(Last modified: Monday, April 28, 2008, 10:31 AM )

I'll use Maple syntax for mathematical notation on this page.
All section and page numbers refer to sections from Smith & Minton's Multivariable Calculus, 3rd edition.

Due Monday 4/28 at 9am

Section 14.4: Green's Theorem

Reading questions:

1. What surprises you about Green's Theorem?

   JE, CN, and EL say (combining their answers):
   

   What surprises me about Green's Theorem is that the (line) integral of Mdx + Ndy is the same as the (double) integral of N_x -M_y.

   That is, M is paired with dx on the line integral side but it's being partialed with respect to y on the double integral side, and the + in the line integral becomes a - on the double integral side.

   CG says:
   

   That you change a single (line) integral to a double integral is surprising.

   BM adds:
   

   What surprises me about Green's Theorem is that a baker with only two years of schooling can teach themselves mathematics and discover such a useful tool.

2. Give an example of a region R in the plane where Green's Theorem does not hold.

   EP says:
   

   An example of a region R that where Green's Theorem doesn't hold is the region bounded by x=2y^2-4, x=-2y^2 x=-4, x=4. The boundary is not simple as it is divides the region.

   

---

Due Friday 4/25 at 9am

Section 14.3: Independence of Path and Conservative Vector Fields

Reading questions:


1. Why are conservative vector fields so helpful when evaluating line integrals?

   EL says:
   

   Conservative vector fields are helpful when evaluating line integrals because then the integral is independent of the path so you'll get the same result for any path you choose.

2. Why is Theorem 3.2 called the Fundamental Theorem for line integrals?

   ZH says:
   

   It makes calculating Line integrals very easy, and it is also almost exactly the same as the FT of Calculus.

---

Due Wednesday 4/23 at 9am

Section 14.2: Line Integrals (continued)

Reading questions:


1. Give at least one use for component-wise line integrals.

   DS says:
   

   We can now find work if we use the component-wise line integrals.

2. If you're evaluating a line integral with respect to any of the axes, does orientation matter? If so, what difference does orientation make?

   MB says:
   

   The orientation here does matter. If oriented incorrectly, the sign will change.

---

Due Monday 4/21 at 9am

Section 14.2: Line Integrals

Reading questions:


1. Give two different uses for line integrals with respect to arc length.

   XZ, JE, and NA say:
   

   The line integral (with respect to arclength) measures surface area. We can use it to find the mass of a spring or other thin curved object. It is also useful in center of mass problems.

2. If you're evaluating a line integral with respect to arclength, does the orientation of the curve matter? If so, what difference does orientation make?

   JB says:
   

   The orientation of the curve does not matter (for this type of line integral): the outcome will be the same either way, but the opposite orientation of a curve C is written -C.

---

Due Friday 4/18 at 9am

Section 14.1: Vector Fields
Reading questions:


1. On page 1118, figures 14.3(a) and 14.3(b) are both representations of the vector field H(x,y)=<y,-x>, yet they look very different. Based on the definition of a vector field, which of these two graphs do you think most correctly shows the vector field? Which one do you think gives you the best feel for the behavior of the vector field?

   CN says:
   

   I think figure 14.3a most correctly shows the vector field, but that figure 14.3b gives you the best feel for the behavior of the vector field.

2. Suppose the vector field shown below represents the flow of a small portion of a river, and suppose a leaf were to fall into the water at the point (-2,2). Describe the movement of the leaf.
   fieldplot(f(x,y), x=-2..3, y=-2..3, fieldstrength=maximal(2),color=red);

   MB says:
   

   If the leaf fell in at (-2,2.5), it would likely stay in the right-moving fast water that seems to be the main part of the river. Because it falls in at a higher y-value, it is not initially swept downwards on its journey.

3. Compare this behavior with the path the leaf would follow if it fell in at the point (-2,2.5) instead.

   EP says:
   

   The leaf would flow down to the 3rd quadrant and into the fourth before turning up into the 1st quadrant.

