evaluate x x e t dt x x = d x dx e x x = e4 Solution We poceed as in poblem 7, but we have to use the hain Rule in addition to the Fundamental Theoem of alculus: d x e t dt dx x = d x dx x e x = e

2 Thus x x e t dt x = e p 97, 9 Without evaluating the integals, explain why the following statement is tue fo positive integes n: x n dx + x /n dx = Solution The functions x n and x /n ae inveses of each othe, which implies that the gaph of one is obtained by eflecting the othe ove the line y = x This eflection peseves the unit squae with vetices,,,,,, and,, and takes the egion undeneath y = x n to the egion above y = x /n Thus xn dx is the aea of the egion inside the unit cube above y = x /n Finally, the sum of the given integals is the sum of the aea of the egion above y = x /n in the unit squae with the aea of the egion below y = x /n in the unit squae, so that the sum is p 97, 6 Use the change of vaiables u = x to evaluate the integal Solution We use the chain ule to evaluate the deivative of u: u du = x dx When x =, u =, and when x =, u = Substituting into the oiginal, we find: x x dx x x dx = u u du = u du = u4 4 = 8 4 = 6 p 4, 6 With snow on the gound and falling at a constant ate, a snowplow began plowing down a long staight oad at noon The plow taveled twice as fa in the fist hou as it did in the second hou At what time did the snow stat falling? Assume the plowing ate is invesely popotional to the depth of the snow Solution Let st be the amount of snow on the gound afte t hous, and let pt be the distance taveled by the plow afte t hous Since the snow falls at a constant ate, thee ae some and s so that s t =, and st = t + s Since the plowing ate is invesely popotional to the depth of the snow, we have p t = st = t + s fo some constant Since the plow is plowing at a positive ate, > Thus p t > fo t >, and the distance taveled is the same as the displacement The displacement of the plow in the fist hou is given by p t dt = t + s dt = t + s The last integal can calculated easily with the substitution u = t + s /, so that p t dt = log + s log dt s

3 Similaly, the displacement of the plow in the second hou is given by p t dt = log + s log + s Since the plow taveled twice as fa in the fist hou as in the second hou, we have log + s s log = log + s log + s anceling the / and eaanging, we find log + s log + s s + log = Using popeties of the logaithm, we have + s s log + s = To simplify this expession, let x = s / so that we have leaing denominatos and expanding, we have anceling, this becomes and the quadatic fomula says that + x x + x = x + 4x + 4x = x + x + x + x + x =, x = ±, of which only + / is positive Note that the initial amount of snow s and the ate of snowfall ae both positive, so x = s / is positive We conclude that + s = Thus st = t + + The snow stated falling when st =, namely when + t =, so that t = 68 Namely, the snow stated falling aound 68 h 6 min/h 7 min befoe noon, which is : am p 49, 64 Suppose a datboad occupies the squae {x, y : x, y } A dat is thown andomly at the boad many times meaning it is equally likely to land at any point in the squae What faction of the dat thows land close to the edge of the boad than the cente? Equivalently, what is the pobability that the dat lands close to the edge of the boad than the cente? Poceed as follows a Ague that by symmety, it is necessay to conside only one quate of the boad, say the egion R: {x, y : x y } b Find the cuve in this egion that is equidistant fom the cente of the boad and the top edge of the boad

4 c The pobability that the dat lands close to the edge of the boad than the cente is the atio of the aea of the aea of the egion R above to the aea of the entie egion R ompute this pobability Solution a OPTIONAL The squae is composed of fou copies of the egion R Fo each of these copies of this egion, the pobability of a dat falling inside it is /4, and the pobability that a dat falls close to the bounday than the cente is equal fo each of these tiangula egions This means the pobability of falling close to the bounday than the cente, in the whole squae, is equal to /4 + /4 + /4 + /4 = times the pobabilty the dat falls close to the bounday than the cente inside R Thus it s enough to calculate the this pobability only inside R Solution b Suppose x, y is a point in R The distance fom x, y to, is x + y, and the distance fom x, y to the line y = is y Thus the set of points that ae equidistant fom, and the line y = is given by the solutions to x + y = y, which is given by anceling and eaanging, we obtain x + y = y = y y + y = x / Solution c Let R be the pat of R with x The intesection of with the line y = x occus when x = x /, o x + x =, with solutions x = ± 8 = ± Thus the intesection of with the line y = x in the fist quadant is given by the point, We daw a vetical line though this point, foming a tiangula egion inside R with height = and width =, whose aea is given by / = / = The emaining pat of R is between the lines y = and the line y = x /, with x between and We compute this with the definite integal: x + x dx = dx = x + x = + = = 8 = 4 The aea of R can now be computed as the sum + 4 = 4, so that the aea of R is 4 The whole egion R is a tiangle with base and height, so that the aea is The pobability of a dat falling close to the bounday is the atio of the aea of R to the aea of R, which is given by 4 78 p 49, 69 Detemine the aea of the shaded egion bounded by the cuve x = y 4 y

5 Solution Solving fo x we have x = ± y 4 y = ±y y Note that the maximum possible value fo y is, o the tem undeneath the squae oot in the calculation of x is negative Thus the aea of the shaded egion can be computed: y y y y dy = y y dy The last integal can be computed using the subsitution u = y, with du = y dy, so that y y dy = udu = u/ / + = 4 9 u/ + = 4 9 y / + Thus the aea of the shaded egion is given by y y dy = 4 9 y / = 4 9 / 49 / = 4 9 p 4, 7 Find the volume of the solid whose base is the egion bounded by the cuves y = x and y = x, and whose coss sections though the solid pependicula to the x-axis ae squaes Solution The cuves y = x and y = x occus when x =, o when x = ± One of the slices, fo a value of x between and, is a squae with base given by the vetical segment between x and x hoosing the value between and, we see that < =, so that y = x is above the gaph of y = x Thus the length of the vetical segment between x and x is given by x x = x The aea of this slice is given by x, so that the volume can be computed by the definite integal x dx = 4 8x +4x 4 dx = 4x 8 x + 4x = = = 64 p 4, 6 Let R be the egion bounded by the cuve y = x + a with a >, the y-axis, and the x-axis Let S be the solid geneated by otating R about the y-axis Let T be the inscibed cone that has the same cicula base as S and height a Show that volumes / volumet = 8/ Solution Note that y = x + a intesects the y-axis when x =, so that y = a Fo a given value of y, the x-coodinate on the cuve is given by y a The solid S is composed of hoizontal slices which ae disks, with adius y a = a y, as y goes fom to a Using the disk method, we may compute the volume of S: a a a vols = πa y dy = π a ay + y 4 dy = π a y a y = π a a a a/ + a/ = π + y a / a/ + a/ = πa / + = 8πa/ When y = on the cuve y = x + a, we have x = a, so that the adius of the cicula base of T is a The height is a, so that the volume of T is given by We compute: volt = πa a vols 8πa/ volt = πa / = πa/ = 8πa/ πa / = 8

Math Section 4.2 Radians, Arc Length, and Area of a Sector

Math Section 4.2 Radians, Arc Length, and Area of a Sector Math 1330 - Section 4. Radians, Ac Length, and Aea of a Secto The wod tigonomety comes fom two Geek oots, tigonon, meaning having thee sides, and mete, meaning measue. We have aleady defined the six basic