1 Vectors and 3-Dimensional Geometry

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1 Calculus III (part ): Vectors and 3-Dimensional Geometry (by Evan Dummit, 07, v..55) Contents Vectors and 3-Dimensional Geometry. Functions of Several Variables and 3-Space Graphing Functions of Variables: Surfaces and Level Sets Graphing Functions of 3 Variables: Level Surfaces Vectors and Scalars The Dot Product The Cross Product Lines and Planes in 3-Space Vector-Valued Functions of One Variable, Derivatives Curves in 3-Space: Velocity and Acceleration; Arclength; Unit Tangent, Normal, and Binormal Vectors; Curvature and Torsion Vectors and 3-Dimensional Geometry From one-variable calculus, we are very used to -dimensional planar geometry: given a function f(x), we know how to produce the graph of the curve y = f(x) and use calculus to identify some of its important features such as local minima and maxima. We also know how to graph parametrically-dened functions such as the cycloid x = t + sin(t), y = + cos(t). And we even know, more or less, how to graph implicitly-dened functions, like the circle x + y =. In this chapter, we introduce 3-dimensional space (often called 3-space for short) and study the geometry of lines and planes, as well as more general curves and surfaces, using functions of several variables. Along the way, we will introduce and study vectors, which will help to clarify a great deal of the concepts.. Functions of Several Variables and 3-Space A function of several variables is, as the name indicates, a function that takes in several variables and outputs a value associated to the inputs. Example: The function f(x, y) = x + y takes in two values x and y and outputs their sum x + y. Example: The function g(x, y, z) = x y z takes in three values x, y, and z, and outputs the product x y z. Example: The function d(l, w) = l + w gives the length of a diagonal of a rectangle as a function of the rectangle's length l and its width w. Example: The function V (r, h) = 3 πr h gives the volume of a (right circular) cone as a function of its base radius r and its height h. As with functions of one variable, a function of several variables has a domain and a range: the domain is the set of input values and the range is the set of output values. For a function of two variables f(x, y), the domain is now a subset of the -dimensional plane rather than a subset of the real line. Because of this, domains of functions of more than one variable can be rather complicated.

2 In general, unless specied, the domain of a function is the largest possible set of inputs for which the denition of the function makes sense. We generally adopt the conventions that square roots of negative real numbers are not allowed, nor is division by zero. Example: For f(x, y) = x + y, the domain is the rst quadrant of the xy-plane, dened by the inequalities x 0 and y 0. Example: For f(x, y) = x + y, the domain is the set of points in the xy-plane which satisfy x + y : this describes all the points of the plane except for those lying strictly inside the unit circle. Example: For f(x, y) =, the domain is the set of points in the xy-plane which satisfy x y 0: x y this describes all points in the plane except those on the line y = x. Points in 3-space are represented by a triplet of numbers (x, y, z). The new coordinate z represents height above the xy-plane. From an algebraic standpoint, 3-space behaves quite similarly to -dimensional space. For example, in 3-space, we can measure the distance between any two points. By a suitable pair of applications of the Pythagorean Theorem, we can compute that the distance between points (x, y, z ) and (x, y, z ) is given by (x x ) + (y y ) + (z z )... Graphing Functions of Variables: Surfaces and Level Sets There are two primary ways to visualize a function f(x, y) of two variables. The rst way is to plot the points (x, y, z) in 3-dimensional space satisfying z = f(x, y). At the point (x, y) in the plane, the graph has the height z = f(x, y); so we see that as (x, y) varies through the plane, the function z = f(x, y) will trace out a surface, called the graph of f(x, y). Some example of simple graphs z = f(x, y) are given below. Example: The graph z = 0 is the xy-plane. Note more generally that any equation of the form a x + b y + c z = d for some constants a, b, c, d (with not all of a, b, c zero) will give a plane. We will study planes in more depth in a later section. Example: The graph z = x + y is a paraboloid (i.e., a parabolic dish). Example: The graph z = x + y is a right circular cone opening upward, with vertex at the origin. The second way is to plot the points (x, y) in the plane on the level sets f(x, y) = c (for particular values of c), as implicit curves. For a given function f(x, y) and a particular value of c, the points (x, y) satisfying f(x, y) = c are called a level set of f. For a function of two variables these sets will generally be curves, so they are also sometimes called level curves. If we graph many of these level curves together on the same axes, we will obtain a topographical map of the function f(x, y).

3 Some examples of level sets are given below: Example: The level sets for the function f(x, y) = x + y are circles in the plane. More specically, the level set x + y = c (for c > 0) is a circle with radius c centered at (0, 0). For c = 0 the level set is just the single point (0, 0), and for c < 0 the level sets do not contain any points at all. The rst graph is the level set x + y = ; the second graph contains level sets for c =,, 3,, 8, Example: Here are two collections of level curves for the function f(x, y) = x y, along with the 3D graph z = f(x, y). The rst set of level curves is f(x, y) = c for c = 4, 3,,, 0, and the second set of level curves is f(x, y) = c for c = 0,,, 3, The graph of z = x y is called a hyperbolic paraboloid (or more colloquially, a saddle) since it curves upward along the x-direction but downward along the y-direction. [The hyperbolic paraboloid is called that because it looks like a hyperbola in one cross-section, and a parabola in two others.] Example: Here are some level curves for the function f(x, y) = (x y ) e x y, along with a 3D graph of z = f(x, y), both plotted on the region 3 x 3 and 3 y 3: 0 3 As one can see from comparing the plots, the level curves indicate where the function is changing in value: many level curves grouped closely together indicates that the function is changing quickly so that the function's graph will be steep, while having level curves grouped far apart means that the function's graph will be fairly at. Here are a few more graphs z = f(x, y) for functions f(x, y): 3