---

Due Wednesday 4/16 at 9am

Section 13.5: Triple Integrals

Reading questions:


1. Why might you need to compute a triple integral?

   EL says:
   

   Triple integrals are real handy if you're dealing with hypersurfaces for some reason, or if you're (probably) just working with three dimensions, you can use them to find the mass, volume, center of mass, etc. of an object.

2. What is often the hardest part of calculating a triple integral?

   ZH and KB say:
   

   There are six different ways to write the triple iterated integrals!!! Also, of course, finding the limits of integration -- there are so many regions that you have to define.

---

Due Monday 4/14 at 9am

Section 13.4: Surface Area

Reading questions:


1. Give a real-world example where you would want to compute the surface area.

   Combining answers from AM, NGP & BM:
   

   In E & M physics you often calculate surface area of objects. You may want to compute surface area for the distribution of charge on a sphere.

   Engineers use SA all the time, for instance to use the least amount of material to make something.

2. After partitioning the region R, what objects are used to approximate the surface area over each subregion R_i?

   NGP says;
   

   Parallelograms

3. Where does the cross-product come in?

   LD says:
   

   The cross-product is used to find the area of the parallelogram that caps the function that we are trying to take the surface area of.

4. What formula from earlier in the semester does the formula for surface area remind you of? Why might these two quantities have similar formulae?

   MB says:
   

   The formula for surface area reminds me of the formula for arc length. Extending arc length to 2 dimensions does seem to be surface area.

---

Due Friday 4/11 at 9am

Section 13.3: Double Integrals in Polar Coordinates

Reading questions:


1. Why might we want to convert a double integral from rectangular coordinates to polar coordinates?

   XZ says:
   

   If it's in rectangular coordinates, the integration can cause lots of frustrations, if the region we want to integrate over is circular (in some way).

2. When we use iterated integrals to evaluate an integral in polar coordinates, what is dA?

   DS says:
   

   dA = r*dr*dθ when integrating polar coordinates.

---

Due Monday 4/7 at 9am

Section 9.4: Polar Coordinates
Section 13.3: Double Integrals in Polar Coordinates

Reading questions:


1. Find one polar coordinate representation for the rectangular point (5,-2).

   MK says:
   

   r=(sqrt(5^2+-2^2) = sqrt(29)

   arctan(-2/5)= -0.38 rads

   So the polar coordinate is (sqrt(29), -0.38)

2. Find rectangular coordinates for the polar point (1,Pi/3).

   NGP says:
   

   x=rcos(theta), y=rsin(theta)

   x=1(cos(pi/3))=1/2 y=1(sin(pi/3))≈(.866)

   Rectangular coordinates are roughly (1/2,.866)

3. The graphs of (a) r=sin(7θ) and of (b) r=sin(7θ)+5 are shown below. Match the functions to the graphs.

   NA says:
   

   sin(7θ)+5 corresponds to the first graph, and sin(7θ) cooresponds to the second graph.

Note: To get these, I used the commands
• polarplot(sin(7*theta),theta=0..Pi);
• polarplot(sin(7*theta)+5, theta=0..2*Pi);

Due Friday 4/4 at 9am

13.2: Area, Volume, and Center of Mass

Reading questions:


1. If R is the circle of radius 3 centered at the point (5,1), what is ∫∫_R dA? (You should be able to do this without setting up any iterated integral)

   SR says:
   

   ∫∫_R dA is the area of R ... 9Pi.

2. Is M_y used to compute the x- or the y- coordinate of the center of mass? And in a related question -- why do you think it's called M_y?

   GW says:
   

   M_y is used for the x coordinate of the center of mass.

   It is called M_y because it is the moment about the y-axis.Ę

---

Due Wednesday 4/2 at 9am

Section 13.1: Double Integrals (continued)


Reading questions:
In the double integral ∫∫_Rf(x,y)dA,

1. What does dA represent?

   JH says;
   

   The dA represents that the double integral is taken with respect to area.

