#$&* course MTH 277 6/16 6 pm Question: Find an explicit relationship between x and y by eliminating the parameter in the following equations: x = e^-t, y = e^t. Sketch the corresponding curve for -inf <= t <= inf. (inf stands for infinity).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: We could solve the first equation for t, taking natural log of both sides to get -t = ln(x). Substituting this into the second equation we would get y = e^t = e^(- ln(x) ) = 1 / e^(ln(x) ) = 1 / x. Or we could obseved that since e^-t is 1/ e^t, we have x = 1 / e^t and y = e^t, implying that x = 1 / y, which is equivalent to y = 1 / x. You should be able to easily sketch this curve. If necessary substitute +- 1/2, +- 1 and +- 2 for x and think about where the horizontal and vertical asymptotes should be (think also about where the function is undefined). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: Find the parametric and symmetric equations for the line passing through the point (-1,-1,0) and parallel to the line (x-3)/4 = (y-1)/3 = (z+3)/2 The given equations describe a line through (3, 1, -3) parallel to the vector 4 `i + 3 `j + 2 `k. Our line will not be through (3, 1, -3), but will be parallel to the same vector 4 `i + 3 `j + 2 `k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Our line is through (-1, -1, 0) so its symmetric equations are (x + 1) / 4 = (y + 1) / 3 = z / 2 . Let t be the parameter, and set t equal to each of these expressions, so that (x + 1) / 4 = (y + 1) / 3 = z / 2 = t. The parametric equations are thus x = 4 t - 1 y = 3 t - 1 z = 2 t. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:ok ********************************************* 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: 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The xy plane is the z = 0 plane, so our parametric equation for z yields 2 t - 7 = 0, with solution t = 7/2. For this value of t we get x = 3 * 7/2 + 4 = 29/2 = 14.5 y = 1 - 3 * 7/2 = -19/2 = -19.5. So the intersection with the xy plane is (14.5, -19.5, 0). The xz plane is the y = 0 plane, giving use 1 - 3 t = 0 so that t = 1/3. Our resulting point is (5, 0, -19/3), approximately (5, 0, -6.33). The intersection with the y z plane is found similarly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: Show whether the line represented by the parametric equations x = 2-t, y = 3t , z = 3 - 2t and the line represented by x = 5-t, y = -1-3t, z = -3 +4t intersect, are parallel, or if they are skew. If they intersect, give the point of intersection. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We don't want to use the same parameter for both lines, so let's express the second line as x = 5 - s, y = -1 - 3 s, z = -3 + 4 s. The lines then intersect provided there are values of s and t such that all three coordinates are the same for both lines. That condition is 2 - t = 5 - s 3 t = -1 - 3 s 3 - 2 t = -3 + 4 t Eliminating t between the first two equations we get 6 = 14 - 6 s so that s = 4/3 and t = - 5 / 3. So if there is an intersection, it must be for these values of s and t. Plugging these values into the third equation does not lead to an identity, so no simultaneous solution for s and t exists and the lines do not intersect. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: Determine whether the vector v = -(7/3)i - (4/3)j - k is orthogonal to the line passing through the points P(-2,2,7) and Q(1/2,-1/2,9/2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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. confidence 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): ok ------------------------------------------------ Self-critique rating: ok "