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An overview of multivariable calculus, focusing on parametrized curves and surfaces in two and three dimensions. It covers functions in R2 and R3, vector fields, partial derivatives, and integrals. The text also explains how to translate between parametrized curves and equations.
Typology: Lecture notes
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JAMES MCIVOR
So far, you have probably primarily studied functions of one variable, usually written f (x). Here f is the name of the function, and x is the name of the independent variable. We usually have to say at some point what values x may take. For instance, x may be allowed to be any real number, or just a number between 0 and 1, etc. The possible values for x form what is called the domain of f. The important thing to notice about your previous calculus experience is that the domain was always either R or a subset of R, such as [0, 1]. Can we have a function whose domain is R^2 instead of R? Certainly, and since the points of R^2 consist of two real numbers, the input to such a function will be two real numbers, usually written f (x, y). For example, the function f (x, y) = x^2 + y^2 is a function whose domain is all of R^2. The function g(x, y) = (^) xy^1 is a function which takes two inputs x and y, but its domain is not all of R^2 , since it is not defined when either x or y is zero. This expression really defines a function whose domain is all of R^2 except the x- and y-axes. In similar fashion we can describe functions whose domain is R^3 , for instance h(x, y, z) = xyz^2 + y sin(xz). Thus we have extended the domain of functions from R to R^2 and R^3. That is, we have given ourselves greater flexibility with the “inputs” of our functions. But what about the outputs? All the functions above output just a number, or scalar. We would like functions which output points in R^2 or R^3. We will see that points of R^2 or R^3 are the same thing as vectors, and so we will call a function whose outputs are in R^2 or R^3 a vector function, or vector-valued function, or sometimes a vector field. Often we use capital letters to name vector fields, or sometimes bold-face type. For example, F (x) = (2x, x^3 ) is a vector field in R^2 whose domain is R. The vector field H(x, y, z) = (x^2 y, sin yz, ey^ ) is a vector field in R^3 whose domain is also R^3. 1
2 JAMES MCIVOR
So our goals will be to first understand what various objects in R^2 and R^3 look like, and then to study scalar functions on those objects, and finally vector functions. Here’s the most important thing to remember about this class: if you can keep the different functions straight, the calculus part will be easy; but if you get the functions muddled, and get mixed up between scalar and vector functions, then all the calculus we do can seem much more complicated than it needs to be. Once you are comfortable with the idea of functions having many variables, you can ask how to differentiate them. Once you get used to it, this is very easy. Since you have more than one variable, you have more than one derivative: you can differentiate with respect to each different variable. Say we have a function f (x, y) of two variables. To differentiate with respect to x we pretend that y is a constant (sometimes I just replace y by 17 or some other obscure number to prevent confusion). Then take a normal derivative with respect to x. This is called the partial derivative of f with respect to x, and denoted ∂f∂x.
Example 1.1. If f (x, y) = x^2 y + x + xey^ , then ∂f∂x = 2xy + 1 + ey^ and ∂f∂y = x^2 + xey^.
The concept of integration is slightly more subtle. Let’s say we’re taking a definite integral of a single-variable real-valued function. Then the information we need is not just a function, but an interval, or region, over which to integrate it. When we allow functions with more than one variable, we must also allow different regions to integrate over. The reason some students find the later parts of this course confusing is that there seem to be many types of integrals. But there are really only three types: line integrals, surface integrals, and volume integrals. The different types just depend on the dimension of the region over which we integrate. Line integrals are integrals over a one-dimensional region, such as the x-axis, or some curve. Surface integrals are integrals over a two-dimensional region, maybe the xy-plane, or a rectangle in the plane, or even a strange curved surface, like part of an ellipse, for instance. Both types of integrals involve scalar functions - in fact, it only makes sense to integrate scalar functions. Whenever you see an integral involving a vector field (and you will see plenty of these at the end of the course), notice that we always do something to reduce the vector field to a scalar before performing the integration. Usually this involves vector operations like the dot product.
