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course phy 242
5/25/13 around 2:48 pm
1)Between points ( 3,5) and (7,17), the slope is rise/ run = (17-5)/(7-3) = 12/4 = 3Between points ( 3,5) and ( 10,29) , the slope is rise /run = (10-3)/(29-5)= 7/24
Between points (7,17) and (10,29) , the slope is rise/run = (29-17)/(10-7) = 12/3= 4
The higher the slope, the steeper the curve between those point.
Therefore between the points (7,17) and (10,29).
2) x y
2.1 1/.1=10
2.01 1/.01=100
2.001 1/.001= 1000
2.0001 1/.0001= 10000
1. As the x gets closer to 2 , the value of the
function increases by factor of 10.
2. yes the value of the function will exceed a
billion and trillion.
3. Yes it will exceed the number of particles in the
known universe.
4. no
5. At x=2 there is an asymptote,. To the left of the
asymptote , the function approaches
negative infinity, but to the right of the
asymptote the graph approaches positive infinity.
Self-critic: I did not mention that as the x value changed
in small increments the function changed by a greater value.
3) A= ½*(b1+b2)*h
Trapezoid one ( contains points (3,5) and (7,9))
Use the distance formula to find the length of the bases and the height.
B1=sqrt((3-3)^(2) +(5-0)^(2)) = sqrt(25) = 5
B2= sqrt((7-7)^(2) +(9-0)^(2)) = sqrt(81) = 9
h= sqrt((3-7)^(2) + (0-0)^(2)) = sqrt(4) = 2
A= ½*(5+9)*2= ½*13*2= 13
Trapezoid two (contains points (10,2) and (50,4))
B1=sqrt((10-10)^(2) +(2-0)^(2)) = sqrt(4) = 2
B2= sqrt((50-50)^(2) +(4-0)^(2)) = sqrt(16) = 4
h= sqrt((50-10)^(2) + (0-0)^(2)) = sqrt(1600) = 40
A= ½*(2+4)*40= ½*6*40= 3*40= 120
Therefore, the second trapezoid is bigger in area than that of the first trapezoid by about 9 times.
Self-critique : I actually did their areas as opposed to estimate the values as the solutions demonstrate.
4) The greater the slope, the steeper the line segment.
Slope of points (2,4) and (5,25) is rise/run = (25-4)/(5-2)= 21/3= 7
Slope of the points (-1,1) and (7,49) is rise/run = (49-1)/(7+1) = 48/8= 6
Therefore, the line segment that connects x=2 to x=5 on the graph is steeper.
5) 1. The graph of the number of weeks versus grams of gold would be
of a level straight line .
2. The graph would be a rising straight line due to the constant
adding of 1 gram.
3. The graph will then be a line which rises slowly.
6) The same question as number 5.
7) Depth= 100 - 2 t + .01 t^2
At t = 30
Depth = 100- 2*(30) + 0.01*(30^2)
=100-60+0.01*900= 40+9=49 cm
At t=40
Depth = 100- 2*(40) + 0.01*(40^2)
=100-80+0.01*1600= 20+16=36cm
At t=60
Depth = 100- 2*(60) + 0.01*(60^2)
=100-120+0.01*3600= -20+36=16 cm
First time interval
40s - 30s =10 s
49cm - 36cm = 13 cm
13cm/10s= 1.3cm/s
Second time interval
60s -40s = 20 s
36cm -16cm = 20cm
20cm/20s = 1 cm/s
8) rate= 10-.1t
At t = 10
rate= 10 - .1(10)
rate = 10-1 = 9 cm/s
At t =20
Rate= 10- .1(20) = 10- 2= 8 cm/s
x/10 = 8cm/s
x= 80 cm
x/10 = 9cm/s
x= 90 cm
The expected rate of change during the 10 s interval is (8+9)/2 = 17/2= 8.5cm/s
Now the question wants the water level change within 10 s , therefore x/10 = 8.5cm/s , x=85cm.
9) h(x) = f(x)*g(x)
First we find the f(x) at x=6 , which is 4 and then we find g(x) at x=6 which is 1.
H(6) = f(6)*g(6) = 4*1= 4
f(x)= 2x-8
g(x)=(-3/4)x+8
yes H(x) value is greater than at x=6 at x=5.
When plugging in x=5 in to f(x) we get 2 and when plugging in x=5 in g(x) we get 4.25.
Now, h(x)= f(x) *g(x) = 2*4.25 = 8.5, which is greater than the value at x=6.
H(x) = f(x)*g(x) = (2x-8)*((-3/4)x +8)= -(3x^2/2) +22x-64
This equation is a quadratic equation which entails
that the graph is a quadratic graph.
Given the interval 2<=x<=6, within this interval
the graph of h(x) increases at a decreasing
rate due to first derivative becoming negative eventually.
10) The first trapezoid with no curves runs the same as the second trapezoid with the curve which is from x= 3 to x=7, which is 4 units. They differ in the altitude due to the second trapezoid having a curve. The altitude of the first trapezoid is 5 and 9 so the average altitude is 7. The altitude of the second trapezoid is also 5 and 9 but in this case there is a concave down curve which does not allow us to take the average, but since the curve segment is concave down this adds on to the area and so therefore the second trapezoid with the curve has a greater area.
11) 1. Since the car is gaining the same amount of speed, this means
constant slope and so therefore, the graph of car position versus
time will be of an increasing line at a constant rate. The rate of
change graph would be a horizontal line.
2. The graph of the car’s position versus time will be of a concave
up curve which increases at an increasing rate. The rate of change
graph would be the same as the position versus time due to the e^x
shape of functionality of the graph.
3. The graph of the car’s position versus time will be of a concave
down curve which increases at a decreasing rate. The rate of
change graph would be of a concave up curve which is decreasing
from in the x direction of positive infinity.
12) First we must find the liters in each time.
x/100 = 1.4 liters/s
x = 140 liters
x/150 = 1 liters/s
x= 150 liters
(150 liters + 140 liters)/2 = 145 liters
145 liters/ 50s= 2.9 liters/second water flowed out during the 50 second interval.
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