4 Example: The graph z = x 3 3xy is sometimes called the monkey saddle, as it has three depressions rather than the two for the regular saddle (one for each leg, and one for the tail). Example: The graph z = e 3 ( ) x +y / cos x + y produces a surface that looks like ripples in a pool of water... Graphing Functions of 3 Variables: Level Surfaces We can also represent some information graphically for functions f(x, y, z) of three variables. Unfortunately, we cannot really produce proper graphs of these functions, since plotting a graph w = f(x, y, z) would require drawing a 4-dimensional picture. However, we can still talk about level sets of functions of 3 variables these are the points (x, y, z) satisfying a relation f(x, y, z) = c: these will (in general) give rise to level surfaces in 3-dimensional space. Any graph z = g(x, y) is an example of a level surface for the function f(x, y, z) = g(x, y) z: the points (x, y, z) with z = g(x, y) are the same as those with f(x, y, z) = 0. Example: The level surfaces for the function f(x, y, z) = x + y + z are spheres centered at the origin: by the distance formula, the points satisfying x + y + z = c (for c > 0) are precisely those which are at a distance of c from the origin but this is just another way of describing the sphere of radius c centered at (0, 0, 0). A graph of the sphere x + y + z = is below. Example: The level surfaces for the function f(x, y, z) = x + z are (right circular) cylinders running along the y-axis: again, from the distance formula, points satisfying x + z = c (for c > 0) are those which are at a distance of c from the y-axis but this is just another way of describing a right circular cylinder oriented along the y-axis. A graph of the cylinder x + z = is above. Example: Let us examine the level surfaces for the function f(x, y, z) = x + y z. The set of points satisfying x +y z = c for c > 0 forms a surface called a hyperboloid of one sheet, so named because the shape is a hyperbola in two cross-sections and an ellipse in the third, and the graph is connected (one sheet). 4

5 Remark: A hyperboloid of one sheet whose cross-sections are circles is an example of what is called a ruled surface: through each point on the surface pass two lines which are contained in the surface. Such surfaces can therefore be (physically) constructed using materials which are not curved. The hyperboloid of one sheet, in particular, is a common design for cooling towers. The set of points satisfying x + y z = 0 forms a (right circular) double cone whose axis is the z-axis and whose center point at the origin (0, 0, 0). The set of points satisfying x + y z = c for c < 0 forms a surface called a hyperboloid of two sheets, so named because the shape is a hyperbola in two cross-sections and an ellipse in the third, and the graph consists of two pieces (two sheets). Here are plots of the level surfaces for c =, c = 0, and c = respectively:. Vectors and Scalars A vector is a quantity which has both a magnitude and a direction. This is in contrast to a scalar, which carries only a magnitude. We denote the n-dimensional vector from the origin to the point (a, a,, a n ) as v = a, a,, a n, where the a i are scalars. We use the angle brackets rather than parentheses ( ) so as to emphasize the dierence between a vector and the coordinates of a point in space. We will, however, view coordinates of vectors and coordinates of points as essentially interchangeable. We also write vectors in boldface (v, not v), so that we can tell them apart from scalars. When writing by hand, it is hard to dierentiate boldface, so the notation v is also sometimes used. The typical way to think of vectors is as directed line segments: the length of the line segment gives the magnitude of the vector, and the direction the segment is pointing gives the direction of the vector. Note/Warning: Vectors are a little bit dierent from directed line segments, because we don't care where a vector starts: we only care about the dierence between the starting and ending positions. Thus: the directed segment whose start is (0, 0) and end is (, ) and the segment starting at (, ) and ending at (, ) represent the same vector,,. This distinction is rarely necessary in most applications, however. We can add vectors (provided they are of the same dimension!) in the obvious way, one component at a time: if v = a,, a n and w = b,, b n then v + w = a + b,, a n + b n. We can justify this using our geometric idea of what a vector does: v moves us from the origin to the point (a,, a n ). Then w tells us to add b,, b n to the coordinates of our current position, and so w moves us from (a,, a n ) to (a +b,, a n +b n ). So the net result is that the sum vector v+w moves us from the origin to (a + b,, a n + b n ), meaning that it is just the vector a + b,, a n + b n. 5

