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course Mth 173
9/19 15:00
Text section 1.1PROBLEM 1-4
Given points ( 0 , 0 ) and ( 1 , 1 ).
The rise for the line = 1 - 0 = 1
The run for the line = 1 - 0 = 1
The slope for the line = rise / run = 1 / 1 = 1
A line passing through ( 0 , 0 ) and having a slope of 1 is given by
( y - 0 ) / ( x - 0 ) = 1
y = x is the line passing through ( 0 , 0 ) and ( 1 , 1 )
PROBLEM 1-6
Given points ( -2 , 1 ) and ( 2 , 3 )
The rise for the line = 3 -1 = 2
The run for the line = 2 - ( -2 ) = 4
The slope of the line = rise / run = 2 / 4 = 0.5
A line passing through ( -2 , 1 ) and has a slope of 0.5 is given by
( y - 1 ) / ( x + 2) = 0.5
y - 1 = 0.5x + 1
y = 0.5x + 2
PROBLEM 1-10
Given equation -4y + 2x + 8 = 0
Simplifying the equation we get 4y = 2x + 8
y = 0.5x + 2
Comparing it with the standard straight line equation which is y = mx + c , where m is the slope and c is the y intercept.
On comparing we obtain m = 0.5 which is the slope of the line and c = 2 which is the y intercept of the line.
PROBLEM 1-12
Given we have 6 graphs and 6 equations to match them.
The (I) graph is a straight line parallel to x axis, thus it must have a constant y value. The equation (c) which is 5 = y represents the same.
The (III) graph is a straight line passing through origin, thus its y intercept must be 0 that is c must be zero ( considering the basic form of a straight line y = mx + c ). The equation (f) which is y = x/2 represents the same.
Now graphs (II) and (V) are straight lines with positive slope but (II) has a positive y intercept and (V) has a negative y intercept. The equation (a) which is y = x - 5 represents a positive slope but a negative y intercept, thus it must correspond to graph (V) whereas equation (e) y = x + 5 also represents a positive slope but a positive y intercept and the graph (II) represents the same.
The graphs (IV) and (VI) represents graphs with negative slopes but graph (IV) has a positive y intercept and graph (VI) has a negative y intercept. The equations (b) y = -3x + 4 and (d) y = -4x - 5 both have negative slopes but (b) has a positive y intercept thus it resembles (IV) graph and (d) has a negative y intercept thus it resembles (VI) graph.
Thus in brief
(I) - (c)
(II) - (e)
(III) - (f)
(IV) - (b)
(V) - (a)
(VI) - (d)
PROBLEM 1-17
Given equation : y = 5x - 3. Comparing the equation to the standard line equation which is y = mx + c, we know that the slope m = 5. Let slope of the line perpendicular to this line be m1.
Since the lines are perpendicular we know m*m1 = -1
m1 = -1/m = -1/5
the line passes through the point ( 2 , 1 )
equation of line passing through the point ( 2 , 1 ) and with the slope = -1 / 5 can be obtained as
( y - 1 ) / ( x - 2 ) = -1/5
5y - 5 = -x + 2
5y = -x + 7
y = (-1/5)x + 7/5 - this is the equation of the perpendicular line.
PROBLEM 1-22
The domain of the function is the set of all input numbers. As we can see that graph takes input values from -2 to 2. Thus the domain of the function is -2 < x < 2 or x belongs to [ -2 , 2 ]
The range of the function is the set of resulting output values. The graph is bounded between -2 and 2. And the graph is continuous, thus it takes all the values between -2 and 2. Thus the range of the function is -2 < y < 2 or y belongs to [ -2 , 2 ]
PROBLEM 1-23
The domain of the function is the set of all input numbers. As we can see that graph takes input values from 0 to 5. Thus the domain of the function is 0 < x < 5 or x belongs to [ 0 , 5 ]
The range of the function is the set of resulting output values. The graph is bounded between 0 and 4. And the graph is continuous, thus it takes all the values between 0 and 4. Thus the range of the function is 0 < y < 4 or y belongs to [ 0 , 4 ].
PROBLEM 1-24
Given function y = x^2 + 2.
Domain is the set of input values to x. we can see that x can take any input value, thus x belongs to the set of all real numbers, which is the domain of the function x belongs to ( - infinity , + infinity ).
Now since we have x^2 in the function, the resulting value of x^2 for all input values of x will belong from 0 to + infinity. Thus x^2 + 2 will range from 0 + 2 till +infinity + 2. Thus the range of the function will be 2 < y < +infinity or [ 2 , infinity ).
