Vector Fields and Line Integrals


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1 Vector Fields and Line Integrals 1. Match the following vector fields on R 2 with their plots. (a) F (, ), 1. Solution. An vector, 1 points up, and the onl plot that matches this is (III). (b) F (, ) 1,. Solution. An vector 1, points up, and the onl plot that matches this is (I). (c) F f, where f is the scalarvalued function f(, ) Solution. f(, ) 2, 2. If we draw the vector 2, 2 with its tail at (, ), then it points awa from the origin. In addition, as and get bigger, the vectors 2, 2 get longer, which describes plot (IV). (d) F (, ) 2 +, Solution. The onl choice left is (II). The vectors there all appear to be the same length, and indeed, 1 for all (, ) (ecept for (, ), since F (, ) is undefined) (I) (II) (III) (IV) 1
2 (The vectors in each picture have been uniforml rescaled so that the pictures are more clear.) 2. Match the following vector fields on R with their plots. z z z (I) (II) (III) (a) F (,, z),, 1. Solution. This is a constant vector field; that is, at ever point (,, z), we draw the same vector,, 1. This matches (II)., 2 + z 2,. (b) F (,, z) z 2 + z 2 Solution. If we draw the vector, 2 + z 2, z 2 + z 2 with its tail at the point (,, z), then it points toward the ais. This matches (III). (c) F (,, z) ( z 2 ), /2 ( z 2 ), z. /2 ( z 2 ) /2 Solution. If we draw the vector ( z 2 ), /2 ( z 2 ), z /2 ( z 2 ) /2 with its tail at the point (,, z), then it points toward the origin. This matches (I).. Vector fields are used to model various things. For each of the following descriptions, decide which of the vector field plots in #2 (I, II, or III) gives the most appropriate model. (a) Force of gravit eperienced b a fl in a room. More precisel, F (,, z) is the force due to gravit eperienced b a fl located at point (,, z) in a room. (Remember that force is a vector.) Solution. No matter where the fl is in the room, it eperiences the same force due to gravit the magnitude of the force is just the fl s mass multiplied b acceleration due to gravit, or 9.8 m/s 2 (this is reall the fl s weight), and the direction of the force is alwas straight toward the ground. Therefore, we want a constant vector field with all vectors pointing toward the ground, which means (II) is the best model. (b) Force of Earth s gravit eperienced b a space shuttle. More precisel, F (,, z) is the force that 2
3 Earth s gravitational field eerts on a space shuttle located at the point (,, z). In the picture ou ve chosen, where is the Earth? Solution. No matter where the space shuttle is, the force eerted b Earth s gravitational field will be a vector pointing toward the Earth. The force should be stronger near the Earth (vectors of greater magnitude or length) and less strong awa from the Earth (vectors of smaller magnitude). This matches (I), with the Earth at the origin. (c) f, where f(,, z) is the temperature in a room in which there is a heater along one edge of the floor. In the picture ou ve chosen, where is the heater? (Hint: The gradient of a function f alwas points in the direction in which f is?) Solution. We know that the gradient of a function f alwas points in the direction in which f is increasing the most (instantaneousl). In this case, f represents temperature, so the gradient at an point points towards the warmest direction. This matches (III), with the heater being positioned along the ais. 4. Let F be the vector field on R 2 defined b F (, ) 1,. (We saw this vector field alread in #1.) (a) Let be the bottom half of the unit circle (in R 2 ), traversed counterclockwise. Evaluate F d r. Solution. First, we must parameterize the curve. One possible parameterization is r(t) cos t, sin t with t 2. (1) Then, we can simpl compute, using the definition of the line integral: 2 F d r F ( r(t)) r (t) dt F (cos t, sin t) sin t, cos t dt 1, cos t sin t, cos t dt ( sin t + cos 2 t) dt ( sin t + ) 1 + cos 2t 2 cos t + t sin 2t t t dt (b) We write to mean the same curve as (in this case, the bottom half of the unit circle) but oriented in the opposite direction (so clockwise instead of counterclockwise). What is F d r? (1) One wa to arrive at this parameterization is to think about a particle traveling along the curve. In polar coordinates, the unit circle is just r 1. The particle starts at θ and moves toward θ 2 (with θ increasing). So, in polar coordinates, we could parameterize the curve just b taking r 1, θ t with t 2. Translating this back to artesian coordinates, cos t and sin t, so we have the parameterization r(t) cos t, sin t.
