course Mth 163
Week 4 quiz 1Version 4
If water depths of 7.2, -11.3, -20.3 and -20.1 cm are observed at clock times 29.8, 44.7, 59.6 and 74.5 sec, then at what average rate does the depth change during each time interval?
Sketch a graph of this data set and use a sketch to explain why the slope of this graph between 44.7 and 59.6 sec represents the average rate at which depth changes during this time interval.
If f(x) = x2, give the vertex and the three basic points of the graphs of f(x- 1.75), f(x) - .35, .25 f(x) and .25 f(x- 1.75) + .35. Quickly sketch each graph.
The average rate at which the depth changes during each time interval is -1.2416,-0.6040, -0.0134 respectively.
Include unites, and indicate the details of how you obtained these results. For example, you might say that during the first interval depth changed by `dy = -11.3 cm - 7.2 cm = -18.5 cm, clock time changed by 44.7 s - 29.8 s = 14.9 s, so the average rate of change is `dy / `dt = -18.5 cm / 14.9 s = -1.24 cm/s.
Having explained the process for one step, you could then simply say something like 'using the same reasoning on the remaining intervals we obtain rates of -0.6040 and -0.0134 respectively, both in cm/s.'
The slope of the graph between 44.7 and 59.6 represents the average rate because the slope is rise/run or `dy/`dx which is the set up to find the average rate, therefore, average rate is slope.
The average rate of change of y with respect to t is defined as `dy / `dx. Since `dy is the rise and `dx is the run, the slope represents the average rate of change of y with respect to x.
f(x- 1.75)=(x-1.75)^2 = x(x-1.75)-1.75(x-1.75) = x^2-1.75x-1.75x+3.0625 = x^2-3.5x+3.0625
xVertex = 3.5/2(1) = 3.5/2 = 1.75
yVertex = 1.75^2-3.5(1.75)+3.0625 = 3.0625-6.125+3.0625=0
Vertex (1.75,0); (2.75,1)(.75,1)
It's good that you can do this by using x = -b / ( 2 a) and evaluating the function, but the intent here is to use shifts and stretches starting witht the basic points of y = x^2, which is much simpler and gives much more insight into the construction of the function.
From the fact that (0, 0) is the vertex of y = f(x) = x^2 you can figure out that the vertex of f(x - 1.75) is (1.75, 0). This is because the graph of f(x - 1.75) is horizontally shifted 1.75 units from the graph of f(x); a horizontal shift of 1.75 units on (0, 0) gives you (1.75, 0).
The same shift on the other two basic points (-1, 1) and (1, 1) gives you the points (2.75, 1) and (.75, 1).
f(x)-.35 = x^2-.35 Vertex (0,-.35); (1,.65)(-1,.65)
.25f(x)=.25x^2 Vertex (0,0); (1,.25) (-1,.25)
.25 f(x- 1.75) + .35 = .25(x- 1.75)^2 + .35 = .25 (x^2-3.5x+3.0625) + .35 = .25x^2-0.875x+0.765625+3.5 = .25x^2-0.875x+4.265625
xVertex= 0.875/2(.25) = 0.875/.5 = 1.75
yVertex = .25(1.75)^2-0.875(1.75)+4.265625 = .25(3.0625) - 0.875(1.75) + 4.265625 = 0.765625-1.53125+4.265625 = 3.5
Vertex (1.75,3.5); (2.75,3.75)(.75,3.75)
Good, but just take the basic points of y = x^2, vertically stretch by factor .25 (which you already did), then vertically shift -.35 units and finally horizontally shift 1.75 units and you get the same result without ever 'touching' the equation of the function.
Week 4 quiz 2
version 37
Given the depth vs. clock time function y = f(t) = .016 t2 + -2.18 t + 93, with depth in cm when clock time is in seconds, find the clock time t such that f(t) = 56.74375 cm and find f(t) when t = 37 cm. Find the clock time when water depth is 25.74375 cm. Using the same function determine the depth at clock time t = 13 sec.
56.74375 = .016 t^2 + -2.18 t + 93
0 = .016t^2-2.18t+36.25625
t= [2.18+-sqrt(2.18^2)-4(.016)(36.25625)]/2(.016)
t= [2.18+-sqrt(4.7524-2.320208)]/0.032
t= [2.18+-sqrt(2.432492)/0.032
t= 116.8609sec
f(t)=.016(37)^2-2.18(37)+93
=.016(1369)-2.18(37)+93
=21.904-80.66+93
=34.244cm
25.74375cm=.016t^2-2.18t+93
0=.016t^2-2.18t+67.25625
t= [2.18+-sqrt(2.18^2)-4(.016)(67.25625)]/2(.016)
t= [2.18+-sqrt(4.7524-4.3044)]/0.032
t= [2.18+-sqrt(0.448)/0.032
t= 47.2085sec
f(t)=.016(13)^2-2.18(13)+93
=.016(169)-2.18(13)+93
=2.704-28.34+93
=67.364cm
Good.