SUMMARY:
2.1. The Idea of Cutting Out. There are two fundamentally different ways to describe a curve or a surface. One is by giving equations, and this will be our focus today. Tomorrow we will discuss the other method, using parameters. I like to call the first method “cutting out” curves and surfaces, for reasons that will become clear shortly. Let’s work in R^2 for the moment. If I go through and pick out all the points (x, y), with no equations, then what I get is all of R^2. But if I instead go through and pick out only those points (x, y) which satisfy the equation y − x = 0, then what I get is a line, namely the line y = x. So by picking out only the points satisfying a certain equation, I have focused my attention on a smaller part of R^2. While R^2 itself is two-dimensional (it’s a plane), the line I picked out by my equation y − x = 0 is a one-dimensional object. We will see that as a general rule, each equation you impose cuts down the dimension by one. What if I instead use the equation y + x = 0? I just get a different line, this one with slope -1 instead of 1. But still with this one new equation I have gone from the two-dimensional R^2 to a one-dimensional object.
Now suppose I play a new game. Starting with R^2 as above, I now go through and pick out only those points (x, y) which satisfy both equations y − x = 0 and y + x = 0. What sort of object have I picked out now? To find out, we simply solve both equations simultaneously. The first equation forces y = x, while the second equation forces y = −x. The only choices of x and y that work are (0, 0), often called simply the origin. Thus by starting with all of R^2 and imposing two equations,
4 JAMES MCIVOR
only time when the dimension drop does not work is when the extra equation imposed is redundant. For example, the equation x = 0 cuts out a line in R^2 , namely the y-axis. But so does the equation 5 x = 0, since multiplying both sides by 5 doesn’t change the solution set to the equation. So if we first impose the equation x = 0, and then impose a second equation 5x = 0, the dimension does not drop - in fact, nothing at all changes. Imposing one equation is the same thing as imposing both. The second equation was redundant. Here is another example:
Example 2.2. Analyze the figure cut out by x^2 + y^2 + z^2 = 25, y = 3, and x^2 + z^2 = 16
If we just obey the dimension-drop rule, we expect something zero-dimensional. Let’s see that this is not so: first impose x^2 + y^2 + z^2 = 25 and y = 3, then impose x^2 + z^2 = 16. Geometrically, the first equation describes a sphere of radius 5 in R^3 , and the equation y = 3 describes a plane. To see what they cut out together, we intersect the sphere and the plane, giving us a circle which lives in the plane y = 3. If you do a little algebra, you can see that the radius of this circle is 4. Thus the third equation tells us nothing new. We started with R^3 (3-dimensional), cut out a sphere, which is a type of surface (2-dimensional), then cut out a plane and intersected it with our sphere, giving a circle (1-dimensional). But when we imposed the third equation, we did not drop down to a 0-dimensional object, which is just a set of points, but stayed where we were, since the third equation followed from the first two. Notice, however, that what’s cut out by the third equation isn’t the same thing as that cut out by the first two, since the third by itself cuts out a cylinder, whereas the first two determine just a circle.
2.5. SUMMARY. - Equations cut out geometric objects in R^2 and R^3.
3.1. Parametrizations. Today we discuss the second of the two main approaches to describing curves and surfaces: parametrization. Roughly speaking, a parametrization is a point in R^2 or R^3 which changes as the values of the parameter(s) change. By varying the parameter, we can pick out all the points on the curve or surface in which we are interested. As usual, we should specify at the outset whether we are working in R^2 or R^3. First let’s consider only curves in R^2. Curves are one-dimensional, and require only one parameter (for us, this is basically the definition of “one-dimensional”). Since we’re in R^2 , we will write down a point (x(t), y(t)) in R^2 , where the x- and y- coordinates depend on a parameter t. For example, (t, t^2 ) is a parametrization of a curve. So is (cos t, sin t). We think of t as being a time parameter, so as time changes, the point in R^2 varies. It is useful to imagine the coordinates (x(t), y(t)) describing the location of a moving particle at time t. Sometimes a parametrization is just given by expressing x and y as functions of t, for instance, x(t) = 2t, y(t) = 5t describes a line of slope 5/2 through the origin. This is equivalent to putting the two functions together as an ordered pair and referring to the parametrization (2t, 5 t).
