( 1) = 9 = 3 We would like to make the length 6. The only vectors in the same direction as v are those


 Darcy Aron Cummings
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1 1.(6pts) Which of the following vectors has the same direction as v 1,, but has length 6? (a), 4, 4 (b),, (c) 4,, 4 (d), 4, 4 (e) 0, 6, 0 The length of v is given by ( 1) We would like to make the length 6. The only vectors in the same direction as v are those of the form λv where λ is a positive number. Now λv λ v λ v. So we let λ..(6pts) Which line below is an equation of the line of intersection of the two planes x + 3y 4z 10 and 3x 4y + z. Note ( 1, 0, ) is a point on both planes. (a) t 10, 16, , 0, (b) t, 3, 4 + 1, 0, (c) t 10, 16, 16 +, 3, 4 (d) t 3, 4, + 1, 0, (e) t 10, 16, 17 Actually ( 1, 0, ) is a point on neither plane since ( 1) + 3(0) 4() 10 and 3( 1) 4(0) + () 1. However if you do what you should you should never notice. The two normal vectors are, 3, 4 and 3, 4,. Hence a vector parallel to the line is , 16, Hence t 10, 16, , 0, is an equation of the line. 3.(6pts) Given two lines t 1, 3, + 1, 1, 1 and t 3,, 1 + 7, 8, 8, find the point of intersection, if any? (a) (4, 10, 7) (b) (3, 7, 5) (c) (4, 10, 1) (d) (4, 7, 10) (e) There are no points of intersection. The equations are t + 1 3s + 7 3t + 1 s + 8 t + 1 s + 8 Then t 3s + 6 so (3s + 6) + 1 s + 8 and 6s + 13 s + 8 or 5s 5 or s 1. Then t so t 3 and the point is (4, 10, 7). 4.(6pts) Compute the curvature of the curve r(t) t sin(t), 1 cos(t), t at t π 3. (a) κ 10 8 (b) κ (c) κ 5 4 (d) κ 5 (e) κ 1
2 r 1 cos(t), sin(t), 1 and r sin(t), cos(t), 0 so κ r r r 3 cos(t), sin(t), cos(t) 1 1 cos(t), sin(t), (cos(t) 1) ( 3 cos(t) ) 3 At t π/3, cos(t) 1/ and so κ(π/3) 5/ (6pts) Given three vectors u, v and w in R 3, which of the following statements is not necessarily true? (a) If u v 0, then we must have either u 0 or v 0. (b) u v v u. (c) (u + v) w u w + v w. (d) u u 0. (e) u v is perpendicular to u and v. All the listed properties are true, except for u v implying u 0 or v 0. We could have, e.g., u v 1, 0, 0. 6.(6pts) Suppose that two vectors u and v R 3 are such that u 7, v 3 and the angle between them is θ 90. Compute u v. (a) 1. (b) 3 7. (c) 7. (d) 1π. 3 (e) There is not enough information. Using the identity u v u v sin(θ), we get u v 1. 7.(6pts) If f(x, y) x + e xy and x s + t, y sin(t), compute f at (s, t) (1, 0). (a) 3 (b) 0 (c) 1 (d) (e) 1
3 The chain rule says that f f x x + f y y. In this case, f (x + yexy ) 1 + xe xy cos(t) At (s, t) (1, 0), we have x 1, y 0, so that f (1, 0) OR f x + ye xy, xe xy and if r(t) s + t, sin(t) and r 1, cos(t). When s 1, t 0, x 1, y 0 so f(1, 0) r (1, 0), 1 1, (6pts) The plane S contains the points (0, 1, 3), (,, ), and (3,, 1). Which of the following is an equation for S? (a) x y + z (b) x + y + z 6 (c) x + 3y + z 6 (d) y + z 4 (e) x 4y + 3z 5 There are two ways to solve this problem. You can calculate the normal vector as in the worksheet and find a plane with that normal and one of those points and get x y + z. Alternately, you can test each equation to see if all three given points satisfy it, and the only one which works for all three is x y + z. 9.(6pts) Let f(x, y, z) e xy + z y. Find the sum of partial derivatives f x + f y + f z. (a) (x + y)e xy + yz + z (b) e y + e x + z + zy (c) ye xy (d) e xy + z + zy (e) e x + e y + z + 1 f x + f y + f z ye xy + xe xy + z + zy. 10.(6pts) Which one of the following functions has level curves as concentric circles? (a) f(x, y) e (x +y ) (d) f(x, y) e 4x y (b) f(x, y) 8 x y (c) f(x, y) x y (e) f(x, y) sin(x + y) The contour at level c is given by f(x, y) c. If f(x, y) e (x +y ) the contour at level c is x + y ln c, which is a circle centered at (0, 0). (Note c (0, 1]) For f(x, y) 8 x y the level curves are x+y 8 c, which are lines. For f(x, y) x y the level curves are x y c, which are hyperbolas and two lines. For f(x, y) e 4x y the level curves are 4x y ln c, which are parabolas. For f(x, y) sin(x + y) the level curves are x + y arcsin c which are lines. 11.(10pts) Find the area of the triangle with vertices P (1, 4, 6), Q(, 5, 1) and R(1, 1, 1).
