Midterm 1 Review. Distance = (x 1 x 0 ) 2 + (y 1 y 0 ) 2.

Midterm 1 Review Comments about the midterm The midterm will consist of five questions and will test on material from the first seven lectures the material given below. No calculus either single variable or multi-variable will be needed to answer the questions. You will be expected to know the basic rules of algebra for simplifying expressions and trigonometric identities. Also, a good geometric intuition can be helpful on some problems. You might need to draw a picture to answer some questions, but such pictures will be simple to draw. Points and distance Before we get into multi-variable calculus we need to first need to develop the tools and terminology to handle multiple variables i.e., we need to shed off our two dimensional limitations and embrace higher dimensions. In two dimensions we represent points by a, b where a indicates our x-coordinate i.e., how far we move in the x-direction and b indicates our y- coordinate i.e., how far we move in the y-direction. Using the Pythagorean Theorem we can find distance between points by measuring the length of the line segment connecting the two points. In particular, given x 0, y 0 and x 1, y 1 then Distance = x 1 x 0 2 + y 1 y 0 2. In three dimensions we represent a point by a, b, c where a and b are as before and c is the z- coordinate i.e., how far we move in the z-direction. Instead of using the axis to split up space we use planes, i.e., the xy-plane z = 0, the xz-plane y = 0 and the yz-plane x = 0 split up three dimensional space into eight octants. By using the Pythagorean Theorem bootstrapping our way up we can again measure the distance between points as the length of the line segment connecting the points. In particular, given x 0, y 0, z 0 and x 1, y 1, z 1 then Distance = x 1 x 0 2 + y 1 y 0 2 + z 1 z 0 2. Once we can do distance we can also do spheres. A sphere is the set of points that are distance r called the radius from a central point x 0, y 0, z 0, by the distance formula this is the set of points x, y, z with x x 0 2 + y y 0 2 + z z 0 2 = r 2. Sometimes we might be given an equation that is not written in such a form but still represents a sphere, to help identify the sphere we use algebraic techniques such as completing the square to put it into the appropriate form. Also we note that the above equation gives us the points on the surface of the sphere, if we wanted to get all points inside the sphere we would use inequalities. In particular the points on the sphere and the interior of the sphere sometimes called the ball will be x, y, z with x x 0 2 + y y 0 2 + z z 0 2 r 2. In general inequalities are useful to help describe volumes and regions in space. Vectors Once we know how to get to a point the next important thing to know is how to move in a certain direction. In two dimensions we tended to use a vague notion of slope to help give some indication about which way to go. In general a better tool is to use vectors. A vector is something which has both a distance and a magnitude or length. A useful way to think about vectors is to tell us how to get from one point to another. Graphically we can represent vectors by arrows where the length is the length of the arrow and the direction is indicated by which way the arrow is pointing. We can add vectors pictorially by putting the initial point on one vector on the terminal point of the other vector and we can scale vectors which changes the length, when scaling by a negative we reverse direction. In particular once we can scale and add we can also subtract. A special vector is the 0- vector, denoted 0 which has zero length and no direction, intuitively 0 is saying to stay put. Dealing with vectors graphically has limitations especially for those who art artistically challenged!. So we want to develop a way to deal with vectors algebraically. We do this by splitting a vector into components. Namely if a vector is telling us how to move, then we break it up by asking how much we move in each direction. So in two dimensions a vector will be written as a = a 1, a 2 where a 1 is how much we move in the x-direction and a 2 is how much we move in the y-direction. In three dimensions a vector will be written as a = a 1, a 2, a 3 where a 1, a 2 are as before and a 3 is how much we move in the z-direction. Given two points A = x 0, y 0, z 0 and B = x 1, y 1, z 1 the vector that starts at A and ends at B, denoted AB, is found by looking at how much we change in each direction, i.e., AB= x 1 x 0, y 1 y 0, z 1 z 0. The way we deal with vectors algebraically is to work component by component, for example addition of two vectors is done as follows a 1, a 2, a 3 + b 1, b 2, b 3 = a 1 +b 1, a 2 +b 2, a 3 +b 3 subtraction similarly is done by a 1, a 2, a 3 b 1, b 2, b 3 = a 1 b 1, a 2 b 2, a 3 b 3