3.2. Going From Parameters to Cutting Out. If you’re given a parametrization and want to know what it looks like, you can just plot some points for various values of t. This is often good enough to give you an idea of what the curve looks like. Usually you should put a little arrow on the curve to indicate the direction of increasing t. Having spent an entire lecture on describing curves (and surfaces) by equations, it would also be nice to know how to translate between a parametrized curve and an equation. This is easy to do. The parametrization involves a new variable, t, namely x(t) = ... and y(t) = ..., whereas the equation which cuts out the curve will just be an equation involving x and y. So to get from the parametrization to the equation involving x and y only, one just eliminates t from the two equations. When performing the algebra, it’s handy to just write x for x(t) and y for y(t).
For instance, to express the parametrization x(t) = 2t and y(t) = 5t of the line from above as an equation, we must remove t from the system x = 2t, y = 5t. The first equation say that t = x/t,
MULTIVARIABLE CALCULUS NOTES 5
and plugging this into the second equation gives y = 5x/2. Here’s another:
Example 3.1. Write the equation which cuts out the ellipse (2 cos t, 3 sin t) in R^2.
We have to use the equations x = 2 cos t and y = 3 sin t to get one equation which relates x and y. Usually you’ll try to solve for t and substitute. Here it’s easier to isolate the trig functions and use the Pythagorean identity: Since x/2 = cos t and y/3 = sin t, the Pythagorean identity cos^2 t + sin^2 t = 1 yields ( x 2 )^2 + ( y 3 )^2 = 1, which is the equation of an ellipse, as we’ll see in a few lectures.
3.3. Curves in R^3. What about curves in R^3? The same idea applies: we express the three coor- dinates x, y, and z of R^3 each as functions of t, and we imagine this as describing the location of a particle in three-dimensional space for all values of time. Here’s a classic example:
Example 3.2. The helix in R^3 is parametrized by x(t) = cos t, y(t) = sin t, and z(t) = t. Draw a picture of this curve.
If you want to find equations for a parametrized curve in R^3 , remember first that it takes two equations to cut out a curve in R^3 (one equation alone would cut out a surface). So we have to take the three expression for x, y, and z in terms of t and eliminate t to get two equations involving x, y, and z. This can often be surprisingly difficult, and I will only ask you to do it for manageable cases. Here’s an example that illustrates how this can be a little tricky.
Example 3.3. A “slanted” ellipse in R^3 is parametrized by (cos t, sin t, cos t). Find its defining equations.
Intuitively, the x, y-coordinates are going around in a circle, while the z-coordinate (the “height”) is oscillating. But even without getting an idea of what it looks like, we can let algebra do the talk- ing: The x, y-coordinates satisfy x^2 + y^2 = 1, and the y, z-coordinates satisfy y^2 + z^2 = 1. These two equations nearly cut out our ellipse: since combining equations corresponds to intersecting surfaces, the curve cut out by these two equations is the intersection of two cylinders - one directed upwards, along the z-axis, and the other directed horizontally along the x-axis (the direction in which a cylin- der extends is along the axis corresponding to the variable which does NOT appear in the equation cutting out the cylinder). This intersection of two cylinders should be fairly easy to visualize or draw. Then you see that it actually consists of two ellipses! So the most natural two equations for the ellipse cut out not one but two ellipses, and these are NOT the right defining equations. To get the right ones, you have to notice that any ellipse must lie in a plane. The plane containing our el- lipse can be identified by the fact that the x- and z-coordinates of our parametrization are the same. Thus our ellipse lies in the plane cut out by x = z. What is the other equation? Well, we can just use one of the cylinders from before. So two equations that cut out this ellipse are x = z and x^2 +y^2 = 1.
The above example also illustrates two important points from the previous lecture: (1) there are many sets of equations which cut out a given curve or surface, and (2) sometimes what we call a “curve” may really be a union of two objects which are curves, as in the pair of ellipses encountered above.