Week 4 quiz 3
version 16
At clock time t = 15 sec the temperature of an object is 63 Celsius, while at clock time t = 33 sec the temperature is 34 Celsius. Plot the corresponding points on a graph of temperature vs. clock time and determine the slope of the straight line segment connecting these points. Explain why this slope represents the average rate at which the temperature changes over this time interval.
For the exponential function y = f(t) = .015 * 1.016t, determine the average rate of change of y with respect to t, between clock times t = 33 and t = 38.
Slope is defined as rise/run the rise is equivalent to `dy and the run is the same as `dx therefore, rise/run becomes `dy/`dx which is how the average rate is found, so essentially slope is the same as average rate.
slope=-29/20=-1.45
y=f(t)=.015*1.016t
(33,.50292)
(34,.51816)
(35,.5334)
(36,.54864)
(37,.56388)
(38,.57912)
all have an average rate of -0.01524
Good.
ŜˤzƛV̼T鏴
assignment #008
ҶQ̫
Precalculus I
09-29-2006
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23:26:05
Were you able to complete the DERIVE exercise?
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yes
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23:26:13
Did you understand everything in the DERIVE exercise?
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yes
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23:26:20
Do you have any questions about the DERIVE exercise?
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no
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R͓ەsxnki
Student Name:
assignment #008
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23:30:27
`q001. Note that this assignment has 4 questions
For the function y = 1.1 x + .8, what are the coordinates of the x = 2 and x = 9 points? What is the rise between these points and what is the run between these points? What therefore is the slope between these points?
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(2,3)
(9,10.7)
rise = 3-10.7=-7.7
run = 2-9=-7
slope = -7.7/-7 = 1.1
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23:30:33
Evaluating y = 1.1 x +.8 for x = 2 and x = 9 we obtain y = 3 and y = 10.7. The graph points are therefore (2,3) and (9,10.7).
The rise between these points is 10.7 - 3 = 7.7 and the run is 9-2 = 7. Thus the slope is 7.7 / 7 = 1.1.
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23:31:50
`q002. For the function y = 1.1 x + .8, what are the coordinates of the x = a point, in terms of the symbol a? What are the coordinates of the x = b point, in terms of the symbol b?
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(a, 1.1a+.8)
(b. 1.1b+.8)
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23:31:59
If x = a, then y = 1.1 x + .8 gives us y = 1.1 a + .8.
If x = b, then y = 1.1 x + .8 gives us y = 1.1 b + .8. Thus the coordinates of the x = a point are (a, 1.1 a + .8) and (b, 1.1 b + .8).
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23:33:44
`q003. We see that the coordinates of the x = a point are (a, 1.1 a + .8) and (b, 1.1 b + .8). What therefore is the rise between these two points? What is the run between these two points?
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rise = (1.1a+.8)-(1.1b+.8) or 1.1a-1.1b
run = a-b
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23:33:54
The rise between the points is the rise from y = 1.1 a + .8 to y = 1.1 b + .8, a rise of
rise = (1.1 b + .8) -(1.1 a + .8) = 1.1 b + .8 - 1.1 a - .8 = 1.1 b - 1.1 a.
The run is from x = a to x = b, a run of
run = b - a.
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23:35:23
`q004. We see that the rise between the x = a and x = b points of the graph of y = 1.1x +.8 is 1.1 b + .8 - (1.1 a + .8), while the run is b - a. What therefore is the average slope of the graph between these points? Simplify your answer.
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(1.1a-1.1b)/(a-b)
=1.1(a-b)/(a-b)
=1.1
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23:36:03
The slope is
slope = rise / run = (1.1 b - 1.1 a) / (b - a) = 1.1 (b - a) / (b - a) = 1.1.
The significance of this series of exercises is that the slope between any two points of the straight line y = 1.1 x + .8 must be 1.1, no matter whether the points are given by numbers (e.g., x = 2 and x = 9) or by symbols (x = a and x = b). Mostly
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Very good work. See my notes and let me know if you have questions.