course MTH 271 Feb, 5 2010 -- 9:01 p.m. 003. PC1 questions*********************************************
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Given Solution: `aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,17) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Setting it up on paper really does help with the overall problem. ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When x=2, then (2-2)*(2(2)+5) =0 (0) *(9) =0 0=0 When x=-2.5, then ((-2.5)-2)* (2(-2.5)+5) = 0 (-4)*(0) =0 0=0 The only way for the product to be zero in this expression is to make the product be zero in either the (x-2) or (2x+5) expression, so that it can be multiplied by the other expression to always equal zero. confidence rating #$&*: 3!! I know I got this one! I can feel it ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form. STUDENT QUESTION I think I have the basic understanding of how x=2 and x=-2.5 makes this equation 0 I was looking at the distributive law and I understand the basic distributive property as stated in algebra a (b + c) = ab + ac and a (b-c) = ab – ac but I don’t understand the way it is used here (x-2)(2x+5) x(2x+5) - 2(2x+5) 2x^2 + 5x - 4x - 10 2x^2 + x - 10. Would you mind explaining the steps to me? INSTRUCTOR RESPONSE The distributive law of multiplication over addition states that a (b + c) = ab + ac and also that (a + b) * c = a c + b c. So the distributive law has two forms. In terms of the second form it should be clear that, for example (x - 2) * c = x * c - 2 * c. Now if c = 2 x + 5 this reads (x-2)(2x+5) = x * ( 2 x + 5) - 2 * (2 x + 5). The rest should be obvious. We could also have used the first form. a ( b + c) = ab + ac so, letting a stand for (x - 2), we have (x-2)(2x+5) = ( x - 2 ) * 2x + (x - 2) * 5. This will ultimately give the same result as the previous. Either way we end up with 2 x^2 + x - 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): No mistakes for this one. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero? Like the previous problem, when you make one of the expressions equal zero, it doesn’t matter what the product is for the remaining expressions, anything multiplied by zero, is always zero. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the first expression: 3x-6 = 0 -- add 6 to both sides 3x = 6 -- divide both sides by 3 x=3 so when x =2, the expression will end up being zero In the second expression: x+4 = 0 -- subtract 4 from both sides x= -4 so when x = -4, the expression will end up being zero In the third expression: x^2-4 = 0 --x^-2 which translates into x^1/2 so x would have to be 0^1/2 to be zero.
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Given Solution: `aIn order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t get the part with the exponents. ------------------------------------------------ Self-critique rating #$&*: 2 ********************************************* 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'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After sketching the points on my graph paper, it is obvious that the first trapezoid is a little more than twice the size of the second trapezoid. Because of this it has the greater area. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, 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. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step. ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For x we will use x > 0 – eg. 1,2,3, 4,5 For the function: y = 1/x and you plug in the above values, you get: 1,.5,1/3,1/4,1/5. These values are getting smaller, which means they are decreasing each time. Each time they are getting closer and closer to zero. For the function: y =x^2, and you plug in the above values, you get: 1,4,9,16,32. The values are increasing the more you change the number, so it would be understood that this graph is increasing. For the square root of x function: you plug in the values and get: 1, 1.4,1.7,2,2.2. These values are also increasing, confidence rating #$&*: 1 – I’m unsure here. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand why the square roots are at a “decreasing rate” because you increase the x values of 1,2, 3,4,5 and the y values are also increasing, 1, 1.4, 1.7.. etc.????
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Given Solution: `aAt the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I got this one. I really like the last paragraph in the solution to see the common error. ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The values are approaching zero because they continue to get smaller and smaller without actually approaching or crossing through zero. The reciprocals of 10 are endless to approaching zero but never touching it. 1/10, 1/100, 1/1000, 1/10000. It looks that the graph gets steeper and steeper but there is no way to count 1 by 0. confidence rating #$&*: 1 – this was fuzzy to me. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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 . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see that the explanation is different than what I got but I see what you meant about the steeper as it approaches the x-axis. But what I need to refresh my mind on asymptotes. ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When t =5 – v = 3(5) +9 – v=24 E = 800v^2 – 800(24)^2 – 576*800 = 460800 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800. • ********************************************* Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t think that there really is an expression for E in terms of t, because t is part of the v expression and you need v for the E expression. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see what you meant here. ------------------------------------------------ Self-critique rating #$&*: "