3.4. Parametrizing Surfaces. So much for curves in R^2 and R^3. What about surfaces? This is only interesting in R^3 , since the only surfaces that live in R^2 are flat, for instance R^2 itself. So let’s work in R^3 , where we must parametrize the three coordinates x, y, and z. Using only one parameter will give us a curve, but if we use two parameters, say s and t, we get a two-dimensional object, i.e., a surface. So to give a parametrized surface is to give three expressions: one for x in terms of s, t, one for y in terms of s, t, and one for z in terms of s, t. As before, it is possible to “translate” a parametrized surface into one given by an equation: you have to eliminate s and t to obtain one equation relating x, y, and z. Here are a few examples:
Example 3.4. (1) x = s + t; y = s − t; z = 2t. This surface is a plane. The first two equations can be subtracted to give x − y = 2t, hence x − y = z is the equation that cuts out this plane. In general, whenever x, y, and z are linear functions of s and t, you get a plane.
MULTIVARIABLE CALCULUS NOTES 7
describe? [Hint: Just draw each part one at a time.] This is another one you can’t cut out with equations.
SUMMARY
4.1. Qualitative Characteristics. Firstly, if the values of the parameter t are bounded, you can locate the start and end points of the curve. If not (e.g., if t is allowed to be any real number), you can see what happens in the long run by considering the behavior for very large values of t (or very negative values of t). Sometimes you can also observe that the curve lives only in certain quadrants, by seeing whether the x- and y-coordinates are either always positive or always negative.
Example 4.1. The curve x(t) = (^1) t cos t, y(t) = (^1) t sin t, π ≤ t < ∞ starts at the point (− (^1) π , 0) when t = π; it has no end point since t can be arbitrarily large. Nevertheless, we can determine the long term behavior by seeing what happens to each of the coordinates as t goes to infinity. The sine and cosine factors just oscillate back and forth, but the factor of (^1) t in each coordinate goes to zero, so as t → ∞, the curve approaches the origin. This curve is an inward spiral.
4.2. Colliding Particles. The next thing you might want to do with parametrizations is deter- mine whether two different curves intersect. This is much easier to do when the curves are given by equations, since, as we have seen, the intersection can be found by solving both equations si- multaneously. So if given two parametrized curves, and asked to find their intersection, I suggest you either translate into equations and solve the equations, or else just sketch the curves. There is another interesting question which does involve the parametrization, however: If two particles are moving in the plane (or in space), how can you determine whether they collide? I illustrate with an example.
Example 4.2. Two particles, A and B, are moving in the plane. Particle A’s motion is described by the parametrization (2 + 2t, −t), and particle B’s motion is described by (2 cos t, sin t). Determine when and where, if at all, they collide.
For the particles to collide, two things must happen: (a) the curves along which they move must intersect; and (b) the particles arrive at the intersection points at the same time. For (a), we convert to the equations which cut out each curve. Particle A moves along a line whose slope is -1/2 and whose y-intercept is y = 1, which has equation y = 1 − 12 x. Particle B’s parametrization you may
recognize as that of an ellipse, whose defining equation is x
2 4 +^ y
(^2) = 1. First we determine where
the line and ellipse meet by solving these equations simultaneously. This gives two points: (0, 1) and (2, 0). These points represent where the curves intersect, but that does not guarantee that the particles collide - they might each pass through these points at different times. So for each point, we must determine at what times particles A and B arrive there. Particle A passes through (0, 1) at time t = −1 and through (2, 0) at time t = 0. Particle B passes through (0, 1) at t = π/ 2 , 5 π/ 2 , So the two particles never meet at (0, 1). But particle B passes through the point (2, 0) at t = 0, 2 π, etc. So the particles do collide at (2, 0), at time zero, and this is the only point at which they collide.