PROBLEM 1-27
The volume of a sphere is proportional to the cube of its radius, r.
Let V be the volume of the sphere
Thus V = f(r) = k(r^3)
Where k is the constant of proportionality.
PROBLEM 1-31
Given number of animal species, N
Length of the certain body = l
Since number of animal species is inversely proportional to length, thus we have
N = k*(1 / l)
Where k is the constant of proportionality.
PROBLEM 1-32
Given function, S = f(t) where S is the average annual sea level.
f(0) is the average annual sea level in 2008. The year 1983 is 25 years before 2008.
Average annual sea level in 1983 = 7.019 meters.
The function thus would be
f(25) = 7.019 meters
PROBLEM 1-36
(a) we know that the snowfall in inches is a function of time because , the graph sketched has date on the x axis and snowfall in inches on the y axis, which tells that date is the independent value and snowfall in inches is the dependent variable and is a function of date
(b) From the graph the snowfall on December 12 is slightly below the 5 inches line, so approximately it must be 4.9 inches.
(c) there is only one day when the snowfall was higher than 10 inches and that is on December 11, 2014.
(d) during the December 10 and December 11 interval we see the highest increase in the snowfall in inches from about 1-2 inches to 16-17 inches.
PROBLEM 1-37
Given function V = f(a), where V is the value of the car in thousands of dollars and a is the age of the car.
(a) the statement f(5) = 6 means that after 5 years of the car the value of the car is $6000.
(b) a possible graph of V against a would be a decreasing graph with decreasing slope. The graph would start from the y axis at some point f(0) which would be greater than 6. The graph would be decreasing passing through ( 5 , 6) and finally towards x - axis. After great number of years the car would definitely come to a point where it has no value and has to be dumped, that year would be the ending point of the graph on the x - axis at a point beyond ( 5 , 0 ) definitely. The graph would only be located in the first quadrant.
(c) the vertical intercept , that is the y intercept on the graph is the actual , starting value of the car at which it initial cost. The horizontal intercept, that is the x intercept is the number of years the car actual could run before becoming valueless, worth dumping.
PROBLEM 1-39
Given function H = f(t) where H is the temperature in ℃ and t is time in min
(a) the statement f(30) = 10 indicates that after putting the object for 10 min outside the temperature of the hot body reduced to 30℃.
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H is the temperature, so the temperature would be 10 degrees.
t is the time, so f(30) refers to the temperature 30 minutes after being placed outside.
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(b) the vertical intercept, a that is the y intercept is the initial temperature in ℃ when the object was placed outside at t = 0.
The horizontal intercept, b that is the x intercept represents the time in min that the object will take to reduce its temperature to 0℃.
PROBLEM 1-40
Given function s = f(t) where s is the height of the rock above the ground in meters and t is time in seconds.
(a) The graph would be a decreasing graph with increasing slope.
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The slope would not be increasing.
If the rock is falling, then the graph is indeed decreasing, but since it is falling at a faster and faster rate, the steepness will be increasing. The slopes are negative, and negative slopes that are getting steeper and steeper are decreasing slopes.
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(b) the statement f(7) = 12 tells about the rock that 7 seconds after the fall of the rock , it would be at a height of 12 meters above the ground.
(c) the y-intercept of the graph is the initial height at which the rock initial was before throwing it to the ground and the x-intercept of the graph is the time the rock takes to reach the ground from the initial height.
PROBLEM 1-43
(a) company 1 rents cars at $40 a day and 15 cents a mile and company 2 rents cars at $50 a day and 10 cents a mile.
Let C(m) be the function of the cost of renting a car in dollars where m is the number of miles travelled.
Thus for company 1 the function of renting a car is C(m) = 40 + 0.15m
Thus for company 2 the function of renting a car is C(m) = 50 + 0.1m
(b) the graph for both the lines is a straight line. The straight line for company 1 has a slope of 0.15 and has a y intercept of 40 and the second straight line for company 2 has a slope of 0.1 and a y intercept of 50.
(c) In order to decide which company is cheaper, we can have a look at the graphs. the 2 straight line intersect at the point ( 200 , 70 ) . The line of 40 + 0.15m lies below the 50 + 0.1m until the point of intersection. Thus if you want to travel less than 200 miles company 1 is cheaper, whereas the straight line 50 + 0.1m lies below the line 40 + 0.15m for x beyond 200 miles. Thus if you want to travel a distance of more than 200 miles company 2 is cheaper. For 200 miles both the companies cost the same so if you want to travel exactly 200 miles either company is fine.
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Very well done, but see my two notes.
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