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course PHY 232
6/29
Question: `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
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Your solution:
To find the steepness of the graph we will concentrate on finding the slope of the graph which is rise/run.
For first straight line:
(17-5)/(7-3) = 12/4 = 3
For second line:
(29-17)/(10-7) = 12/3 = 4
Here the second line has more slope hence it is more steeper.
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Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
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Your solution:
First let x=2
(2-2)*(2*2 + 5) = 0*(9) = 0
Now let x=-2.5
(-2.5-2)*(-2.5*2 + 5) = (-4.5)* (0) = 0
These two values are the roots of the equation, hence when plugged into the equation they turn the expression zero.
`q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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Your solution:
To find the roots of the equation we put values in individual brackets zero:
3x-6 = 0
X=2
X+4=0
X=-4
X^2 -4 = 0
X^2=4
X=+4,-4
These are the values which can make the expression zero.
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Question: `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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Your solution:
The first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area.
`q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:
As we move from left to right the graph increases as its slope increases.
As we move from left to right the graph decreases as its slope increases.
As we move from left to right the graph increases as its slope decreases.
As we move from left to right the graph decreases as its slope decreases.
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Your solution:
For x = 1, 2, 3, 4:
The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate.
The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero.
Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases.
We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate.
For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
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Question: `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
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Your solution:
Initial population = 20 frogs
10% increase every month
10% of 20 = (10/100)*20 = 2
Hence after first three months: 20+2+2+2 = 26 frogs
Since frogs increase at a constant rate,
One month is the time required for 2 frogs to increase.
In 300 months = 300*2 = 600 frogs
Hence final number of frogs = 20 + 600 = 620
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Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
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Your solution:
x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1.
So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc..
Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere.
The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become.
The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis.
This is what it means to say that the y axis is a vertical asymptote for the graph .
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Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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Your solution:
Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here.
For further reference, though, note that this expression could also be expanded by applying the Distributive Law:.
Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get
E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800
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Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero?
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Your solution:
The given expression is zero for values of x like
X=0
X=+5,-5
X=-3
Question: One straight line segment connects the points (3,5) and (7,9) while another connects the points (3, 10) and (7, 6). From each of the four
points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area?
Any solution is good, but a solution that follows from a good argument that doesn't actually calculate the areas of the two trapezoids is better.
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Your solution:
Here the co-ordinates are given. We can judge the value of the areas of the two trapezoids by drawing the figures on a page.
We notice that because of the given co-ordinates, both trapezoids will have the same base length. The area will be judged by the more height of the co-ordinates. Hence the line connecting (3,10) and (7,6) forms a trapezoid with more area.
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Question:
Suppose you invest $1000 and, at the end of any given year, 10% is added to the amount. How much would you have after 1, 2
and 3 years?
What is an expression for the amount you would have after 40 years (give an expression that could easily be evaluated using a calculator, but don't bother to actually evaluate it)?
What is an expression for the amount you would have after t years?
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Your solution:
10% of 1000 = 100
After 1 year = $1010
After 2 years = $1020
After 3 years = $1030
Expression:
1000 + 10x = y
Where x are the number of years and y is the total amount.
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