8 JAMES MCIVOR
4.3. Tangents. The next thing we’ll do is find tangent lines. If we’re given a parametrized curve (x(t), y(t)), then the slope of the tangent line at the point at time t is given by y′(t)/x′(t) (assum- ing x′(t) 6 = 0). This is consistent with the childhood catchphrase “slope is rise over run”, since the derivatives of the y- and x-coordinates are just the instantaneous changes in the vertical and horizontal directions, respectively. If you prefer to remember the formula in Leibniz’ notation:
dy dx
dy dt dx dt
which is easy to remember if you imagine that we are just “cancelling out” the dt. Although the formula is only valid when dxdt is nonzero; when dxdt is zero (and dydt isn’t), we have a vertical tangent. Here’s an example:
Example 4.3. The circle of radius one at the origin is parametrized by (cos t, sin t). At t = π/4 the slope of the tangent line is dy dx
dy dt (π/4) dx dt (π/4)
cos π/ 4 − sin π/ 4
which is clear from drawing a picture. Notice that dxdt is zero when t = 0, π, etc: these are the points with vertical tangent lines. We saw yesterday that there are many parametrizations of the same curve - these correspond to particles moving around the curve at different speeds. Intuitively, the slope of the tangent lines should not depend on the speed. Let’s reparametrize the circle and see. Another perfectly good parametrization of the circle is (cos(t^2 ), sin(t^2 )). This one goes round a little slower than the original parametrization until t = 1, and a little faster than the original after that. At t = π/4, the previous particle was at the point (
2 /2). The new particle reaches that point at t =
π/2. At that time, we compute dy dx
dy dt (
π/2)^2 dx dt (
π/2)^2
cos((
π/2)^2 )(
π/2) − sin((
π/2)^2 )(
π/2)
For surfaces, there is also a notion of tangent plane. Since a surface has “two degrees of freedom”, i.e., requires two parameters to describe, the tangent plane involves finding slopes “in two directions”. This will make sense once we study partial derivatives. It is also much easier to describe planes using the terminology of vectors. So our discussion of tangent planes to surfaces will be postponed for the moment, but be aware that they exist.
4.4. Arc Length. The final thing we’ll learn to do with a parametrization is compute the length of a curve. Needless to say, this is a fundamental notion in geometry. Moreover, we’ll see later how to generalize this calculation to define line integrals, which are one of the fundamental tools of this course. In fact, if you understand the definition of the arc length integral, you can probably guess how to define a line integral. Say we have a curve C in R^2 which is parametrized by (x(t), y(t)), and we want to compute the length of that portion of the curve between t = a and t = b. Intuitively, we just add up the lengths of infinitesimal shreds of the curve. The “length of an infinitesimal shred of the curve”, is denoted ds, and is related to the parametrization by the formula
ds =
dx dt
dy dt
dt,
which may be familiar from computations of surface area of surfaces of revolution in a single-variable calculus course. The arc length of the curve C is
L(C) =
C
ds
which is just a shorthand - it just says the length is computed by adding the arc length element ds up along the entire curve C. The actual calculation is made by the formula
∫ (^) b
a
dx dt
dy dt
dt.
We’ll do an example in a moment, but let’s have a brief philosophical aside. Why did I bother writing the formula L(C) =
C ds^ at all, if we use the formula immediately above to do all the calculations? The formula L(C) =
C ds^ is like the skeleton of the actual formula. As long as you remember the definition of ds above, and you know that the subscript C means we have to integrate over all of C, and therefore take t from a to b, you can easily write down the longer, more useful
10 JAMES MCIVOR
How do we parametrize the graph of f (x, y)? Since it’s a surface, we will need two parameters. Just like in the one-variable case, we just set each of the variables equal to the parameters. Thus the parametrization of the graph of f (x, y) is
(s, t, f (s, t))
You will usually have to specify what values of s, t you are using. Also note that some people will just use the variables x and y themselves as the parameters, but I like to consistently use the variables s, t for a parametrized surface.