   I add:

   The dA can be thought of in two ways. In one sense, it simply tells you that you're integrating with respect to area. In another sense, however, you can think of it as an infinitesimally small area (coming from the region R). That multiplied by the height of the function f(x,y) give us the volume of an infinitesimally thin piece of our total volume. This is the infinitesimal analogy of the boxes we add up the volume of in the Riemann sum that lead to the definition of the double integral.

2. Do the two integral signs go together -- that is, are they, in a sense, a new symbol? Or do they represent first one integral and then another?

   JE says:
   

   They are their own symbol. Unlike the symbol used in Calc 1 and Calc 2, this symbol indicates the area under a function and above a region R. To represent this, we use a new symbol, rather than two of the old symbol, which indicate taking the area under a curve.

   I add:

   While Fubini's theorem tells us that one way we can actually calculate a double integral is to do it as nested integrals (one inside the other), that is a consequence, or a result ... not the definition of what a double integral is. A double integral is (signed) volume under a surface, and the double integral symbol is a single symbol, just as dA represents a tiny piece of area, NOT dxdy or dydx (although Fubini's theorem tells us we can replace dA with dxdy or dydx).

---

Due Monday 3/31 at 9am

Section 13.1: Double Integrals

Reading questions:


1. If f(x,y) is a positive function of two variables, what does _R f(x,y)dA measure?

   SR says:
   

   It measures the volume lying beneath the surface z=f(x,y)and above the region R.

2. Explain Fubini's Theorem in your own words. What is its importance?

   XZ says:
   

   When taking the double the integral of f(x,y) over the region R={(x,y)|a≤x≤b and c≤y≤d}, we can either first integrate with respect to x followed by integrating with respect to y OR first integrate with respect to y followed by integrating with respect to x.

   ZH adds:
   

   Its purpose is to turn a double integral over a region into two integrals like one does in Calc 1.

---

Due Friday 3/28 at 9am

Section 12.7: Extrema of Functions of Several Variables (continued)

Reading questions:


1. Describe the idea behind the method of steepest ascent in your own words.

   ED says:
   

   Pick a point. Move in the steepest direction (the direction of the positive gradient) until the slope stops being positive. Compute new gradient at that point. Move in the steepest direction until slope stops being positive. Keep doing this process until the difference between the starting point's and the gradientŐs steepness is insignificant.

---

Due Wednesday 3/26 at 9am

Section 12.7: Extrema of Functions of Several Variables
Reading questions:


1. If the partials f_x and f_y exist everywhere, at what points can f have a local max or a local min?

   EP says:
   If the partials f_x and f_y exist everywhere, then f can have a local max or min when f_x and f_y=0.

2. Suppose f_x(x₀, y₀)=f_y(x₀, y₀)=0, and at that same point both f_xx and f_yy are positive (so that in both the x and y direction, f is concave up). Must f have a local min at the point (x₀, y₀)?

   BM says:
   

   If f_x(x₀, y₀))=f_y(x₀, y₀)=0, and at that same point both f_xx and f_yy are positive, then we still need to know f_xy to classify the critical point.

   NGP adds:
   

   If D(x₀, y₀) is greater than zero, then there is a local min at (x₀, y₀).

3. Suppose that f is a "nice" function, and that the following are true:
   • f_x(2,-1)=0=f_y(2,-1)
   • f_xx(2,-1)=-1
   • f_yy(2,-1)=-3
   • f_xy(2,-1)=2.5
   Will the point (2,-1) be a local max, local min, neither, or do you not have enough information to tell? In any case, justify your answer.

   JB says:
   

   (2,-1) will be a saddle point because: D= f_xxf_yy-(f_xy)^2=3-6.25=-3.25, and if D<0 then f(a,b) is a saddle point.

---

Due Friday 3/21 at 9am

Section 12.6: The Gradient and Directional Derivatives (continued)

Reading questions:


1. Toward the top of page 988, the book says:

   ..notice that a zero directional derivative at a point indicates that u is tangent to a level curve.

   Why is this true?

   DS says:
   

   This statement is true because level curves are curves in the xy-plane on which f is constant.

