#$&* course Mth 277 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: We could solve the first equation for t, taking natural log of both sides to get -t = ln(x). Substituting this into the second equation we would get y = e^t = e^(- ln(x) ) = 1 / e^(ln(x) ) = 1 / x. Or we could obseved that since e^-t is 1/ e^t, we have x = 1 / e^t and y = e^t, implying that x = 1 / y, which is equivalent to y = 1 / x. You should be able to easily sketch this curve. If necessary substitute +- 1/2, +- 1 and +- 2 for x and think about where the horizontal and vertical asymptotes should be (think also about where the function is undefined). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the parametric and symmetric equations for the line passing through the point (-1,-1,0) and parallel to the line (x-3)/4 = (y-1)/3 = (z+3)/2 The given equations describe a line through (3, 1, -3) parallel to the vector 4 `i + 3 `j + 2 `k. Our line will not be through (3, 1, -3), but will be parallel to the same vector 4 `i + 3 `j + 2 `k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The symmetric equations of the line through (-1,-1,0) and parallel to 4i+3j+2k are (x+1)/4=(y+1)/3=z/2. The parametric equations are x=4t-1, y=3t-1, and z=2t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 z=0, so 0=2t-7 and t=7/2. When t=7/2, x=14.5 and y=-19.5 so the point is (14.5,-19.5,0) The xz plane is y=0, so 0=1-3t and t=1/3. When t=1/3, x=5 and z=-19/3 so the point is (5,0,-19/3). The yz plane is x=0, so 0=3t+4 and t=-4/3. When t=-4/3 y=5 and z=-11/3 so the point is (0,5,-11/3). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The xy plane is the z = 0 plane, so our parametric equation for z yields 2 t - 7 = 0, with solution t = 7/2. For this value of t we get x = 3 * 7/2 + 4 = 29/2 = 14.5 y = 1 - 3 * 7/2 = -19/2 = -19.5. So the intersection with the xy plane is (14.5, -19.5, 0). The xz plane is the y = 0 plane, giving use 1 - 3 t = 0 so that t = 1/3. Our resulting point is (5, 0, -19/3), approximately (5, 0, -6.33). The intersection with the y z plane is found similarly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Show whether the line represented by the parametric equations x = 2-t, y = 3t , z = 3 - 2t and the line represented by x = 5-t, y = -1-3t, z = -3 +4t intersect, are parallel, or if they are skew. If they intersect, give the point of intersection. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To confirm an intersection of the lines, the parameter t is replaced with s in the second line and simultaneous equations are made: 2-t=5-s 3t=-1-3s 3-2t=3+4s In the second equation, t is solved to be t=-1/3-s, plug this into the first equation to solve for s: 2+1/3+s=5-s and s=4/3. Then t is found to be -5/3. These solutions work for the first two simultaneous equations but not the third, so the lines do not intersect. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: PQ=2.5i-2.5j-2.5k. The dot product of PQ and v is -35/6+20/6+2.5= 0. The vectors are only orthogonal if the dot product between them is zero, so these 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 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!