Example 5.2. The “northern hemisphere”, i.e., the top half of the sphere of radius 1, can be expressed as the graph of the function f (x, y) =
1 − x^2 − y^2. Thus it can be parametrized by (s, t,
1 − s^2 − t^2 ). Notice, though, that we have to be very careful with the values of s, t allowed, or else the square root is undefined. The allowed values are those s, t which satisfy s+t^2 ≤ 1. This corresponds to a disc of radius one in the xy-plane.
5.3. Level Sets. Surfaces tend to be harder to visualize than curves. The rest of the day we’ll look at a very useful technique for visualizing surfaces, that of level sets. Roughly speaking, a level set is a “slice” or “cross-section” of a surface.
Say you’re given a surface cut out by some equation. The equation will involve x, y, and z. Now set z equal to fixed number (usually an easy one like 0,1,2, etc.). Let’s say for example we set z = 2. This gives an equation involving only x and y. Often we will know what this looks like - it will be some curve. This curve is what the surface looks like at the height z = 2, and is called the level set for z = 2 (some people call it a level curve). If you draw a few level sets together in the same plane, and label each one with the appropriate z-value, you get what is called a contour map.
Example 5.3. (1) A sphere is cut out by x^2 + y^2 + z^2 = 1. Let’s look at some level sets. The level set for z = 1 is a point (x = y = 0). The level set for z = 0 is a circle of radius one x^2 + y^2 = 1 (this is just the part of the sphere that lives in the xy-plane). The level set for z = − 1 /2 is given by x^2 + y^2 = 3/4 - it’s another circle, but with slightly smaller radius than the one for z = 0. If you stack all these level sets above one another, ordered by the associated z-values, you can start to see how the sphere is built out of these horizontal cross-sections. (2) Paraboloid z = x^2 + y^2
(3) Ellipsoid
x^2 4
y^2 9
z^2 16
(4) The graph of f (x, y) = x^2 y^3. [Hint: try plotting level sets for y instead. That is, fix various y-values and see what the cross-sections in the xz-directions look like.]
6.1. Defintions. Polar coordinates r, θ are a new way of locating points in the plane. They replace the Cartesian coordintes x, y. The point (r, θ) is the point in the plane which is r units away from the origin, in a direction which makes an angle θ with the positive x-axis (the angle is measured counterclockwise, as usual). The usual trigonometric formula gives us the following relationships between Cartesian and polar coordinates:
Cartesian to Polar
(1) r =
x^2 + y^2 ; θ = tan−^1
y x Polar to Cartesian
(2) x = r cos θ; y = r sin θ
Example 6.1. Express the points whose Cartesian coordinates are (1, 1) and (− 2 , 0) in polar coor- dinates.
6.2. The Problem - Ambiguity. The annoying thing about polar coordinates is that many pairs of coordinates represent the same point. This is not true for Cartesian coordinates: every point has a unique pair of Cartesian coordinates. For polar coordinates there are two things to watch out for:
(1) There are many choices of the angle θ, each differing by 2π (2) You may use as the distance from the origin either a positive number r, or −r, but if you choose −r you must change the angle by π.
Example 6.2. Find all the ways of writing the point (−
MULTIVARIABLE CALCULUS NOTES 11
6.3. Functions in Polar Coordinates. Given a function f (x), we can obtain a picture of the function by setting y = f (x), and seeing what curve this cuts out in the plane. For polar coordinates, the idea is the same, but the pictures end up looking quite different. Given a function f (θ) of the variable θ, the polar equation r = f (θ) cuts out a curve in the plane. To graph it, you usually think of starting at angle θ, and as the angle increases, you track how close or far from the origin you are by finding r from the equation r = f (θ). Let’s look at some examples:
Example 6.3. (1) r = 2 (this is a function of θ - a constant function) (2) θ = π/3 (this is not a function of θ, but it still cuts out a curve) (3) r = θ (4) r = sin θ (contrast with Cartesian curve) (5) r = 1 + sin θ (6) r = 2 + 3 cos 2θ
6.4. Cartesian Equations for Polar Curves. You can use equations (1) and (2) above to trans- late between the Cartesian and Polar equations which cut out a given curve. Let’s do this for the examples above.
