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by: Addison Beer


Addison Beer
GPA 3.76


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This 4 page Study Guide was uploaded by Addison Beer on Wednesday September 9, 2015. The Study Guide belongs to MATH 126 at University of Washington taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/192096/math-126-university-of-washington in Mathematics (M) at University of Washington.




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Date Created: 09/09/15
Math 126 Winter 2009 Midterm 1 Review SOLUTIONS PROBLEM 1 Note There are a number of ways of solving this problem so your method might be valid even if it isn t the same as the one presented here But the solution should be the same This is a 2step problem We rst nd a point where the two planes intersect and the direction vector for their line of intersection This is the same as nding the equation for the line of the intersection of the two planes The second step is using that to nd the normal vector for the plane STEP 1 The direction vector for the line of intersection is the cross product of the normal vectors which is alt111gtx 3271lt73471gt1 A point in the intersection of the two planes can be found by setting 2 0 and solving the two equations which gives P8 7 STEP 2 Now you have a line and a point Q1 2 1 not on the line or either of the original planes This is solved by taking the vector between a point on the line and the point not on the line and taking the cross product of that with the direction vector for the line Since both vectors are in the plane and based at the same point the cross product is the normal vector to the plane Vector PQ from point on line to point not on line is m17827713170lt771511 Cross product is i i E 77 15 1 lt71971017 73 4 71 and this is the normal vector to the plane So the nal solution is 719 710117 95 y 2 7 lt1 2 1 0 which simpli ed is 71995 7103 172 22 01 PROBLEM 2 We must use two parameters one for each of the lines since the lines could intersect for di erent values of t and s This gives us a system of 3 equations in 2 unknowns to solve corresponding to the x y and 2 components 4 7 t 1 23 6 2t 14 7 83 71 3t 7 7 43 We solve the rst two equations Adding 2gtlt the rst equation to the second gives 14 16 7 43 so 3 12 Substituting into the equations for x y and 2 in 3 gives the point of intersection as lt2 5 1 Important Note On tests use the equations in t and s to check your answer 3 12 implies t 2 since t 4 7 1 23 Then putting s 12 and t 2 into the third unused equation should give the same number on both sides On the other hand if you were trying to check if the lines intersected or not then equality or nonequality for the nal equation would say whether the lines did or did not intersect PROBLEM 3 a The distance from a point Px y 2 to the plane is ly 7 5 and the distance to the line is y 7 12 2 7 2 The easiest way to see this is to note both distances are independent of x and instead think about the problem on the yzplane where the line becomes the point 1 2 and the plane becomes the line y 5 Setting the two distances equal gives ly 7 5 y 7 12 2 7 2 Then square both sides and simplify W75 2712222 22710y2572f72y12722 722 3y The nal equation is the equation of the surface It is a parabolic cylinder so it looks like a piece of paper bent so it s pro le is a parabola b The curve is found by setting 2 6 so y 76 7 228 3 72 3 1 So the curve is the line y 1 2 6 PROBLEM 4 a NO 415t24tgt so r1 lt4154gt Then the de nition of T means T1 lt4154gt 4 15 4 1 WIN v 4215242 lt l 1 V257 3 b The equation for a tangent line to a curve rt at t a is sr a ra where for the tangent line we use the parameter s to emphasize that it is di erent from the parameter t for Mt So the equation for the tangent line at t 1 is sr 1 r1 slt4154gt lt452gt lt43 4153 543 2 PROBLEM 5 This problem is very tricky The curve de ned by the polar equation r sec 9 tan9 is not valid for all values of 9 Speci cally thinking about only angles 9 between 0 and 27139 it doesn t make sense if 9 7r237r2i Important Note it suf ces to consider 9 between 0 and 27r because sec 9 and tan 9 are both 27r periodici As 9 gets close to 7r2 from below the radius goes to 0 Now as rcos9 sec9 tan9cos9 1sin9 so as 2 if 1sin9 2 If 0 S 9 S 27r then 1 sin9 2 only if 9 7r2 So the curve doesn t intersect 7r2 Graphing r sec9 tan9 veri es that the curve has an asymptote at x 2 getting in nitely close to the line without intersecting You will get this problem wrong though if you immediately convert to x 1 sin9 and don t think about what values of 9 are valid in the original equation PROBLEM 6 We want to use the formula W t X r t Hmlo 5 WW but that formula is only valid for curves in R3 So we instead consider the curve as lying in the xyplane in R3 by setting rt lt9 7 21 t2 t2 0gti Then r t 7 lt2t 7 3 2t 2 0 THU 120 rt gtlt r t lt0 0 710 lmportantly that means lrt gtlt r tl 10 which is constant So mmwmxfm WW is largest when lrt g is smallest So it suf ces to nd t so lr tl3 is minimum That the same as nding t so lr tl2 is minimum since 952 and x3 are both increasing functions wt 7 217 32 213 22 7 8t 7 413 13 To minimize take a derivative with respect to t get 161 7 4 0 so t 14 is the solution PROBLEM 7 m lt12t7ti2gt Wt lt02t33gt The component of acceleration in the direction of the tangentvelocity vector at t i a compr UV1 T i 7 Ul v1 4 1 V5 b We solve Tt T t 4t 7 2t 5 0 for all values of ti Simplifying the equation is t6 12 So UWl lt1271gtlt022gt 2 W 1 55 l The curve is only de ned for t gt 0 or at least t y 0 so we say for t gt 0 so the only time is t 2 19 ti


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