Assignment_1_Solutions0

Assignment_1_Solutions0 - S e ction 1.1 P a ge 1 S e ction...

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Unformatted text preview: S e ction 1 .1 P a ge 1 S e ction 1 .1 P a ge 2 S e ction 1 .2 P a ge 1 S e ction 1 .3 P a ge 1 Assignment 1-additional question solutions ! ! Calculus I with Review Math 150 (Summer 2011) Assignment 1- additional question solution ! Assignment 1- additional question solution Quiz date: Friday, September 17 Calculus I with Review Math 150(Spring 2010) Complete this assignment by Wednesday in your homework journal. This will give you plenty of time to make sure you Calculus I with Review understand the material : the end of Friday’s class. Quiz A1. Solution before the quiz at Math 150(Spring 2010) questions will be taken from items 1, 2 or 3 below. −3−(5) −8 a) Solution : SomeA1. The slope of theyour homework 3−1 = are: = −4 and so the equation of the line is: suggestions for using line is m = journal 2 • • • • • − Do roughThe slope of thepaper. m = −3y (5)(5) −8 −4(x4− 1) so the equation of the line is: − a) work on scratch line is 3−1 = = = − and 2 If you find one solution, try to find another (a simpler solution may reveal itself). y without + −. 1) little details. = −4x all9 When you find a solution, try to see it as a whole = −4(x the y − (5) Do questions in orderpointclearly label question and of the line y = −4x + 9 and the parabola and of intersection, (x, y ), section numbers. To find the y = − Catch . Grade your own 2 assignment when solutions are posted. 4x + 9your mistakes now when the stakes are low rather y = −2x + 4x + 1 we first solve: than making find the exams.of intersection, (x, y ), of the line y = −4x + 9 and the parabola To them on point − y = −2x + 4x + quiz, your answer should − in form + 1 To obtain maximum marks on the 1 we first solve:4x + 9 =be 2x a + 4x that another student could understand without 2 undue effort: a poorly expressed but correct 2x − 8xnot 8 = 0 result is + sufficient. 2 2 2 −4x + 9 = −2x + 4x + 1 4 x 4) = 0 2x −x8+ + 8 = 0 2)2 = 0 1. Online Questions: (from LONCAPA:− + 4) = 0 2(x22(x4xhttps://loncapa.sfu.ca): − 2 2(x2 − men t com typ e 1 yfi nal stud yM T stud ecte corr che cked e don stio que ion sect text n d 2(x − 2)2 = 0 Questions in folders: 5.1, 5.2 so x = 2. The corresponding y value is found by plugging x into either equation. Using the textbook: y= = 1. The intersection is 2. Questionsxfrom linear equation we find value −4(2) + 9by pluggingpoint of either equation. so = 2. The corresponding y is found x into therefore (2, 1). Using the linear equation we find y = −4(2) + 9 = 1. The point of intersection is therefore (2, 1). 5 22 2 2 2 CD using Riemann sums to estimate area 13 22 2 2 2 CD using Riemann sums from a table of values St. 5.1 16 22 2 2 2 CD the distance problem from a graph 19 22 2 2 2 CD expressing area as a limit 20 22 2 2 2 RE deconstructing a limit 2 22 2 2 2 RE right-hand Riemann sums 6 22 2 2 2 RE using Riemann sums to estimate area from graph 12 22 2 2 2 RE the Midpoint rule to approximate an integral 17 22 2 2 2 CD recognizing a limit as a definite integral 22 22 2 2 2 CD computing an integral from the definition c) ( −4 , 11) is the only point we2 find on yCDg (x). To see why, we only know f as a limit can 2 = recognizing a definite integral at 3 so 22 2 3 29 St. 5.2 evaluate g (x) we must have 3x + 7 = 3, hence x = −4/3 and g (−4/3) = f (3) = 11. to −4 34 2 2 2 c) ( 3 , 11) is2 only point we can find on RE g (use facts from basiconly know f at 3 so the 2 y = x). To see why, we geometry 38 know(x2we must have 3x 2 h(=) 3, hence x = −x/3 and g (−4/3) interpreting as an area 2g f )at 5 2 to evaluate 7 x RE must have 42 an3integral by = 2f= 8 = 11. 