(1) x^2 + y^2 = 4 (2) y =
3 x (3) tan(x^2 + y^2 ) = yx (yuck) (4) r =
y r
, so y = r^2 = x^2 + y^2
(5) r = 1 +
y r
, so y = r^2 − r = x^2 + y^2 −
x^2 + y^2 (6) 2x = r^2 − 3 r =...
6.5. Tangents to Polar Curves. Many students get polar coordinates confused with parametriza- tions. They are different - here’s how, explicitly. A parametrization gives each of the two coordinates in terms of a third quantity, t (the parameter). Polar coordinates are just a different style of coor- dinate.
The source of the confusion is that there is a useful way to get a parametrization of a curve if you’re given a polar equation that cuts it out. Let’s say we want to parametrize the polar curve r = f (θ). We might try to turn it into Cartesian coordinates and see whether we recognize a famil- iar parametrization. But polar curves can often be strange and you probably won’t know how to parametrize it by looking at the Cartesian equation that cuts it out.
Instead, to make the polar coordinate θ into a parameter, use the following equations:
(3) x(θ) = f (θ) cos θ; y(θ) = f (θ) sin θ
This is a parametrization of the curve because it gives the two (Cartesian) coordinates x and y as functions of the parameter θ (which used to be a coordinate, but now we think of it as a parameter for x and y). These equations are just the earlier formulae x = r cos θ and y = r sin θ, but we’ve plugged in f (θ) instead of r. This is useful for finding tangents, since we saw two lectures ago how to find the tangents to a curve given a parametrization: use the formula
dy dx
dy dθ dx dθ
The product rule lets us compute
dx dθ
= f ′(θ) cos θ − f (θ) sin θ,
dy dθ
= f ′(θ) sin θ + f (θ) cos θ.
Example 6.4. Find the horizontal tangents of the cardioid of example 4.10(5)
Solution Horizontal tangents when d dyθ = 0. Setting d dyθ = (^) ddθ (1+sin θ) sin θ = cos θ+2 sin θ cos θ = 0 gives cos θ = 0 or sin θ = − 1 /2. This first equation gives solutions (2n + 1)π/2; the second gives θ = 7π/6, 11π/6, etc. Question: look at the picture of the cardioid. What’s happening at the origin? Our calculation says there should be a horizontal tangent there, but it doesn’t seem right. What went wrong?
MULTIVARIABLE CALCULUS NOTES 13
Anyway, we’ll focus on the main three in this lecture. There is a huge amount to say about conic sections - Apollonius wrote 8 books about them! For this reason, I’ll try to organize in the lecture the main things you need to know. For each type, I will give the general Cartesian equations, describe the curve from a geometric point of view, and then describe it in terms of eccentricity and polar coordinates.
7.1. Parabolas. These are the sections which occur when the plane is parallel to the edge of the cone. A parabola in the xy-plane can be described as follows. Fix a point F with coordinaes (0, p), called the focus, and a line, say y = −p, called the directrix. Then a parabola is the set of points which are the same distance from the focus as they are from the directix. Given a point (x, y) in the plane, its distance from the focus is
x^2 + (y − p)^2 and its distance from the directrix is |y + p|. If you set these two distances equal to one another and solve for y, you get the equation
y =
4 p
x^2.
[This doesn’t hold if p = 0, but in this case the focus lies on the directrix, and the only point on the parabola is the origin - not much of a parabola, really.] You can also calculate the equation x = (^41) p y^2 for the case when the directrix is vertical.
Often you’ll be asked to find the focus and directrix, given an equation of a parabola. If the equation looks like the formula above, you can just read off the value of p and this tells you where the focus and directrix are. But this equation only works for a parabola through the origin. What if it’s shifted?
Example 7.1. Find the focus and directrix of the parabola cut out by 4x = y^2 − 2 y − 3.
Solution: Notice that since y is squared, the parabola opens in the horizontal direction. Do some algebra to write the equation as x + 1 = 14 (y − 1)^2. This looks like the equation x = 14 y^2 , where p = 1, but it has been shifted to the left by one and up by one. The focus and directrix of x = 14 y^2 are (0, 1) and x = −1, respectively, so the focus and directrix of the shifted version are (− 1 , 2) and x = −2, respectively.
