MATH 251 504 Practice Problems for Examination 1 Fall 2006 1 For which values of a are the vectors ha 3 9i and ha 4a 4i orthogonal 2 Determine whether or not the points P 3 7 1 Q 4 2 1 R 1 3 1 and S 0 0 2 lie in the same plane 3 Find symmetric equations for the line that is contained in both the plane z x and the tangent plane to the surface z 2x2 5y 2 at the point 1 1 7 4 Give an example of three dimensional vectors a b and c such that a b a c but b 6 c 5 Determine whether the following statement is true or false For every function f R2 R which is continuous at 0 0 the function g x y x 1 y 1 f x y is continuous at 1 1 6 Show that the limit doesn t exist in each of the following two cases a x y 3x x y 0 0 5x2 y 2 lim b x2 2x y 2 1 x y 1 0 x2 2x y 2 1 lim 7 Find fxx fxy and fyy a f x y 1 x2 1 2x3 b f x y ln x2 e2y c f x y ye 2x cos y 2 8 Find an equation for the tangent plane to the surface z xyex at the point 1 1 e 9 Find the partial derivatives and z 4t w s and w t where w 5xyz ln x y and x st y t2 10 Give an example of a function f R2 R such that fx 0 0 and fy 0 0 exist but f is not continuous at 0 0 1 11 Use differentials to approximate the number 35 3 215 12 In which direction does the function f x y z zey xy 2 decrease the fastest at the point 1 1 1 13 Classify the quadric surfaces x2 y 2 2y z 2 5 and y 4 z 2 x2 4x 14 Find an equation for the normal line to the surface z cos y sin x at the point 2 4 0 Solutions 1 The two vectors are orthogonal precisely when the dot product ha 3 9i ha 4a 4i a2 12a 36 a 6 2 is zero that is when a 6 2 It is equivalent to determine whether or not the vectors P Q h1 5 0i P R h 4 4 0i and P S h 3 7 1i are coplanar They are not coplanar because the scalar triple product P Q P R P S 1 5 0 4 4 0 3 7 1 1 4 5 4 0 24 is not equal to zero 3 Set f x y 2x2 5y 2 Then fx x y 4x and fy x y 10y and in particular fx 1 1 4 and fy 1 1 10 Therefore the equation of the tangent plane at 1 1 7 is z 7 fx 1 1 x 1 fy 1 1 y 1 4 x 1 10 y 1 or 4x 10y z 7 and it has h4 10 1i as a normal vector The plane z x has h 1 0 1i as a normal vector and so the line of intersection of the two planes has direction vector h4 10 1i h 1 0 1i h10 3 10i Setting x z 0 we see from 7 the equation 4x 10y z 7 that 0 10 0 is a point on the line of intersection of the two planes Thus symmetric equations for this line of intersection are 7 y 10 x z 10 3 10 2 4 Take for example a b i and c i 5 The statement is false Consider for example the function 1 if x y 6 1 1 x 1 y 1 f x y 0 if x y 1 1 2 3x 3 6 a Approaching 0 0 along the x axis we have the limit limx 0 5x 2 5 while approaching 0 0 along the y axis we have the limit limy 0 y02 0 and so the given limit doesn t exist 2 1 while b Approaching 1 0 along the x axis we have the limit limx 1 xx2 2x 1 2x 1 y 2 approaching 1 0 along the line x 1 we have the limit limy 0 y2 1 and so the given limit doesn t exist 7 a fxx x y 2 12x 40x3 fxy x y 0 fyy x y 0 b fxx x y 2 x2 e2y 4x2 x2 e2y 2 fxy x y 4xe2y x2 e2y 2 fyy x y 4e2y x2 e2y 4e4y x2 e2y 2 c fxx x y 4ye 2x fxy x y 2e 2x fyy x y 2 sin y 2 4y 2 cos y 2 8 Set f x y xyex Then fx x y x 1 yex and fy x y xyex and in particular fx 1 1 2e and fy 1 1 e Therefore the equation of the tangent plane at 1 1 e is z e fx 1 1 x 1 fy 1 1 y 1 2e x 1 e y 1 or 2ex ey z 2e 9 We have w w x w y w z s x s y s z s 1 5yz t x y t 20t4 t s t and 3 w x w y w z w t x t y t z t 1 1 s 2 5xz t 20xy 5yz x y x y s 2t 80st3 t s t 10 An example is the function f x y 0 if xy 0 1 if xy 6 0 The partial derivatives fx 0 0 and fy 0 0 are both zero However f is not continuous at 0 0 since approaching 0 0 along the x axis we have zero as the limit while approaching 0 0 along the line y x we have 1 as the limit so that lim x y 0 0 f x y doesn t exist 11 Define the function f x y x1 2 y 1 3 Then fx x y 21 x 1 2 y 1 3 and fy x y 1 1 2 2 3 1 x y and in particular fx 36 216 12 and fy 36 216 18 Then 3 3 35 215 f 36 216 dz f 36 216 fx 36 216 dx fy 36 216 dy 1 1 5 36 1 1 36 2 18 9 12 We have f x y z hy 2 zey 2xy ey i and in particular f 1 1 1 h1 e 2 ei So f is decreasing the fastest at the point 1 1 1 in the direction of the vector h1 e 2 ei 2 2 2 z4 1 and y 8 z 2 x 2 2 The 13 Completing the square we have x4 y 1 4 first is a hyperboloid of two sheets and the second is a hyperbolic paraboloid 14 Define the function f x y …
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