6 Geometrically, we can think of vector addition using a parallelogram whose pairs of parallel sides are v and w and whose diagonal is v + w: We can also 'scale' a vector by a scalar, one component at a time: if r is a scalar, then we have r v = ra,, ra n. Again, we can justify this by our geometric idea of what a vector does: if v moves us some amount in a direction, then v should move us half as far in that direction. Analogously, v should move us twice as far in that direction, while v should move us exactly as far, but in the opposite direction. Example: If v =,, and w = 3, 0, 4 then w = 6, 0, 8, and v+w =,,. Furthermore,v w = 7,, 0. At this point, we will restrict ourselves to talking just about -dimensional space and 3-dimensional space. Our primary reason for this is that most of the immediate applications of vectors (e.g., to physics) happen in 3-dimensional space. It will be useful to have a way to denote the unit coordinate vectors of 3-dimensional space. So we denote i =, 0, 0, j = 0,, 0, and k = 0, 0,. Example: The vector 3,, 5 is equal to 3i + j 5k. Note: Some authors primarily use ijk-notation when working with vectors. We will generally use angle bracket notation, except in a few cases when there are useful mnemonics that are easier to remember using ijk-notation..3 The Dot Product We now analyze lengths and angles between vectors. Denition: We dene the norm (length, magnitude) of the vector v = a,..., a n as v = (a ) + + (a n ). This is just an application of the distance formula: the norm of the vector a,..., a n is just the length of the line segment joining the origin (0,..., 0) to the point (a,..., a n ). Example: For v =,, and w = 3, 0, 4, we have v = ( ) + + = 3, and w = ( 4) = 5. If r is a scalar, we can see immediately from the denition that r v = r v, since we can just factor out a r = r from each term under the square root. From any nonzero vector we can nd a unit vector (that is, a vector of length ) in the same direction of v just by scaling v by over its length. In other words, the vector u = v is a unit vector in the same v direction as v. This vector u is sometimes called the normalization of v. Example: For v =,,, since v = 3, we see that u = direction as v. 3, 3, 3 is a unit vector in the same 6

7 Example: For w = 3, 0, 4, since w = 5, we see that u = direction of w. 3 5, 0, 4 5 is a unit vector in the If we have two vectors, we now know how to nd their lengths. But another thing we might want to know about two vectors is the angle θ between them. This motivates the denition of the dot product: Denition: The dot product of two vectors v = a,..., a n and v = b,..., b n is dened to be the scalar v v = a b + a b + + a n b n. Note: The dot product of two vectors is a scalar, not a vector! (For this reason, the dot product is sometimes called the scalar product of two vectors.) Example: The dot product, 3, 4 is ()(3) + ()(4) =. Example: The dot product,, 3, 0, 4 is ( )(3) + ()(0) + ()( 4) =. The dot product obeys several very nice properties reminiscent of standard multiplication. For any vectors v, v, v, w, and any scalar r, we can verify the following properties directly from the denition: The dot product is commutative: v w = w v. The dot product distributes over addition: (v + v ) w = (v w) + (v w). Scalars can be factored out of a dot product: (r v) w = r (v w). The dot product of a vector with itself is the square of the norm: v v = v. There is a nice relation between the dot product and the angle between two vectors: Theorem (Dot Product): For vectors v and v forming an angle θ between them, we have v v = v v cos(θ). Proof: To prove this statement, we use the Law of Cosines in the triangle formed by v, v, and v v, which states that v v = v + v v v cos(θ). Now we know that the square of the norm is the dot product of a vector with itself so we can apply this and the other dot product properties to see that v v = (v v ) (v v ) = (v v ) (v v ) (v v ) + (v v ) = v (v v ) + v. Now by comparing to the Law of Cosines expression, we see that everything cancels and leaves us precisely with the result we wanted. Using the Dot Product Theorem, we can compute the angle between two vectors: Example: Compute the angle between the vectors v =,, 3 and w = 0, 3,. We compute v w = ()(0) + ()( 3) + ( 3)() = 3, and v = w = ( 3) =. + + ( 3) = 3 and Then by the Dot Product Theorem, the angle θ between the vectors satises 3 = ( ) 3 cos(θ), meaning that θ = cos = π 3. Example: Compute the angle between the vectors v =,, and w = 3, 4, 0. We compute v w = ()(3) + ()(4) + ( )(0) = 4, and v = + + ( ) = 3 and w = ( 4) = 5. 7

8 Then by the Dot Product Theorem, the angle θ between the vectors satises 4 = 3 5 cos(θ), so ( ) 4 θ = cos. 5 Using the Dot Product Theorem, we can see that the sign and magnitude of the dot product is roughly measuring whether the vectors are pointing in the same direction: If v and v are nonzero vectors, then both v and v are positive, so by the theorem above, the value v v = v v cos(θ) will have the same sign as cos(θ). If 0 θ < π, the dot product v v will be positive. Furthermore, the smaller θ is, the larger the value of v v will be. Thus, a large positive value for the dot product indicates that the vectors are pointing in roughly the same direction. Inversely, if π < θ π, the dot product v v will be negative, and the larger θ is, the larger negative v v will be. Thus, a large negative value for the dot product indicates that the vectors are pointing in roughly opposite directions. We have a special name for the case where the angle is π : Denition: We say two vectors are orthogonal if their dot product is zero. ( π ) From the Dot Product Theorem, since cos = 0, we see that two nonzero vectors are orthogonal if the angle between them is π, which is to say, if they are perpendicular. Example: The vectors,, 4 and 3,, are orthogonal, since their dot product is ()(3)+( )()+ (4)( ) = 0. Remark: Since the dot product of the zero vector with any vector is zero, by our denition above, the zero vector is orthogonal to every vector. Another basic question about vectors is: given a vector v and another vector w, how much of w is in the direction of v, and how much of w is orthogonal to v? (This is a problem that often arises in Newtonian physics: one often needs to separate the vector representing a force into orthogonal components.) In other words: we want to write w = av + y, where y is orthogonal to v. The goal is to determine the value of the scalar a. Since y = w av, and y is orthogonal to v, taking the dot product with v gives 0 = y v = (w av) v = w v a(v v). Solving for a gives a = w v v v. Denition: The vector projection of w onto the nonzero vector v is the vector Proj v (w) = The vector projection of w onto v gives the piece of w in the direction of v. Example: Find the vector projection of w = 3, 4, onto the coordinate vectors i, j, and k. ( w v ) v. v v We have Proj i (w) = w i i i i = 3 i = 3i = 3, 0, 0. Next, Proj j (w) = w j j j j = 4 j = 4j = 0, 4, 0. Finally, Proj k (w) = w k k k k = k = k = 0, 0,. These results are consistent with our intuition about what the vector projection should be: for example, the component of 3, 4, in the direction of, 0, 0 should logically be 3, 0, 0, and this is indeed what we got. 8