and scaling by a constant k is done by k a 1, a 2, a 3 = ka 1, ka 2, ka 3. The zero vector is now simply 0 = 0, 0, 0. Similar statements hold for vectors in other dimensions. We said that vectors have length and direction. To find length again we can measure the line from the initial to the terminal point. Again by the Pythagorean Theorem we have that the magnitude of the vector a, b, c is a, b, c = a 2 + b 2 + c 2. Here the is denoting magnitude, it also looks like the absolute value which is intentional since absolute value also is a measurement of magnitude. For direction we will use unit vectors which are vectors with length 1 i.e., unit length. To find a unit vector that goes in the same direction as a we simply scale a so that it has magnitude 1, this is easy to do by scaling by the reciprocal of the magnitude, i.e., unit vector in direction of a = 1 a a. Vectors behave as we would expect, i.e., all of the following properties hold. a + b = b + a a + b + c = b + a + c a + 0 = a a + a = 0 ka + b = ka + kb c + da = ca + da kla = kla 1a = a Three important unit vectors are i = 1, 0, 0, j = 0, 1, 0 and k = 0, 0, 1 which point in the x, y and z direction respectively. The basic idea is the same as breaking into components, and so as before we break a vector into each of the three directions and so a, b, c = ai + bj + ck. Finally, we should mention that vectors are useful in physics where many ideas incorporate both a direction and a magnitude, these include velocity and force. In that case the problem is to think about how vectors combine and use that to solve for information about vectors. A simple example is that if an object is at rest then the sum of the forces acting on it must be 0. This can make for some fun word problems! The dot product Adding and subtracting is great but it would also be nice to be able to multiply. There are two general products that act like multiplication but both have limitations. The easier of the two is the dot product, which we denote by a dot. This works by a 1, a 2 b 1, b 2 = a 1 b 1 + a 2 b 2 a 1, a 2, a 3 b 1, b 2, b 3 = a 1 b 1 + a 2 b 2 + a 3 b 3 and similarly for higher dimensions, but we will be spending most of our time in dimensions two and three so this will suffice. Note that for a dot product we have to have two vectors of the same length and the output is not a vector but a number, i.e., vector vector = number. So for example the expression a b c is meaningless. The dot product behaves nicely not surprising given the definition, in particular the following properties hold. a b = b a a b + c = a b + a c 0 a = 0 ka b = ka b = a kb But the most important property of the dot product is what happens when we take a vector and dot it with itself, namely, a a = a 2. Using this along with the law of cosines we have that a b = a b cos θ or cos θ = a b a b. Here θ is the angle between the two vectors when the initial point of the vectors are together if the tails are not together, you should put them together to get your vector. In particular the dot product is useful for finding the angle between two vectors where we express the angle as an arc-cosine. An important angle is 90, or π/2. When the angle between two vectors is 90 we call the vectors perpendicular or orthogonal. By the above formulation we have that two vectors a, b are orthogonal if and only if the a b = 0 one of the most useful tools of the dot product. In particular, 0 is perpendicular to all vectors. Dot products make it easy to find projections. Given two vectors a and b we might ask how much of a is in the direction of b. We have already seen this by breaking a vector up into its component pieces and looking at how much we change in x or in y or in z. To answer this question we need to know two things,

a length and a direction. The length, also called the component projection, is given by comp b a = a b, b the direction is found by the unit vector pointing in the direction of b, i.e., 1/ b b. Therefore the projection of a onto b is given by a b proj b a = b b b The cross product We now turn to the more mysterious way to multiply two vectors, namely the cross product which works only in three dimensions. This is defined by a 1, a 2, a 3 b 1, b 2, b 3 = a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 This is difficult to remember in this format so we can also write it in terms of determinants which make it slightly easier to remember, i.e., a 1, a 2, a 3 b 1, b 2, b 3 = i j k a 1 a 2 a 3 b 1 b 2 b 3 = a 2 a 3 b 2 b 3 i a 1 a 3 b 1 b 3 j + a 1 a 2 b 1 b 2 k It is important to see that the dot product and the cross product are very different, in particular the output of the cross product of two vectors is again a vector, i.e., vector vector = vector. The cross product has some unexpected properties. For instance a a = 0 a b = b a a + b c = a c + b c a b + c = a b + a c so now the order you take the cross product makes a difference! The last two properties indicate that we can still distribute. The two most important properties for cross product are: a b is orthogonal to a, a b is orthogonal to b, a b = a b sin θ. The first one says that a b gives a vector orthogonal to both a and b useful as we saw in working with planes; the exact choice of which direction there are essentially two is done by use of the right hand rule you are not going to be tested on the right hand rule. The second can be interpreted by saying that the magnitude a b is the area of a parallelogram formed using the vectors a and b. In particular, the triangle formed by the vectors a and b has an area of 1 2 a b. Also we note that two non-zero vectors are parallel if the angle between them is 0 or 180, or in other words if the two vectors are parallel multiples of one another. By the above rule we see that two non-zero vectors are parallel if and only if their cross product is 0. Usually this is not an efficient way to test for parallel, i.e., it is easier to just see if we can scale one to get the other. Applying the various rules to the vectors i, j and k we have i i = 0 i j = k i k = j j i = k j j = 0 j k = i k i = j k j = i k k = 0 Using these rules along with the distribution property of the cross product we can recover the definition of the cross product. We should point out that the cross product is useful in physics, for instance torque is defined in terms of a cross product. There are a few other rules for cross product. You will not be tested on the following two rules, a b c = a b c a b c = a c b a b c. The first equation just follows by noting that the expression is the determinant of a 3-by-3 matrix and in both cases you get the same determinant. In particular the value that you recover is the area of the parallelogram formed using the three vertices a, b and c. Lines and planes The idea behind calculus is that while we understand flat things lines, squares, etc. really well, the world around us is filled with things which are not flat; so what we will do is to study things which aren t flat by things which are flat usually done by looking at everything close-up. Of course we need to make sure we understand our flat things really well. We already have seen lines in two dimensions, usually in the form y = mx + b. The idea is that b is the y-intercept and that can be used to help position our line and m is the slope which gives us the direction in which to draw the line. In general, to find a line what we need are two items, one is a point on the line and the other is the

