CALCULUS 3 February 6, st TEST

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1 MATH 400 (CALCULUS 3) Spring 008 1st TEST 1 CALCULUS 3 February, 008 1st TEST YOUR NAME: 001 A. Spina...(9am) 00 E. Wittenbn... (10am) 003 T. Dent...(11am) 004 J. Wiscons... (1pm) 005 A. Spina...(1pm) SHOW ALL YOUR WORK No calculats allowed. No cheat-sheets allowed. Make sure you write an arrow on top of vect quantities to differentiate them from scalar quantities (numbers). A wd-process and boldface fonts were used in writing test, but you are writing by hand! Remember that, within the same context, v (with the arrow) is a vect ( v v) and v (without the arrow) is the nm of the previous vect (v v v ). If a vect is the null vect, write an arrow on top of the zero! DO NOT WRITE INSIDE THIS BOX! problem points 1 10 pts 10 pts 3 10 pts 4 10 pts 5 10 pts 10 pts 7 10 pts 8 10 pts 9 10 pts pts TOTAL 100 pts sce

2 MATH 400 (CALCULUS 3) Spring 008 1st TEST 1. [10 pts] Find a parametric equation of the line L through the points P 0 (1, 3, ) and P 1 ( 4, 3, 0). A parametric equation of a line through P 0 (x 0, y 0, z 0 ) and parallel to v =< v 1, v, v 3 > is given by r(t) = r 0 + tv =< x 0, y 0, z 0 > +t < v 1, v, v 3 > x(t) = x 0 + t v 1 y(t) = y 0 + t v z(t) = z 0 + t v 3 A vect v parallel to the line L is given by v = OP 1 OP 0 =< 4, 3, 0 > < 1, 3, >=< 5, 0, > therefe the equation we are looking f is r(t) =< 1, 3, > +t < 5, 0, > x(t) = 1 5t y(t) = 3 z(t) = t. [10 pts] Find the equation of the plane through the points P 0 (0, 1, 1), P 1 (1, 0, 1), and P (1, 1, 0). 1st Way. The equation of the plane through P 0 (x 0, y 0, z 0 ) and with nmal n =< n 1, n, n 3 > is given by n (r r 0 ) = 0, Two vects, v and w parallel to the plane are n 1 (x x 0 ) + n (y y 0 ) + n 3 (z z 0 ) = 0. v = w = P 0 P 1 = < 1, 0, 1 > < 0, 1, 1 > = < 1, 1, 0 > P 0 P = < 1, 1, 0 > < 0, 1, 1 > = < 1, 0, 1 > A third vect perpendicular to these two vects (and therefe perpendicular to the plane) is given by i j k n = v w = =< 1, 1, 1 > The equation of the plane is therefe x + (y 1) + (z 1) = 0 x + y + z = nd Way. Replacing the codinates of P 0, P 1, and P in the general equation of a plane ax + by + cz = d we obtain the following system of equations b + c = d a + c = d a + b = d Solving this system we get a = b = c = d

3 MATH 400 (CALCULUS 3) Spring 008 1st TEST 3 Therefe d x + d y + d z = d Since d 0 in der f a solution to exist, we divide all the equation by d and multiply by, thus obtaining x + y + z = 3. [10 pts] Let v =< v 1, v >= v 1 i+v j be a vect in D with v 0. Find v, a vect perpendicular to v. Write it in terms of the components v 1 and v of the iginal vect. Note: Even if you already know the answer to this problem, show how you obtain the components of your v from the given condition it must satisfy. Let v be the vect perpendicular to v. We have,, in terms of their components, the easy solutions to this equation are given by v v = 0 v v =< v 1, v > < v 1, v >= v 1 v 1 + v v = 0 v 1 = v, v = +v 1, which represents a counter-clockwise rotation by π/ of the vect v, v 1 = +v, v = v 1, which represents a clockwise rotation by π/ of the vect v. Therefe v = v i ± v 1 j =< v, ±v 1 > 4. [10 pts] The following statements are either true false. If true, then say so and explain why. If false, then say so and give a simple counter-example to show why the statement is false. (a) (a + b) (a b) = a b ; (b) (a + b) (a b) = a b; (c) If a 0 and b 0, then (a b) a is parallel to b; (d) The plane given by x y + z = intersects the plane given by 3x y z = 1. Note: Just one lucky example doesn t prove a statement right, but it can prove it false! (a) True. The expression on the left can be easily expanded, yielding (a + b) (a b) = a a a b + b a a a b b b, b therefe (a + b) (a b) = a b (b) False. The left-hand side is a vect, the right hand side is a scalar. The crect answer to the expansion of the expression on the left is (a + b) (a b) = a a 0 a b + b a a b b b 0 therefe (a + b) (a b) = a b