9 Example: Find the vector projection of w = 6, 3, 0 onto the vector v =,,. We have Proj v (w) = ( w v ) (6)() + ( 3)( ) + (0)( ) v = v v + ( ) + ( ) v = 8 v = v = 4, 4,. 9 As a sanity check, note that w Proj v (w) =,,, and indeed, this vector is orthogonal to v, because,,,, = ()() + ()( ) + ()( ) = 0..4 The Cross Product In addition to the dot product, we have another type of product dened for vectors in 3-space, called the cross product: Denition: The cross product of v = x, y, z and v = x, y, z is dened to be the vector v v = y z y z, z x z x, x y x y. It is orthogonal to both v and v. Important Note: The cross product is only dened for vectors with 3 components, and outputs another vector with 3 components. In contrast, the dot product is dened for vectors of any length, and outputs a scalar. A way to remember the cross product formula (aside from memorization) is the determinant formula i j k v v = det x y z x y z = y z y z i x z x z j + x y x y k, where i, j, k are the standard unit vectors: i =, 0, 0, j = 0,, 0, and k = 0, 0,. It may seem a little unusual to have vectors inside a determinant, but it works out to the correct answer. Don't forget the minus sign on the middle term! A central property of the cross product v v is that it is orthogonal both to v and to v. To verify this, we can just evaluate the dot products v (v v ) and v (v v ) and check that they are both zero. For example, we have v (v v ) = x (y z y z ) + y (z x z x ) + z (x y x y ), which some algebra will conrm is equal to zero. Example: If v =,, 3 and v =,,, nd v v and v v and verify that v v is orthogonal to both v and v. i j k First, v v = 3 = 3 i 3 j + k = 4, 5, 3. i j k Also, v v = 3 = 3 i 3 j + k = 4, 5, 3. To check: we have v (v v ) =,, 3 4, 5, 3 = = 0, and also v (v v ) =,, 4, 5, 3 = = 0. Here are a few other algebraic properties of the cross product: Unlike the dot product, the cross product is NOT commutative! Indeed, we can see from the denition (and the example above) that v v = (v v ). In particular, if we set v = v in this denition, we see that v v = 0 for any vector v, where 0 = 0, 0, 0 denotes the zero vector. The cross product does have a distributive property, like with the dot product: it is fairly easy to check from the denition that (v + v ) w = (v w) + (v w). Scalars can also be factored out of a cross product: (r v) w = r (v w) = v (r w). 9

10 The cross product is also not associative. It does satisfy the so-called Jacobi identity: u (v w) + v (w u) + w (u v) = 0. Like with the dot product, there is a relationship involving the cross product and the angle between two vectors: Theorem (Cross Product): If θ is the angle between v and v, then v v = v v sin(θ) = A, where A is the area of the parallelogram formed by v and v. Proof: We just need to show that v v +(v v ) = v v, because we know that v v = v v cos(θ) from the Dot Product Theorem. To check this, we simply multiply it out. So we need to see that (y z y z ) + (z x z x ) + (x y x y ) + (x x + y y + z z ) is equal to [ (x ) + (y ) + (z ) ] [(x ) + (y ) + (z ) ]. When we expand the rst quantity, we get each of the 9 possible square terms ( ) where and are each one of x, y, or z, and the cross terms like x x y y will all cancel out. We get exactly the same sum of 9 square terms when we expand the second quantity: thus, they are equal. For the statement about the area, it is a standard fact of geometry that the area of the triangle with sides v and v is v v sin(θ). The parallelogram's area is twice this. Remark: This quite nice property is one reason we chose the denition we did for the cross product. Example: Find the area of the triangle whose vertices are the points A(,, ), B(, 3, ), and C(,, ). By the Cross Product Theorem, the area of the triangle is A = v w, where v and w are vectors representing two sides of the triangle. We can take v = B A =,, and w = C A = 0, 5,. We then compute v w = i j k 0 5 = 5 i 0 j k = 3,, 5. Hence the area of the triangle is 3,, 5 = 3 + ( ) + 5 = Lines and Planes in 3-Space Among the many useful applications of vectors is that they can be used to give simple descriptions of lines and planes in 3-space. In the -dimensional plane, the general equation of a line is ax + by = d for some constants a, b, d (with a and b not both zero). The 3-dimensional version of this would be ax + by + cz = d (with a, b, c not all zero), which will describe a plane. To describe lines in 3-space, we instead need to describe them as parametric curves, in the form x = x(t), y = y(t), z = z(t) for some functions x(t), y(t), z(t). As t varies, the set of points (x(t), y(t), z(t)) will form a curve: the goal is to choose the functions so that the curve is a line. Proposition (Parametrization of a Line): Given distinct points P = x, y, z and P = x, y, z, the points x, y, z on the line l through P and P are given parametrically by x, y, z = P + t v, as t varies through the real numbers, and v = P P = x x, y y, z z. Proof: There is a unique line between two points, by the axioms of geometry. So we just need to check that this is a line, and that it goes through P and P. The parametric equation for l explicitly tells us that x = x + t(x x ), y = y + t(y y ), and z = z + t(z z ), and these are all linear equations. So it's a line. 0