direction that the line can go in, which for us will be done using vectors. Whenever we are given a problem about finding a line it will always reduce to finding a point and a directional vector of the line. For example, if we are given two points it is easy to find the direction, just find the vector between the two points, and then finally pick one of the two points. Given a point x 0, y 0, z 0 and a direction vector on the line a, b, c we can express the line in different ways. One way to express the line is in vector form which is to say that it is all points x, y, z so that x, y, z = x 0, y 0, z 0 + t a, b, c where t is a parameter. We could also solve for each component on each side and that gives us our parametric form of the line x = x 0 + at, y = y 0 + bt, z = z 0 + ct. Finally, if we solve each expression for t we get the symmetric form of the line my personal least favorite t = x x 0 = y y 0 = z z 0. a b c We note that the equation for the line is not unique. In particular, we could have chosen any point on the line and we could also have chosen any directional vector which means that we can scale our directional vector by a constant, also we note that given any one of the forms above that we can easily recover the point and the directional vector. Lines are parallel if they have the same directional vector. However, unlike in two dimensions just because two lines are not parallel they still may not intersect, non-intersecting lines which are not parallel are sometimes called skew. A plane is a two-dimensional analogue of a line, basically imagine a flat surface i.e., a tabletop or piece of paper in three dimensions i.e., the room around you. Since the plan is two-dimensional there are many different vectors in the plane. But we still want to associate a direction with the plane. The way we do this in three dimensions! is to pick a vector which is perpendicular to the plane, we call such a vector a normal vector and will usually denote it by n = a, b, c. Then if x 0, y 0, z 0 is a fixed point in the plane and x, y, z is any arbitrary point in the plane then by orthogonality conditions we have 0 = n x x 0, y y 0, z z 0 = a, b, c x x 0, y y 0, z z 0 We can rearrange this to = ax x 0 + by y 0 + cz z 0. ax + by + cz = d where d = ax 0 + by 0 + cz 0. This is the form of the plane which we will most commonly use. Note in particular that given this form of the line it is easy to read off the coefficients and recover the normal vector. In any problem where we are asked to find a plane it will always boil down to finding a point and a normal vector a point is usually very easy, the normal vector usually takes a while the trick is to identify two non-parallel vectors which have to be in the plane for example if we have three points then we can find two such vectors. Two planes are parallel if their normal vectors are multiples of each other. When two planes are not parallel the angle of intersection between the planes is found by using the angle between the corresponding normal vectors. Parametric curves A curve in either two or three dimensional space can be described by saying what is happening in each coordinate. For example either xt, yt or xt, yt, zt. Recall the story of the fly, we can think of t as a time parameter saying where the current position of the fly is at time t. This has some advantages. For example the unit circle cannot be written as y = fx because it miserably fails the vertical line test. But as a parametric curve it is easily described by cost, sint. Given a parametric curve we can sometimes recover the curve y = fx that the points on the parametric curve will be on. This can be done by solving for t or using various identities trigonometric, etc.. For example xt, yt = t, t 2 corresponds to points on the parabola y = x 2 since the y value is the square of the x value. But we also have xt, yt = 2t, 4t 2 also corresponds to points on the parabola y = x 2 ; both give the same curve but they have different behavior if we look at them as t progresses, namely the second goes twice as fast. We also have that xt, yt = sin t, 1 cos 2 t also will lie on y = x 2 but will not be all points on the parabola, in particular as time moves forward this will only oscillate back and forth on the portion of the parabola between 1, 1 and 1, 1. Given a parametric curve it is helpful to try and identify a curve that the points will lie on. Another option is to plot values for fixed values of t and connect the dot a popular choice among computers. We can also specify an interval for the curve, in which case we will specify a starting time and a stopping

time which will correspond to a starting point and a stopping point. Another way to deal with parameterized curves is to plot each coordinate on its own axis, i.e., plot x as a function of t and y as a function of t and so on then get the general behavior of which direction the curve is going in the plane, i.e., northeast x increasing, y increasing northwest x decreasing, y increasing southeast x increasing, y decreasing southwest x decreasing, y decreasing and of course plot points where behavior changes. This works fine when xt and yt are simple functions but becomes too tedious when they become complicated. Thank goodness for computers! When we are dealing with a parameterized function and we want to know what the curve is approaching at a certain point, we can pass the limit inside assuming that each limit exists, i.e., lim ft, gt, ht = t t 0 lim ft, lim gt, lim ht. t t 0 t t0 t t0 But this idea seems a little calculus-esque and so we will not test on this during this midterm of course it might come back later!. Finally we will note that we can also have a parameterize vector equation as compared to parameterizing a curve. The difference between them in not important right now and we will only deal with parameterized curves on the midterm.