4 MATH 400 (CALCULUS 3) Spring 008 1st TEST 4 (c) False. As a counter-example take a = i + j and b = j, then ( ) (i + j) j (i + j) = (i j + j j) (i + j) which is not parallel to j. = (k + 0) (i + j) = k (i + j) = k i + k j = j i (d) True. The nmals to the planes, n 1 =< 1, 1, > and n =< 3, 1, 1 > are not parallel, therefe the planes intersect. 5. [10 pts] Find the angle of intersection of the planes given by the equations Π 1 : x y + z =, and Π : x + y + z = 1. The angle of intersection θ of two planes is the angle between their nmals. A vect perpendicular to the first plane Π 1 is given by n 1 =< 1,, 1 >, and a vect perpendicular to the second plane Π by We know that n =<, 1, 1 >. n 1 n = n 1 n cosθ = cosθ = n 1 n n 1 n = 1 where the absolute value bars around the dot product are to ensure that we obtain the smallest of the two angles. Therefe ( ) 1 θ = arccos. [10 pts] Find a parametric equation of the line of intersection of the the planes Π 1 : x y + z =, and Π : x + y + z = 1. The parametric vect equation of a line L, that goes through P 0 and is parallel to v is given by r = r 0 + tv. A vect perpendicular to the first plane Π 1 is given by n 1 =< 1,, 1 >, and a vect perpendicular to the second plane Π by n =<, 1, 1 >. Since the line L is going to be perpendicular to n 1 (because it lies on Π 1 ), and also perpendicular to n (because it lies on Π ), it will be parallel to the cross product of n 1 with n. Therefe we can take i j k v = n 1 n = 1 1 =< 3, 1, 5 > 1 1

5 MATH 400 (CALCULUS 3) Spring 008 1st TEST 5 F r 0 we need to find one point of intersection of the planes, that is we look f a solution of the system of equations, x y + z = x + y + z = 1 any solution will do, so we set y = 0 and solve the resulting system x + z = x + z = 1 = x = 1, z = 3. Therefe r 0 = 1, 0, 3 will be the position vect of a point on the line (the one of its intersection with the xz plane). A parametric equation of the line can therefe be written as r = 1, 0, 3 + t < 3, 1, 5 > x = 1 3t y = t z = 3 + 5t 7. [10 pts] Given the equation z x y = 0 that represents a quadric surface, in the grid below draw the traces on the xz plane, the yz plane and on the planes z = 9, z = 4 and z = 0 (beware, these planes are not drawn); with the above curves as a guide complete a sketch of the quadric making sure that you do not spoil the traces you already draw. and sketch it x 1 y The traces yield x = 0 { z = y x = 0 parabola 8 y = 0 { z = x y = 0 parabola z 4 z = k { x + y = k z = k circles 0 8. [10 pts] Write an equation f the ellipsoid 4x + 4y + z = 1 in (a) cylindrical codinates; (b) in spherical codinates. (a) therefe 4x + 4y + z = 1 = z = 1 4 ( x + y ) z = 1 4r

6 MATH 400 (CALCULUS 3) Spring 008 1st TEST (b) 4x + 4y + z = 1 = 4 ( x + y + z ) = 1 + 3z = 4ρ = 1 + 3ρ cos φ therefe ρ ( 4 3 cos φ ) = 1 9. [10 pts] Find a rectangular equation f the surface whose cylindrical equation is r = cosθ and identify the surface it describes. r = cosθ = r = r cosθ = x + y = x = (x 1) + y = 1 This is a cylinder with axis parallel the z axis that intersects the xy plane on a circle of radius 1 with center (1, 0, 0). Therefe (x 1) + y = [10 pts] Find a rectangular equation f the surface whose spherical equation is ρ = sin θ sinφ and identify the surface it describes. ρ = sin θ sinφ = ρ = ρ sin θ sin φ = x + y + z = y Therefe ( x + y 1 ) + z = 1 4 which is the equation of a sphere with center (0, 1/, 0) and radius 1/.