11 Setting t = 0 yields P, while setting t = yields P + (P P ) = P. Thus, l passes through both P and P, so it is the desired line. Note: This procedure works to nd the parametrization of a line in a space of any dimension, not just 3-space. Remark: We call the vector v = P P the direction vector for the line l: it tells us in which direction the line is pointing. The term P in the sum P + t v species which particular line we want, of all possible lines in that direction. Example: Find a parametrization of the line through the points (,, 3) and (,, ). First, we nd the direction vector: we get v =,,,, 3 =, 0, 4. Then the line is given parametrically by x, y, z =,, 3 + t, 0, 4 = t,, 3 4t. More explicitly, this line is described by x = t, y =, z = 3 4t. As t ranges through the real numbers, the points (x, y, z) will range along the desired line. Next, we tackle descriptions of planes. Proposition (Planes and Normal Vectors): The plane dened by a x + b y + c z = d is orthogonal to its normal vector n = a, b, c. In other words, every line lying in this plane is orthogonal to a, b, c. Conversely, given a nonzero vector n = a, b, c, there is a unique plane normal to that vector passing through a given point (x 0, y 0, z 0 ), and its equation is a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. Proof: Suppose l is a line in the plane. All we need to show is that its direction vector is orthogonal to n. So suppose the direction vector is v = P P, where both of the points P = x, y, z and P = x, y, z lie in the plane. Then P n = a x + b y + c z = d since P lies in the plane, and similarly P n = d. But then we have v n = (P P ) n = P n P n = d d = 0, so that v is orthogonal to n as claimed. For the converse statement, clearly if n = a, b, c then the equation of the plane must be a x+b y+c z = for some value of, by the previous argument. But if we are given a point that lies in the plane, we can plug in to see that = ax 0 + by 0 + cz 0, and so we have uniquely determined the equation of the plane, and hence the plane. We can rewrite the equation as a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0, so this is the desired plane. Remark: This proposition says that a plane is specied by its normal vector along with a point that the plane passes through. Example: The plane with normal vector, 3, passing through (,, 7) has equation (x ) + 3(y ) (z 7) = 0. Using these two results we can solve a wide variety of problems involving lines and planes. The fundamental ideas to remember when working with lines and planes are: specifying a line requires knowing its direction vector and a point it passes through, and specifying a plane requires knowing its normal vector and a point it passes through. Example: Find an equation for the plane containing the vectors v =,, and v =,, 3 and passing through the point P = (,, ). The normal vector to the plane is orthogonal to both v and v, so we can nd it by taking their cross product. i j k We get n = v v = 3 = 3 i 3 j + k =,,. Sanity check: we have n v = ()()+( )()+()() = 0 and n v = ()()+( )()+()(3) = 0.

12 Now we see that the equation for the plane is (x ) (y + ) + (z ) = 0, which we can rewrite as x y + z = 4. Example: Find an equation for the plane passing through the three points P = (3, 0, ), P = (,, ) and P 3 = (,, 4). We need the plane's normal vector n. To nd it, we need to nd two vectors lying in the plane, and then take their cross product to get n. We can get two such vectors as v = P P and v = P 3 P. We have v = P P =,, 3 and v = P 3 P = 5,, 5. Then n = v v = 3 5 i j + 5 k = 7, 5, 8. Sanity check: n v = (7)( 5) + ( 5)() + (8)(5) = 0 and n v = (7)( ) + ( 5)() + (8)(3) = 0. Thus the equation of the plane is 7(x 3) 5(y 0) + 8(z + ) = 0, or 7x 5y + 8z = 3. For an extra error check, we verify that all three points do lie in this plane: we have 7(3) 5(0) + 8( ) = 3, 7() 5()+8() = 3, and 7( ) 5()+8(4) = 3. Hence this is the correct equation. Example: Parametrize the line of intersection of x y + z = 3 and x + y z = 0. The normal vector to each plane will be orthogonal to the line of intersection. Therefore, we can get the direction vector of the line by taking the cross product of the two planes' normal vectors. We have n =,, and n =,,. Then the cross product is v = n n =, 5, 3. Sanity check: v n = () + 5( ) + 3() = 0 and v n = () + 5() + 3( ) = 0. Now we need to nd a point in both planes (since we need a point on the line). We try looking for one with x = 0: this requires y + z = 3 and y z = 0. Solving yields y = z = 3, so (0, 3, 3) is in both planes and thus on the line l. Applying the line parametrization formula gives l : x, y, z = t, 5t + 3, 3t + 3. Another common problem is to compute the distance from a point to a plane. It turns out there is a reasonably nice formula for this distance: Proposition (Point-to-Plane Distance): The distance from the point P = (x 0, y 0, z 0 ) to the plane ax+by+cz = d is equal to ax 0 + by 0 + cz 0 d a + b + c. The shortest vector connecting P to the plane will be in the same direction as the normal vector to the plane. In theory, we could use this information to parametrize the line joining P to the plane, nd the intersection of the line and plane, and nally compute the distance we seek. However, there is a less messy way using vector projections. Proof: Let Q be any point on the plane ax + by + cz = d and let n be the normal vector to the plane. Then the vector v connecting P to the plane is given by the vector projection of w = Q P onto the normal n. ( n w ) n w n w Explicitly, v = Proj n (w) = n. Then the length of v is v = n n n =. n n From our earlier results, we can take n = a, b, c : then n w = n P n Q = (ax 0 + by 0 + cz 0 ) (d), and n = a + b + c. (The fact that n Q = d is just a restatement of the fact that Q lies in the plane ax + by + cz = d.) Then v = n w n = ax 0 + by 0 + cz 0 d, as claimed. a + b + c Example: Find the distance from the point P = (,, 4) to the plane x + y z = 3. The formula gives ( ) = 6 9 =.

