#$&* course Mth 277 9/17/2011 @ 8:20 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: We could solve the first equation for t, taking natural log of both sides to get -t = ln(x). Substituting this into the second equation we would get y = e^t = e^(- ln(x) ) = 1 / e^(ln(x) ) = 1 / x. Or we could obseved that since e^-t is 1/ e^t, we have x = 1 / e^t and y = e^t, implying that x = 1 / y, which is equivalent to y = 1 / x. You should be able to easily sketch this curve. If necessary substitute +- 1/2, +- 1 and +- 2 for x and think about where the horizontal and vertical asymptotes should be (think also about where the function is undefined). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the parametric and symmetric equations for the line passing through the point (-1,-1,0) and parallel to the line (x-3)/4 = (y-1)/3 = (z+3)/2 The given equations describe a line through (3, 1, -3) parallel to the vector 4 `i + 3 `j + 2 `k. Our line will not be through (3, 1, -3), but will be parallel to the same vector 4 `i + 3 `j + 2 `k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the symmetric equation, we more or less take the first equation and sub in the known points (-1, -1, 0). We also know that the format of the numerator is x - #, so we will swap the sign from all of our points giving us (x + 1)/4 = (y + 1)/3 = z/2. To find parametric, we simply set our symmetic equations parts all equal to a randomly chosen variable (a). (x+1)/4 = a becomes 4a - 1 = x. (y + 1)/3 = a becomes 3a - 1 = y, and finally z = 2a. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our line is through (-1, -1, 0) so its symmetric equations are (x + 1) / 4 = (y + 1) / 3 = z / 2 . Let t be the parameter, and set t equal to each of these expressions, so that (x + 1) / 4 = (y + 1) / 3 = z / 2 = t. The parametric equations are thus x = 4 t - 1 y = 3 t - 1 z = 2 t. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Find the intersection of the line represented by the parametric equations x = 3t + 4, y = 1 - 3t, z = 2t - 7 with each of the coordinate planes (if the line doesn't intersect one or more coordinate plane, specify which one). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the point of intersection of each plane, we must find the value for “t” to plug into each equation to get a coordinate point. To find t, we simply recognize that the value for one of the 3 planes is zero since, for example, the x-y plane is 2D and has no depth. So, for x-y, 2t - 7 = 0, so t = 7/2. We plug that into the other two equations and we get the point of intersection of the x-y plane as (29/2, -19/2). X-z = (5, -19/3), y-z= (-3, -29/3) confidence rating #$&*:my arithmetic holds up, very. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The xy plane is the z = 0 plane, so our parametric equation for z yields 2 t - 7 = 0, with solution t = 7/2. For this value of t we get x = 3 * 7/2 + 4 = 29/2 = 14.5 y = 1 - 3 * 7/2 = -19/2 = -19.5. So the intersection with the xy plane is (14.5, -19.5, 0). The xz plane is the y = 0 plane, giving use 1 - 3 t = 0 so that t = 1/3. Our resulting point is (5, 0, -19/3), approximately (5, 0, -6.33). The intersection with the y z plane is found similarly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I neglected to include the 0 of the plane that I used to solve for t in my solution, but this is just a careless oversight and not an issue of understanding what happened. ------------------------------------------------ Self-critique rating: ********************************************* Question: Show whether the line represented by the parametric equations x = 2-t, y = 3t , z = 3 - 2t and the line represented by x = 5-t, y = -1-3t, z = -3 +4t intersect, are parallel, or if they are skew. If they intersect, give the point of intersection. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We first set the like parts of the equations equal to each other giving us 2 - t1 = 5 - t2, 3-t1 = -1 - 3t2, 3 - 2t1 = -3 + 4t2. We then, using the same technique as the last problem, solve for t1 which gives us t1 = -1/3 - t2. We can then plug this into any equation, but I chose 2 - t1 = 5 - t2. This leaves us with t1 = -5/3 and t2 = 4/3. If we plug these values into the original equations and graph, we can see they do not intersect and are not parallel. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We don't want to use the same parameter for both lines, so let's express the second line as x = 5 - s, y = -1 - 3 s, z = -3 + 4 s. The lines then intersect provided there are values of s and t such that all three coordinates are the same for both lines. That condition is 2 - t = 5 - s 3 t = -1 - 3 s 3 - 2 t = -3 + 4 t Eliminating t between the first two equations we get 6 = 14 - 6 s so that s = 4/3 and t = - 5 / 3. So if there is an intersection, it must be for these values of s and t. Plugging these values into the third equation does not lead to an identity, so no simultaneous solution for s and t exists and the lines do not intersect. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Can I find the identities you are referring to in the book or were they mentioned in a qa question and I just overlooked it? I feel like graphing these is not at all necessary. I think the identity for parallel is just them being multiples of each other, but I’m not sure about intersecting. (Get the same value when I plug in t1 and t2 maybe?) ------------------------------------------------ Self-critique rating:
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Given Solution: The vector PQ is 5/2 i -5/2 j - 5/2 k. The two vectors are orthogonal if and only if their dot product is zero. (-(7/3)i - (4/3)j - k ) dot (5/2 i -5/2 j - 5/2 k ) = -35/6 + 20/6 + 5/2 = -15/6 + 5/2 = -5/2 + 5/2 = 0 so the vectors are orthogonal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!