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 1additional 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 righthand 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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 Summer '08
 MOHOLLUND
 Riemann Sums, Additional Question Solution

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