13 Example: Find the distance from the point P = (,, 4) to the plane z = 0. The plane can be written as 0x + 0y + z = Then the formula gives = = 4. This result agrees with our intuition: the plane z = 0 is the horizontal xy-plane, and so the distance of any point to this plane is simply the absolute value of its z-coordinate. Example: Find the distance between the planes x + y + z = and x + y + z = 7. These two planes are parallel since they clearly have the same normal vector,,. To compute the distance between them, we can pick any point on one plane and compute its distance to the other plane Since the point (0,, 0) lies on x+y+z =, the formula gives the distance as = =. Remark: By the same argument, the distance between the two planes ax+by+cz = d and ax+by+cz = d d d is equal to a + b + c. Example: Find the distance between the lines l : x, y, z = 4t, t, t and l : x, y, z = + s, 3, 5 s. Imagine the two lines l and l as sitting inside two parallel planes. The distance between the lines will then be the same as the distance between the planes, which we can compute by nding the distance of any arbitrary point in one plane to the other plane (using the point-to-plane distance formula). In order to nd these two parallel planes, we just need to nd their (common) normal vector n, which will be orthogonal to the direction vectors of both lines. So to compute n, we can simply take the cross product of the direction vectors. We compute n = 4,,, 0, =,,, meaning that the planes are of the form x+y+z = d for some value of d. To get the plane containing l, choose any point on l. Picking t = 0 gives (0,, ), and so we get d = = 6, so the rst plane is x + y + z = 6. Setting s = 0 in the equation for l gives the point (, 3, 5), which lies in the second plane Then using the point-to-plane distance formula gives the required distance as = = 4..6 Vector-Valued Functions of One Variable, Derivatives Denition: A vector-valued function of one variable r(t) is a function whose output is a vector, each of whose components is a function of the parameter t. Examples: r (t) = t, t, r (t) = t, t, t, r 3 (t) = cos(t ), e t, tan ( t + ). We have already encountered a few vector-valued functions when we studied lines in 3-space: parametrization of a line is an example of a vector-valued function. the We can add and scalar-multiply vector-valued functions in the same manner as normal vectors. Example: For r (t) = e t, cos(t), t and r (t) = t, 0, t we have r (t)+r (t) = e t + t, cos(t), and r (t) = t, 0, t. We will primarily be interested in vector functions of the form r(t) = x(t), y(t) and r(t) = x(t), y(t), z(t), which have a single input parameter t and output a vector with or 3 coordinates. As we vary t, these functions will trace out parametric curves in or 3-dimensional space (respectively). 3

14 Later, we will also be interested in vector functions of the form r(s, t) = x(s, t), y(s, t), z(s, t), for two input parameters s and t. These functions, in general, will describe surfaces in 3-dimensional space as we vary s and t. If we graph a vector-valued function r(t) = x(t), y(t) as t varies, we will obtain a curve in the plane. Example: The curve given by r(t) = cos(t), sin(t) traces around the unit circle as t varies. It traces counterclockwise, and completes one revolution after a time interval of π. Remark: Vector-valued functions in the plane are often studied in single-variable calculus (where they are usually termed parametric curves). If we graph a vector-valued function r(t) = x(t), y(t), z(t) as t varies, we will obtain a curve in space. A parametric curve in 3-space is just the set of points (x(t), y(t), z(t)) for some functions x(t), y(t), and z(t). Reminder: We are thinking of functions giving parametric curves interchangeably with vector functions: instead of getting a point (x(t), y(t), z(t)) as output from a function, we can equally well think of getting the vector output x(t), y(t), z(t). Example: The curve given by r(t) = t, t, t is a line passing through the origin with direction vector,,. More generally, as we have already seen, for any constants a, a, a 3 and b, b, b 3, the curve given by r(t) = (a t + b, a t + b, a 3 t + b 3 ) will be a line passing through the point (b, b, b 3 ) and pointing in the direction of the vector a, a, a 3. Example: The curve given by r(t) = (sin(t), cos(t), t) is a helix wrapping around the cylinder x + y =. We can see that as t increases, the x and y parts just trace around a unit circle at constant speed, while z increases at constant speed. Here is a plot of the curve winding around the cylinder, for 0 t 8π: Example: The curve given by r(t) = cos(t), sin(t), cos(t) is an ellipse. We can see that this curve is an ellipse by observing that it is the intersection of the plane z = x with the cylinder x + y = (above). The next question to ask is: can we take derivatives of vector-valued functions? The answer is yes: Denition: The derivative of the vector function r(t) is given by r r(t + h) r(t) (t) = lim h 0 h limit exists., provided the Note the extreme similarity of this denition with the denition of the derivative of a (scalar) function of one variable f(t), which reads f f(t + h) f(t) (t) = lim. h 0 h For r(t) = x(t), y(t), z(t), then by applying the denition of the derivative, we see that r (t) = x (t), y (t), z (t). Quite sensibly, taking the derivative of a vector function is the same thing as differentiating each component. Example: For r(t) = e t, cos(t), t, we have r (t) = e t, sin(t), t. 4