7.2. Ellipses. Normal description: points P such that the sum of the distances from P to F 1 and P to F 2 are the same. F 1 and F 2 are the foci. Let the foci F 1 and F 2 be at (±c, 0), and call the sum of the distances 2a. Then the ellipse meets the x-axis at (±a, 0) (the vertices), and meets the y-axis at (0, ±b), where c^2 = b^2 − a^2.
The ellipse is cut out by x^2 a^2
y^2 b^2
This description can be thought of as taking a piece of string, and fixing each end to a focus. now take a pencil and use it to pull the string taught. As you move the pencil around, you draw an ellipse. The length of the string is the sum of the distances to the foci.
For an ellipse whose major axis is the y-axis, the equation is the same , but the larger number a is in the denominator of the y^2 term, and the foci in this case are at (0, ±c).
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7.3. Hyperbolae. A hyperbola has equation
x^2 a^2
y^2 b^2
There are two foci F 1 and F 2 at (±c, 0), and points on the hyperbola must be such that the absolute value of the distance to F 1 minus the distance to F 2 is constant, equal to 2a. The points (±a, 0) are where the hyperbola meets the x-axis. The hyperbola is a curve which has two separate components, called branches. There are also two asymptotes; these are given by the formula y = ± ab x.
7.4. Conic Sections with Polar Coordinates. It may seem that the parabola is the odd one out, since its description involves distances to a point (focus) and a line (directrix), whereas the ellipse and hyperbola are describe in terms of two foci. But there is a description of all three in terms of one focus and a directrix, so the parabola is not so unique as it first seemed. This description involves using polar coordinates.
Fix a poinf F and a line L. Then each of the three conic sections above can be described as the set of points P such that the distance from P to F and the distance from P to L always have the same ratio. This ratio is called the eccentricity e of the conic section
(1) If e < 1 the section is an ellipse (2) If e = 1 it is a parabola (3) If e > 1 it is a hyperbola If we put the focus F at the origin, these sections have the nice polar equations
(1) r =
ed 1 + e cos θ
for a vertical directrix x = d
(2) r =
ed 1 − e cos θ
for a vertical directrix x = −d
(3) r =
ed 1 + e sin θ for a horizontal directrix x = d
(4) r =
ed 1 − e sin θ
for a horizontal directrix x = −d
Idea of Proof If P = (r, θ), then |P F | = |P L| means r = d − r cos θ. Solve for r to get formula. To decide which conic, can assume it’s not a parabola. In that case, square equation and translate
to Cartesians to identify the familiar equation of the conic. Get a^2 =
e^2 d^2 (1 − e^2 )^2
and b^2 =
e^2 d^2 1 − e^2
For a parabola, we already know what the directrix looks like. For an ellipse it lies on one side of the curve, and for a hyperbola it lies inbetween the two branches. In both cases the directrix is given by the equation d = a^2 /c.
8.1. Generalities. • Quadric surfaces are 2-D versions of conic sections (their level sets/traces ARE conic sections). They are 2-D objects (hence “surface”), but live in R^3.
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(3) The degenerate ones are called degenerate because some of their cross sections are lines, which are degenerate conic sections. Why are they degenerate? Because a conic section should come from a quadratic equation in two variables. The lines occur when all the degree two terms are missing - so it’s really a linear equation in two variables, which is technically still a quadratic equation, but a degenerate case. (4) The cylinders all arise whe there are only two variables in the equation. To graph them, just sketch the curve in the xy-plane (or whatever plane involves the two variables present), and extend in the direction of the missing variable.
Example 8.1. (1) (Stewart, 12.6.6) Find traces of and sketch the surface cut out by the equa- tion 4 x^2 + 9y^2 + z = 0. (2) (Stewart, 12.6.15) Identify and then sketch the surface cut out by −x^2 + 4y^2 − z^2 = 4