2. What information does the gradient give you about f(x,y)?

   AM says:
   

   It tells you the change in a quanity as you move in some direction from your current position.

   BM adds:
   

   the gradient allows you to find maximum and minimum rates of change of f(x,y).

   And AJ adds:
   

   The gradient is orthogonal to the function, so it is a normal vector, useful when creating such things as tangent planes.

---

Due Wednesday 3/19 at 9am

Section 12.6: The Gradient and Directional Derivatives

Reading questions:


1. In order to find D_uf(a,b), what must be true about our choice of u?

   CN says:
   

   Our choice of u must be a unit vector u₁i+u₂j.

2. What type of quantity is the gradient of f? If we have z=f(x,y) is a surface in 3 dimensions, then what dimension does the gradient live in?

   NA says:
   

   The gradient is a vector-valued function whose components are first order partial derivatives. With a surface in 3 dimensions the gradient will be in 2 dimensions.

3. For f(x,y)=x^2y-4y^3,
   

   (a) Compute D_uf(2,1) for u=<-3/5,4/5>.
   

   CG says:
   

   df/dx = 2xy = 4
   df/dy = x^2 - 12y^2 = -8

   D_uf(2,1)=<4, -8> * <-3/5,4/5> = -12/5 + -32/5 = -44/5

   (b) Compute the gradient of f(x,y)

   LD says:
   

   (2xy,x^2-12y^2)

---

Due Friday 3/7 at 9am

Section 12.5: The Chain Rule

Reading questions:


Suppose that w=f(x,y,z) and that x,y, and z are all functions of r,s, and t.
1. How many partial derivatives do you need to calculate to determine ∂w/∂t? (Using a tree diagram may prove useful.)

   JE says:
   

   Using the tree diagram, there would be three branches coming from w=f(x,y,z), one for each variable's partial derivatives. From here, each variable becomes dependent on t, yielding three derivatives dependent on t. So, two steps of three partial derivatives are necessary to find ∂w/∂t (for a total of 6).

2. What is the expression for ∂w/∂t?

   MK says:
   

   ∂w/∂t= (∂w/∂x)( ∂x/∂t) + (∂w/∂y)(∂y/∂t) + (∂w/∂z)(∂z/∂t)

Due Wednesday 3/5 at 9am

12.4: Tangent Planes and Linear Approximations

Reading questions:

1. If f(x,y) is a well-behaved function and has a local maximum at (a,b), what can you say about the linear approximation to f(x,y) at (a,b)?

   ZH says:
   

   The linear approximation would be horizontal (parallel to the xy plane)

   EL adds:
   

   If a linear approximation were to be taken at (a,b), it would always approximate f(x,y) as f(a,b) because fx(a,b)=fy(a,b)=0.

2. Let f(x,y)=xy^2. Find a vector normal to the plane tangent to z=f(x,y) at the point (2,3,18).

   JH says:
   

   If f(x,y)=xy^2 then fx = y^2 and fy = 2xy,
   so at the point (2,3,18) we have fx = 9 and fy = 12

   so a normal vector is: <9, 12, -1>

3. Let L(x,y) be the linear approximation of f(x,y) at (a,b). What graphical properties of the surface z=f(x,y) would make L(x,y) particularly accurate? particularly inaccurate?

   EP says:
   

   For L(x,y) to be accurate, the surface would have to be sloping very gradually. An inaccurate linear approximation would be one that quickly moves away from the tangent plane.

Due Friday 2/29 at 9am

Section 12.3: Partial Derivatives

Reading questions:

1. For a function z=f(x,y), what information does f_y(1,2) give you?

   DS says:
   

   f_y(1,2) gives us the instantaneous rate of change in the y direction, at the point (1,2) on the function f.

2. For f(x,y)=ye^xy, find f_x(x,y).

   AM says:
   

   For f(x,y)=ye^xy, f_x(x,y)=y²e^xy.

3. How many second-order partial derivatives does a function f(x,y,z) have?

   SR (and several others) say:
   

   There are 6 second-order partial derivatives for f(x,y,z).