15 By the limit part of the denition, we see that the derivative of a vector function tells us in which direction the curve is moving at time t. More specically, r (t) is a vector tangent to the graph of r(t). Compare to the situation of the derivative of a (scalar) function of one variable f(t): the derivative f (t) gives the slope of the tangent line. We can also take higher derivatives: for example, r r (t + h) r (t) (t) = lim, and so forth. h 0 h Example: For r(t) = e t, cos(t), t, we have r (t) = e t, sin(t), t, so r (t) = e t, cos(t), and r (t) = e t, sin(t), 0, and so on. The derivatives of vector-valued functions have the same physical interpretations as we are used to: The rst derivative r (t) gives the velocity of the particle at time t. The second derivative r (t) gives the acceleration of the particle at time t. The magnitude r (t) of the velocity vector gives the speed of the particle. Derivatives of vector functions satisfy rules strongly reminiscent of the product rule with regard to the dot d and cross products: dt [r r ] = r (r ) + (r ) r and d dt [r r ] = r (r ) + (r ) r. These properties can be veried by expanding out the dot and cross products of [r (t + h) r (t)] with [r (t + h) r (t)], applying the limit denition of derivative, and simplifying..7 Curves in 3-Space: Velocity and Acceleration; Arclength; Unit Tangent, Normal, and Binormal Vectors; Curvature and Torsion A particle traveling on a parametric curve r(t) = x(t), y(t), z(t) in 3-space possesses a number of associated vectors and scalar quantities. The velocity of the particle is given by v(t) = r (t). The velocity vector is the direction vector for the tangent line to the curve r(t). The acceleration of the particle is given by a(t) = v (t) = r (t). The arclength of the curve between t = t and t = t is given by s = t t [x (t)] + [y (t)] + [z (t)] dt = t v(t) dt. t The arclength s obeys the relation ds dt = v(t). Example: Find the velocity and acceleration of a particle traveling along the path r(t) = + e t, + t, 3 e t. Also, nd the arclength of the path from t = 0 to t =. The velocity is v(t) = r (t) = For the arclength, rst we compute v(t) = e t,, e t, and the acceleration is a(t) = v (t) = e t, 0, e t. Then the arclength is s = 0 [et + e t ] dt = [e t e t ] t=0 = e e. (e t ) + ( ) + (e t ) = e t + + e t = e t + e t. Example: A projectile is red at time t = 0s from initial position r(0) = 0, 0, 0 m in a vacuum with initial velocity v(0) =, 3, 45 m /s. Assuming that the only force acting on the projectile is the downward acceleration due to gravity of a(t) = 0, 0, 0 m /s, nd the position of the projectile at time t, the time it hits the ground, and set up an integral for the total distance it travels along its path. Since the acceleration is a(t) = 0, 0, 0 m /s, integrating yields v(t) = C, C, C 3 0t m /s for some constants C, C, C 3. 5

16 Setting t = 0 yields v(0) = C, C, C 3, so since v(0) is given as, 3, 45 m /s, we see that v(t) =, 3, 45 0t m /s. Integrating again yields r(t) = C 4 + t, C 5 + 3t, C t 5t m. Since the initial position is r(0) = 0, 0, 0 m, we see that r(0) = C 4, C 5, C 6, and therefore r(t) = t, 3t, 45t 5t m. The height of the projectile is the z-coordinate, so it hits the ground when 45t 5t = 0. Factoring gives 5t(9 t) = 0, so t = 0 or t = 9. Since the projectile is red when t = 0s, it hits the ground at time t = 9s. The total distance travelled by the particle is the arclength of the path between t = 0s and t = 9s. Since v(t) = ˆ (45 0t) m /s, the arclength is given by (45 0t) dt meters. Remark: This integral can be computed using a series of substitutions, although the result is rather messy: ( (45 0t) dt = log 45 + ) When working with curves in 3-space (especially in physics), it is often useful to use normalized tangent vectors, rather than just the velocity and acceleration vectors. There are three standard normalized tangent vectors: the unit tangent vector T(t), the unit normal vector N(t), and the unit binormal vector B(t). The unit tangent vector T(t) = v(t) v(t) 0 measures the direction in which the particle is moving. By the denition, the unit tangent vector satises T(t) =, and T(t) is in the same direction as r (t). Thus, it is a unit vector, in the same direction as the velocity (or tangent) vector. (Whence the name.) The unit normal vector N(t) = T (t), orthogonal to the unit tangent vector, measures roughly the T (t) direction that forces are pulling the particle. By the denition, the unit normal vector satises N(t) =, as long as T (t) 0. To see that N(t) is orthogonal to T(t), since T(t) T(t) =, taking the derivative with respect to t gives T(t) T (t) + T (t) T(t) = 0. But since the dot product is symmetric, this says T (the unit tangent vector) and T (the normal vector) have dot product 0. In other words, the (unit) normal vector is orthogonal to the tangent vector. v(t) a(t) The unit binormal vector B(t) = T(t) N(t) =, orthogonal to the tangent and normal v(t) a(t) vectors, gives the direction perpendicular to the particle's plane of motion. By its denition via the cross product, the binormal vector is orthogonal to both the (unit) normal and (unit) tangent vectors, and since those vectors have length, so does B. Thus, we see that the unit tangent, unit normal, and unit binormal vectors are all orthogonal to each other provided that the normal and binormal vectors are dened (which is not always guaranteed because of the division by T (t) in the denition of N). There are several dierent planes associated to these normalized tangent vectors: The osculating plane is the plane containing T(t) and N(t), passing through the point r(t). It is sometimes called the tangent plane to the curve. The normal plane is the plane containing N(t) and B(t), passing through the point r(t). The rectifying plane is the plane containing T(t) and B(t), passing through the point r(t). Example: Find the unit tangent, normal, and binormal vectors for the curve r(t) = 3 sin t, 5 cos t, 4 sin t. We have v(t) = r (t) = 3 cos t, 5 sin t, 4 cos t, so we get v(t) = 9 cos t + 5 sin t + 6 cos t = 5 cos t + 5 sin t = 5. 6