4 We know that the area of the parallelogram determined by the vectors P Q and QR equals, P Q QR so the area of the triangle is just half this quantity. From here we just compute P Q 3, 1, 7 QR 3, 6, P Q QR , 15, , 3, and then Area of P Q QR (10pts) Let z z(x, y) be the function of x, y given implicitly by the equation Find and y. x + y 3 + z 4 + xyz 1 Let F (x, y) x + y 3 + z 4 + xyz. The function G(x, y) given by G(x, y) F (x, y, z(x, y)) is constant equal to 1, hence both G Rule for G, we may solve for : F F Using the Chain Rule for G, we may solve for y y : y F y F (x + yz) 4z 3 + xy (3y + xz) 4z 3 + xy and G y are zero. Using the Chain 13.(10pts) The line L passes through the point (, 7, 8) and is perpendicular to the plane R whose equation is 3x y + z 14. Find the point where L intersects R. Since L is perpendicular to R, its direction vector is R s normal, 3,, 1. Since we have a direction and a point, we can write an equation for L: r(t) 3,, 1 t +, 7, 8 3t +, t + 7, t + 8 To find the intersection of L and R, we substitute these coordinates into the equation for R. 3(3t + ) ( t + 7) + (t + 8) 14 9t t 14 + t t 14 t 1 Plugging this back into L, we get that the intersection is at (5, 5, 9). A quick check confirms that this point is indeed on R.
5 14.(10pts) A space curve is described by the equation 3 r(t) t, t cos(t) sin(t), t sin(t) + cos(t) for (0, ). (a) Write an equation for the unit tangent vector to the curve at time t. (b) Write an equation for the binormal to the curve at time t. (c) For what values of t (0, π) is the osculating plane at t of the curve perpendicular to the line with equations x 1 z 3 y? 1 3 r 3t, t sin(t), t cos(t), r t, T 3/, sin(t)/, cos(t)/, T 0, cos(t)/, sin(t)/, T 1/, N 0, cos(t), sin(t), B T N 1/, 3 sin(t)/, 3 cos(t)/. OR r (t) 3t, t sin(t), t cos(t) r (t) 3, sin(t) t cos(t), cos(t) t sin(t) r r 3t t sin(t) t cos(t) 3 sin(t) t cos(t) cos(t) t sin(t) t, 3t sin(t), 3t cos(t). and a parallel vector is 1, 3 sin(t), 3 cos(t). Hence a formula for the binormal is B 1/, 3 sin(t)/, 3 cos(t)/. OR A vector parallel to r (t) 3t, t sin(t), t cos(t), is p(t) 3, sin(t), cos(t). p (t) 0, cos(t), sin(t) and p p 3 sin(t) cos(t) 0 cos(t) sin(t) 1, 3 sin(t), 3 cos(t). Hence a formula for the binormal is B 1/, 3 sin(t)/, 3 cos(t)/. The binormal has to be parallel to u 1, 0, 3, the vector of the line. When sin(t) 0 and cos(t) 1, B 1 1, 0, 3 1 u. But in (0, π) the only value(s) of t for which sin(t) 0, cos(t) 1 is t π.
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