2 to only + c) We evaluate so we evaluating − = 5, hence x (3) and √2 2 √ 2 42 2 2 so x = ± 8 = ±2 2. Therefore the onlyCD properties of integrals the graph of g points we can determine on √ c) We 45 know 2 at 52 to2 only √ f √ so evaluate h √ ) have 2 − 3 e 5, hence x2 2 information given are: (2(x2,we and (the result 31 = x dx = e3 − e= 8 and 2 RE must −2 2x5). based on the 5) use , so x49 ± 2 = 2 2 2. Therefore 2 only points we can determine on the graph of g = 8± the √ 2 2 CD properties of integrals √ based on the information given are: (2 CE5) and integral, 5). 2, use (−2 2 properties to estimate value 59 22 2 2 2 1 1 1 See the legend on last page of this assignment for what these acronyms mean. 1 A2. Solution : A2. Solution : A2. Solution : (a) The diagram depicting the situation is drawn on the right. (a) There diagram cases to consider: (i) 0 ≤ h ≤ 5, andon the > 5. The are two depicting the situation is drawn (ii) h right. (a) The diagram depicting the situation is drawn on the right. Therefirst case, caseshto consider: (i) level h ≤ not reachedhthe 5. are ≤ has and > In theThere two two cases to consider:0 (i) 0 ≤ 5, ≤ 5, (ii) (ii) h > 5. are 0 ≤ ≤ 5, the water h and In of first case, the h ≤ 5, end so the water has a reached the has basethe the wall in 0 ≤ shallowthe water levellevel not triangular In the first case, 0 ≤ h ≤ 5, the water has not reached the basesection. Letin denote the length ofthe water has a triangular of the wall the shallow end so the water’s surface and cross base of the wall in the shallow end so the water has a triangular crossh) denoteLet volume of water when the height is h. section. the denote the let V (cross section. Let denotelength of the water’s surface and and the length of the water’s surface let V (h) denote the volume of water when the height is h. let V (h) denote the volume of water when the height is h. h 5 By similar triangles, = , or equivalently, = 8h. The volume of water is the h 405 5 By similarshaded triangle, h , or equivalently, =of h. The volume = 1water is= triangles, = multiplied by the width 8 the pool: V (h) of h · 30 the area of the By similar triangles, 40 = , or equivalently, = 8h. The volume of water is the 2 40 120h2 ,of the ≤ h ≤ 5.triangle, multiplied by the width of the pool: V (h) = 1 h ·130 = area for 0 shaded area of the shaded triangle, multiplied by the width of the pool: V (h2 = 2 h · 30 = ) 120h2 for 0 hand, if h > 2 , for 0 ≤ h On the, other ≤ h ≤ 5. ≤ 5. 5 then the water has a cross section in the shape of a 120h trapezoid. The area of if h shaded region is the sum of the areas of the rectangle and a On the other hand, the > 5 then the water has a cross section in the shape of On the other hand, if h > 5 then the water has a cross section in the shape of a the triangle:The· area of + 1 5 · 40 = region 100. The volume of the of the in this case trapezoid. 40 (h − 5) the shaded 40h − is the sum of the areas water rectangle and 2 trapezoid. The area of the shaded region is the sum of the areas of the rectangle and is then V (h) =40 · (h− 100) · 1 5 ·=11200h h − 100. The volume of the water in this case 30 the triangle: (40h − 5) + 2 40 = 40 − 3000. the triangle: 40 · (h − 5) + 2 5 · 40 = 40h − 100. The volume of the water in this case is then V (h) = (40h − 100) · 30 = 1200h − 3000. is then V (h) = (40h − 100) · 30 = 1200h − 3000. Therefore, 120h2 if 0 ≤ h ≤ 5 2 3000 1200h − 2 if if < ≤ ≤ ≤ 5 50hh8 120h 120h if 0 ≤ h ≤ 5 V (h) = V (h) 1200h − 3000 if 5 < h ≤ 8 = 1200 volume if 5 < h ≤ h (b) The domain is {h ∈ R : 0 ≤ h ≤ 8}. Since h − 3000increases with 8 then the range consists of all values∈ R : 0 ≤Vh ≤ and V (8), which is increases with ,h then the range between (0) 8}. Since volume the interval [0 6600]. (b) The domain is {h (b) The domain is {h ∈ R : 0 ≤ h ≤ 8}. Since volume increases with h then the range consists of all values between V (0) and V (8), which is the interval [0, 6600]. consists of all values between V (0) and V (8), which is the interval [0, 6600]. Therefore, Therefore, V (h) = 2 2 2 ...
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