17 Therefore, T(t) = v(t) 3 v(t) = 5 cos t, sin t, 4 5 cos t. Next, we have T (t) = sin t + cos t =. 35 sin t, cos t, 45 sin t, so T (t) = 9 5 sin t + cos t sin t = Therefore, N(t) = T (t) T (t) = Finally, B(t) = T(t) N(t) = 35 sin t, cos t, 45 sin t. i j k 3 5 cos t sin t 4 5 cos t 3 = 5 sin t cos t 4 5 sin t 4 5, 0, 3. 5 Another quantity associated to the motion of a particle in space is the curvature κ(t) = T (t) v(t) which measures how much the path of the particle is curving. = v(t) a(t) v(t) 3, Note that the curvature is always nonnegative, since the numerator and denominator are nonnegative. Example: Find the curvature of the line r(t) = t, t, t. We have v(t) =,, so v(t) = 3. Then T(t) = 3, 3, and T (t) = 0, 0, 0. Thus, the 3 curvature of this line is always equal to 0. More generally, any line has curvature 0, since for a line T(t) is always constant, implying that T (t) = 0. Conversely, from essentially the same observation, the only curves whose curvature is zero everywhere are lines, since a curve with curvature zero must have T(t) be constant. Example: Find the curvature of the circle r(t) = r cos t, r sin t, 0. We have v(t) = r sin t, r cos t, 0 so v(t) = r. Then T(t) = sin t, cos t, 0 and so T (t) = cos t, sin t, 0. Then T (t) =, so we get κ(t) = r. More generally, for any circle of radius r, the curvature is r. From these examples, we can see that a small curvature indicates that the path is close to a line segment, while a large curvature means the path is turning sharply. For a general curve, the quantity is called the radius of curvature of the curve: it gives the radius κ(t) of the tangent circle to the curve. (For a circle of radius r, the radius of curvature is also r.) The acceleration of a particle can be decomposed into a component in the direction of the unit tangent vector T and a component in the direction of the unit normal vector N. (This decomposition is very frequently used in physical applications.) The tangential and normal components of acceleration are a T = d dt v and a N = κ v, and they give the decomposition a = a T T + a N N. Note that although a T can be positive, negative, or zero, a N is always nonnegative because κ is nonnegative. To see that a = a T T+a N N for these values of a T and a N, observe that v = v T, and then dierentiate both sides using the Product Rule to get a = d v T + v dt = d v T + v T (t) N; then dt dt dt a T = d dt v and a N = v T (t) = κ v. 7

18 From a = a T T + a N N, one has v a = v T (a T T + a N N) = v a T (T T) + v a N (T N) = v a N B. Then one can immediately derive the formulas B = v a v a and v a = v a N = κ v. The nal quantity associated to a curve's motion is the torsion τ = B (t) N(t) v(t) = v a which measures how sharply the path of the particle is twisting out of its plane of motion. x y z x y z x y z The torsion is the least intuitive of all of the quantities we have listed. (Note that it is the only one to involve the third derivatives of the coordinate functions.) Another relation satised by the torsion, which may be easier to use, is τn = db v dt. By observing that ds dt = v, we can rewrite the above relation as db ds Frenet-Serret formulas: the other two say dt ds = κn and dn ds = τb κt. = τn. This is one of the three The dt dn formula follows from the denition of κ, while the formula can be obtained by writing ds ds N = B T and then taking the derivative with respect to s: one obtains dn ds = db ds T + B dt ds = τ(n T) + B (κn) = τb κt. Example: For the curve r(t) = sin(3t), cos(3t), 4t, nd T, N, B, κ, τ, a T, a N. We have v(t) = 3 cos(3t), 3 sin(3t), 4, so that v = 9 cos (3t) + 9 sin (3t) + 6 = 5. 3 cos(3t) Thus, we get T =, 3 sin(3t), Next we compute dt dt = 9 sin(3t), 9 cos(3t), 0 so that dt 5 5 dt = 9 5. Then we see that N = sin(3t), cos(3t), 0. i j k 3 cos(3t) Next, B = T N = 3 sin(3t) 4 4 cos(3t) =, 4 sin(3t), sin(3t) cos(3t) 0 The curvature is given by κ = T (t) v(t) We have db dt = sin(3t), 5 = 9/5 5 = 9 5. db cos(3t), 0, so τ = dt N = 5 v 5. Finally, we have a T = d dt [5] = 0 and a N = κ v = 9. As a sanity check, we compute a(t) = 9 sin(3t), 9 cos(3t), 0, and indeed a(t) = 0 T + 9 N., Well, you're at the end of my handout. Hope it was helpful. Copyright notice: This material is copyright Evan Dummit, You may not reproduce or distribute this material without my express permission. 8

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