#$&* course Mth 277 query_09_5*********************************************
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Given Solution: We could solve the first equation for t, taking natural log of both sides to get -t = ln(x). Substituting this into the second equation we would get y = e^t = e^(- ln(x) ) = 1 / e^(ln(x) ) = 1 / x. Or we could obseved that since e^-t is 1/ e^t, we have x = 1 / e^t and y = e^t, implying that x = 1 / y, which is equivalent to y = 1 / x. You should be able to easily sketch this curve. If necessary substitute +- 1/2, +- 1 and +- 2 for x and think about where the horizontal and vertical asymptotes should be (think also about where the function is undefined). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the parametric and symmetric equations for the line passing through the point (-1,-1,0) and parallel to the line (x-3)/4 = (y-1)/3 = (z+3)/2 The given equations describe a line through (3, 1, -3) parallel to the vector 4 `i + 3 `j + 2 `k. Our line will not be through (3, 1, -3), but will be parallel to the same vector 4 `i + 3 `j + 2 `k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we know x0,y0,z0 and we know a from the line parallel, we can find the equations. (x+1)/4 = (y+1)/3 = (z)/2 We know that each of these equations are not only equal to eachother but also equal to t. (x+1)/4=t x=4t-1 (y+1)/3=t y=3t-1 (z)/2=t z=2t confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: We need to find t for each plane. yz plane;x=0=3t+4 t=-4/3 xy plane;z=0=2t-7 t=3.5 xz plane;y=0=1-3t t=1/3 Then we can find the intersection point for the plane by plugging t into each equation. yz;(0,5,-29/3) x=3(-4/3)+4 x=0 y=1-3(-4/3) y=5 z=2(-4/3)-7 z=-29/3 xy;(29/2,-19/2,0) x=3(7/2)+4 x=29/2 y=1-3(7/2) y=-19/2 z=2(7/2)-7 z=0 xz;(5,0,-19/3) x=3(1/3)+4 x=5 y=1-3(1/3) y=0 z=2(1/3)-7 z=-19/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The xy plane is the z = 0 plane, so our parametric equation for z yields 2 t - 7 = 0, with solution t = 7/2. For this value of t we get x = 3 * 7/2 + 4 = 29/2 = 14.5 y = 1 - 3 * 7/2 = -19/2 = -19.5. So the intersection with the xy plane is (14.5, -19.5, 0). The xz plane is the y = 0 plane, giving use 1 - 3 t = 0 so that t = 1/3. Our resulting point is (5, 0, -19/3), approximately (5, 0, -6.33). The intersection with the y z plane is found similarly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Show whether the line represented by the parametric equations x = 2-t, y = 3t , z = 3 - 2t and the line represented by x = 5-t, y = -1-3t, z = -3 +4t intersect, are parallel, or if they are skew. If they intersect, give the point of intersection. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was unsure how to start this problem considering the same variable was used. After looking at the first line of the solution given and seeing we were suppossed to change the variable, the question became much easier. 2-t=5-s t=3-s 3t=-1-3s t=-1/3-s 3-2t=-3+4s t=3-2s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: First, we need to find the vector through the two points with final-initial PQ=2.5i-2.5j-2.5k Now lets see if the vectors dot product equals zero to test if the vectors are orthogonal. (-7/3)(2.5)+(-4/3)(-2.5)+(-1)(-2.5)=0 Therefore, the vectors are orthogonal to eachother. 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: #*&!