4 Solution. It is simpl the negative of F d r, or 2 2. (c) Now, let be the line segment from (, ) to (, 1). Looking at the picture of F (in #1), do ou think F d r is positive, negative, or zero? Wh? Solution. Here is a plot of the vector field, together with the curve (drawn in red): Along the path, F alwas points to the right, while the path goes up. In particular, F is alwas perpendicular to the path, so the line integral should be zero. To be more precise about it, if we were to find a parameterization r(t) of the curve, then F ( r(t)) is alwas perpendicular to r (t) (which is tangent to ), so the dot product F ( r(t)) r (t) is alwas. This is the thing we integrate to compute the line integral, so the line integral must be as well. (2) (d) What if is instead the line segment from (, ) to (1, 1)? Is the line integral F d r positive, negative, or zero? Solution. Here is a plot of the vector field, together with the curve (drawn in red): Along this path, the vector field generall goes the same direction as the path; that is, the path makes an acute angle with the vectors in the vector field. So, the line integral is positive. (2) To confirm, we could parameterize the curve and compute the line integral; one parameterization of is r(t), t, t 1. 4
5 More precisel, if we were to find a parameterization r(t) of the curve, then F ( r(t)) (the vector field F at a particular point on the path) alwas makes an acute angle with r (t) (the direction the path is going), so the dot product F ( r(t)) r (t) is alwas positive. () Since we integrate this dot product to compute the line integral, the line integral will also be positive. (4) 5. Let f(, ) e + and F f, a vector field on R 2. Let be the curve in R 2 parameterized b r(t) t, t 2, t 1. (a) ompute the line integral F d r. Solution. We compute F (, ) f(, ) e +,. To find the line integral, we just use the definition of the line integral: F d r F ( r(t)) r (t) dt F (t, t 2 ) 1, 2t dt e t + t 2, t 1, 2t dt (e t + t 2 ) dt e t + t t1 e t (b) What is f( r(t))? Did ou use this anwhere when ou computed the line integral in (a)? an ou eplain wh this happened? Solution. f( r(t)) e t + t. We saw this in the secondtolast step of (a): it was what we got from integrating F ( r(t)) r (t) (so we ended up just plugging the starting and ending values of t into this). This can be eplained b the hain Rule, which sas that d dt f( r(t)) f( r(t)) r (t). In this case, f F, so d dt f( r(t)) F ( r(t)) r (t). This was the thing we were integrating with respect to t, and we know b the Fundamental Theorem of alculus that if we integrate d dt (something) with respect to t, we ll just get that something back. (c) Suppose we want to look at a new curve, parameterized b r(t) (sin t)e cos t2 + t, sin t + cos t with t. Find F d r. () Remember that the dot product u v can be written as u v cos θ, where θ is the angle between the vectors u and v. If θ is an acute angle, then cos θ > ; if θ is an obtuse angle, then cos θ <. So, if the angle between u and v is acute, then u v > ; if the angle between u and v is obtuse, then u v <. (4) To confirm, we could parameterize the curve and compute the line integral; one parameterization of is r(t) t, t, t 1. 5
6 Solution. Using what we figured out (b), we can jump directl to the end of the computation: F d r f( r(t)) t t f( r()) f( r()) From the given formula for r(t), r(), 1 and r(), 1, so the line integral is f(, 1) f(, 1). 6
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