#$&* course Mth 277 9pm 9/30/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: We could solve the first equation for t, taking natural log of both sides to get -t = ln(x). Substituting this into the second equation we would get y = e^t = e^(- ln(x) ) = 1 / e^(ln(x) ) = 1 / x. Or we could obseved that since e^-t is 1/ e^t, we have x = 1 / e^t and y = e^t, implying that x = 1 / y, which is equivalent to y = 1 / x. You should be able to easily sketch this curve. If necessary substitute +- 1/2, +- 1 and +- 2 for x and think about where the horizontal and vertical asymptotes should be (think also about where the function is undefined). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): undefined @ x=0 and y=0 ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: Find the parametric and symmetric equations for the line passing through the point (-1,-1,0) and parallel to the line (x-3)/4 = (y-1)/3 = (z+3)/2 The given equations describe a line through (3, 1, -3) parallel to the vector 4 `i + 3 `j + 2 `k. Our line will not be through (3, 1, -3), but will be parallel to the same vector 4 `i + 3 `j + 2 `k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: parametric equations: x = -1 + 4t y = -1 + 3t z = 2t symmetric equations: (x+1)/4 = (y+1)/3 = (z)/2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ 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: x = 3t + 4 y = 1 - 3t z = 2t - 7 t = (x-4)/3 = (y-1)/-3 = (z+7)/2 <3,-3,2> (4,1,-7) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 didn't understand what the question was asking. So, what you were looking for is the intersection of each plane, xy, xz, and yz. I understand now what you were asking. In order to find the intersection of each plane, you set the plane you aren't looking for equal to 0. For instance, you want to find the intersection for yz, you set x= 0. Therefore 3t + 4 = 0 and t=-3/4. Substitute t=-3/4 into -3t +1 = y and 2t -7 =z. This give you the point (0, 13/4, -17/2). At this point, the y and z planes intersect. This is how you would find the intersection of the xz plane as well, setting y=0, solve for t and substitute. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: They do not intersect and they are not parallel. x = 2-t y = 3t z = 3-2t (x-2)/-1 = (y)/3 = (z-3)/-2 <-1,3,-2> (2, 0, 3) second equation: x = 5-t y = -1 -3t z = -3 +4t (x-5)/-1 = (y+1)-3 = (z+3)/4 <-1, -3, 4> (5, -1, -3) 2-t does not equal 5-t 3t does not equal -1 -3t and 3 - 2t does not equal -3 +4t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Did I do it right?????????????? ------------------------------------------------ Self-critique rating: OK
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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): OK ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!