   AJ (and several others) contend:
   

   A function f(x, y, z) has 9 possible second-order partial derivatives: fxx, fxy, fxz, fyx, fyy, fyz, fzx, fzy, fzz.

   So ... which is it?

Monday 2/25 at 9am

Section 12.2: Limits and Continuity

Reading questions:

1. Consider f(x,y)=3x² /(x²+y²). We want to see what we can say about lim_{(x,y) -> (0,0)} f(x,y).

   (a) Find the limit along the path x=0. That is, find lim_(0,y)->(0,0) f(0,y).
   

   JB says:
   

   lim_(0,y)->(0,0)f(0,y)= lim_(0,y)->(0,0)(0/(0+y^2)) =lim_(0,y)->(0,0)(0)=0

   (b) Find the limit along the path y=0. That is, find lim_{(x,0) -> (0,0)} f(x,0).
   

   MK says:
   

   lim_{(x,0) -> (0,0)} f(x,0)= lim _{(x,0) -> (0,0)} (3x^2/(x^2+0) = 3

   (c) What -- if anything -- can you conclude from your results?

   ED says:
   

   Since the limit along the two paths donŐt match each other, the limit does not exist.

2. What is the point of Example 2.5?

   CN says:
   

   This example is important because it shows that even if the limits along straight lines are the same, it does not necessarily mean that the limit exists. You have to choose some path that will cancel out the x's and y's, and then you will be able to see if it is still the same. Using something like the book does using "x=y^2", (which is obviously not a straight line) helps come to the conclusion that the limit does not exist.

3. Why do you suppose we're studying limits now?

   MB says:
   

   Limits are going to be important for understanding how taking derivatives of multivariable functions works, and applying this. The same ideas from Calc 2, like continuity and approximation, are probably equally useful in three dimensions (or higher).

Due Friday 2/22 at 9am

Section 12.1: Functions of Several Variables (continued)

Reading questions:

1. Give the equations of two level curves for the function f(x,y)=x²/4+y²/9. What will these level curves look like?

   NGP says:
   

   In the plane z=c, level curves are x^2/4 + y^2/9=c.

   When c > 0, they would be ellipses.

   One equation could be x^2/4 + y^2/9=4, another x^2/4 + y^2/9=9.

2. Below is a contourplot for the function sin^2(x)+cos^2(y). Roughly where does it have local extrema? How can you tell?
   
   
   (I used the Maple command
   contourplot((sin(x))^2+(cos(y))^2,x=-3..3, y=-1..2, contours=15, coloring=[blue,red]);
   to create this graph. The "coloring=" option allows you to specify which color will be associated with the lowest contours (blue) and which will be associated with the highest (red). )

   LD says:
   

   The graph has local extrema at (-pi/2,0) (0,pi/2) and (pi/2,0). I know this because this is where the centers for the ellipses that are on the graph are, and where other lines ripple out to.

   EL adds:
   

   The lines on the contour plot should be closer together when the derivative of z is large, and far apart when the derivative of z is small. Extrema should appear when the derivative of z equals 0, so where the lines are far apart. This happens at (Pi/2, 0), (-Pi/2, 0), and (0, Pi/2).

Due Wednesday 2/20 at 9am

Section 12.1: Functions of Several Variables

Reading questions:

1. Below are the graphs of 4 surfaces. 2 of them are the graphs of the functions f₁(x,y)=[(x²+y²)-2]³ and f₂(x,y)=y²sin(x). Match the functions with their graphs.

   (a)		(b)
   (c)		(d)

EP says:

f1(x,y)=[(x2+y2)-2]3 matches up with graph d. The x^2 + y^2 means there will be circular cross sections parallel to the xy-plane. In the xz and yz trace, the graphs are cubics, matching up to graph d

CG says:

This is like a normal sine graph except the amount that the sine will equal is increasing exponentially in both directions.

