Vector Fields. Field (II) Field (V)

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1 Math 1a Vector Fields 1. Match the following vector fields to the pictures, below. Eplain our reasoning. (Notice that in some of the pictures all of the vectors have been uniforml scaled so that the picture is more clear. Also notice that there are eight vector fields but onl si pictures. There s probabl a reason behind this.) Here are the possible vector fields: (a) F(, ) = 1, (b) F(, ) =, (c) F(, ) =, (d) F(, ) =, (e) f, where f(, ) = + (f) f, where f(, ) = + (g) f, where f(, ) = (h) f, where f(, ) = Field (I) Field (II) Field (III) Field (IV) Field (V) Field (VI). Recall that the gradient of a function is a vector normal to the level curve of this function. Eplain how this confirms our identification of the pictures for vector fields (e) through (h), above.

2 3. Here are some eamples in three-dimensional space. The vector fields are (a) F =,, z (b) F =, 1, (c) F =, +, + z (d) F =,, ( + +z ) 3/ ( + +z ) 3/ ( + +z ) 3/ 5 5 5

3 Math 1a Flow Lines Let s start with Problem #35 in Section 13.1 of Stewart. The flow lines or streamlines of a vector field are the path of a particle whose velocit vector field is the given vector field. Here s the eample: F(, ) =, = i j F(, ) with flow lines Can ou tell which direction the flow goes in each case?. We d like to find r(t) = (t), (t) for which r (t) = F. This amounts to solving the equations (t) = (t) = Do this to find r(t) = (t), (t).

4 F(, ) =, 1 = i + j F(, ) =, 1 = i + j 5. Above are two sketches of the vector field F(, ) =, 1 = i + j. Sketch the flow lines and repeat the previous problem to find r(t) for which r (t) = F(, ).

5 Math 1a Potential functions 6. Consider two vector fields F 1 (, ) = +, + 1, F (, ) =,. Use Clairaut s theorem to tell which one is the gradient vector field for a smooth function f? Find the corresponding function f. 7. Consider vector field F (, ) =,. Note that the vector field is not defined at + + origin. Does this vector field satisf the Clairaut s test? Is this a gradient vector field for smooth function?

6 Vector Fields Answers and Solutions 1. (I) This is vector fields (d) and (h). (First notice that these two vector fields are the same!) We can see this b noticing that the vectors should point down when > and up when <, and field (I) is the onl one that does this. (II) This is vector field (a). Notice that this vector field alwas has a positive rightward component, which is true onl of Field (II). (III) This is vector field (f). Both (e) and (f) are vector fields that point radiall outward, so the are Fields (III) and (VI). But which is which? Notice that the vector field in (e) is f =,, which has length r = +. On the other hand, the vector field in (f) is f = +, + = r, r, which has length 1. Thus (e) is Field (VI), the field with the vectors that increase in length as the distance from the origin increases, while (f) is Field (III), the field with vectors all the same magnitude. (IV) This is vector field (b). Look, for eample, at the vectors on the aes. On the -ais, the vector field is F(, ) =,, a vector that points verticall up (if > ) or down (if < ). This narrows our choices to Fields (IV) or (V). On the -ais, the vector field is F(, ) =,, a vector that points to the left (if > ) or to the right (if < ). This eliminates Field (V) and confirms Field (IV). (V) This is vector field (c) and (g), b an analsis that is ver similar to the one in Field (IV). (Notice that (c) and (g) are the same!) (VI) This is vector field (e). See (III) for the eplanation.. Four of the vector fields are (eplicitl) gradient fields (we ll be able to tell later that vector fields (a) and (b) are not gradient fields). Since f is perpendicular to the level curves of f, we should be able to see this in the vector field. Here I ve re-drawn the four vector fields in question with some level curves drawn in as well note the perpendicularit! Field (I) and f(, ) = = k for k =, ±, ±, ±6, ±8

7 Field (III) and f(, ) = + = k with k = 1,, 3, Field (V) with f(, ) = = k for k =, ±1, ±, ±3

8 Field (VI) and f(, ) = + = k with k = 1,, 3,, 9, and Here are the answers, with a few tips: (a) This is the top left vector field. The vector at the point (,, z) is the radial vector the same as the vector from the origin to this point (namel,, z ). (b) This is the bottom right vector field. Each vector is the same. (c) This is the top right vector field. Note that there is no z component, so the arrows are all parallel to the -plane. Notice that in clindrical coordinates, this vector field is F = sin θ, cos θ,. This means two things: first, the vectors are all unit vectors (length 1), and second, the vectors are tangent to circles (and perpendicular to the radial vector,, = r cos θ, r sin θ, ). (d) This is the bottom left vector field. Like vector field (a), this vector field is a radial vector field (parallel to,, z ). It is a negative multiple of,, z, so the vector field points toward the origin. The scale is b a factor of ρ 3, so vector fields far from the origin are much shorter than those closer.. We re solving two simple ordinar differential equations: d dt = and d dt =. The solutions are (t) = Ae t and (t) = Be t. (Often we call these constants something like = A and = B to indicate their meaning: the are the values of and at t =.) We get r(t) = e t, e t for some constants and. Notice that we can solve for e t = / = /, so we can determine that these flowlines lie on the curve =.

9 5. Here the two differential equations are d dt = and d dt = 1. The solution to the second is (t) = t +, from which the first differential equation becomes d dt = = t. The solution to this is (t) = 1 t t +, so we get r(t) = 1 t t +, t +, where the constants and have the natural meaning (the are the position at t = : r() =, ). Below we sketch some of these trajectories together with the original vector field: F(, ) =, 1 = i + j with flow lines with (, ) = (n, ) Notice that we can solve for t =, so we can determine that these flowlines lie on the curve = 1 ( ) ( ) +, which is a parabola opening to the left (as we ve seen in the sketches).

10 6. An gradient vector field F = P (, ), Q(, ) must satisf P = Q from Clairaut s theorem. This is called Clairaut s test. Direct computation shows that F does not satisf the Clairaut s test. For F 1, assume f = F 1. Then f = +, f = + 1. From f = +, we know f(, ) = + + C() b taking antiderivative with respect to. So f = + C (), C () = 1. Then we have C() = + C, where C is a constant. So f(, ) = C. 7. Direct computation shows that F satisfies the Clairaut s test. But it s not a gradient vector field. Because it s flow lines are circles, which are closed curves. This can not happen for gradient vector field because the value of the function alwas increases along the flow lines generated b its gradient vector fields.

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