Due Monday 2/18 at 9am

Section 11.6: Parametric Surfaces

Reading questions:

1. Following a process similar to that in Example 1 (not using Maple to graph), identify the surface defined by the parametric equations x=2cos(u)sin(v), y=2sin(u)sin(v), z=6cos(v). (You might want to review the generic equations for quadric surfaces given in Section 10.6).

   ED says:
   

   Eliminate u parameter:
   

   x^2+y^2	=	(2cos(u)sin(v))^2 + (2sin(u)sin(v))^2
   	=	4cos^2(u)sin^2(v) + 4sin^2(u)sin^2(v)
   	=	4sin^2(v) (sin^2(u) + cos^2(u))
   	=	4sin^2(v)

   Next eliminate v parameter:

   x^2+y^2+(z/3)^2	=	4sin^2(v) + (z/3)/^2
   	=	4sin^2(v) + [(6/3)cos(v)]^2
   	=	4sin^2(v) + (2cos(v))^2
   	=	4sin^2(v) + 4cos^2(v)
   	=	4(sin^2(v) + cos^2(v))
   	=	4

   x^2+y^2+(z/3)^2 = 4 => (x^2)/4+(y^2)4+(z^2)/36 = 1.

   Therefore we have an ellipsoid

   JE suggested the following alternative approach:
   

   First step:
   

   Notice that there are going to be some circular cross-sections.
   Think: x^2 + y^2.
   In this case, x^2 + y^2, when simplified, = 4sin^2(v).

   Next Step:
   

   x^2 + y^2 + z^2 = 4sin^2(v) + 36 cos^2(v).

   The RHS of this equation simplifies to 4 + 32cos^2(v).
   Notice that 32 cos^2(v) is the same as (8/9)z^2.

   This can be rewritten as x^2 + y^2 + z^2 = 4 + (8/9)z^2.
   Simplified: x^2 + y^2 + z^2/9 = 4.

   
   

   In Example 6.3, what might make you think to use cosh(u) and sinh(u) in the parametric equations for x and y? That is, what property that's useful in this situation do they have?

   KB says:
   

   cosh^2(u)-sinh^2(u)=1 is a useful property to simplify the equation in order to get a hyperbolic paraboloid.

Due Wednesday 2/13 at 9am

Section 10.6: Surfaces in Space

Reading questions:
Consider the surface x=4y²+4z².

1. Which coordinate plane does the equation z=0 define?

   GW says:
   

   The xy plane.

2. What does the trace of this surface in the xz-plane look like?

   JH says:
   

   " In the x-z plane" -> y = 0 -> x = 4z^2, which is a parabola.

3. What do the traces of this surface in the planes x=k look like?

   ZH says:
   

   k=4y^2+4z^2 => k/4=y^2+z^2 => (sqrt(k)/2)^2=y^2+z^2

   These are circles with increasing radii as k increases of the form: sqrt(k)/2

4. What is this quadric surface called?

   BM says:
   

   The surface is a circular parabloid

Due Monday 2/11 at 9am

Section 11.5: Tangent and Normal Vectors

Reading questions:

1. Suppose you are skiing down a hill along a path that curves left. Describe the direction of the unit tangent, principle unit normal, and binormal vectors to the curve that describes your motion.

   DS says:
   

   The unit tangent is straight ahead of me, in the direction that I am going at that instant. The principle unit normal is 90 degrees to my left, in the direction of the center of the curve.

Due Friday 2/8 at 9am

Section 11.4: Curvature

Reading questions:

1. Explain the idea of curvature in your own words.

   JB says:
   

   Curvature is how sharp a turn on a curve is, or how much a curve turns within a certain unit of length.

2. If the helix in Example 4.5 were changed to r(t)=< 2sin(t), 2cos(t), 4t²>, will the curvature still be constant? Don't actually do the calculation, but give an intuitive justification - think about how changing the z-coordinate to 4t² from 4t will affect the graph.

   MB says:
   

   The change in the z coordinate will affect the curvature of the helix. As t increases, there will be more and more stretching of the helix, as the squared z coordinate will result in a greater height increase per spiral of the helix.

Due Wednesday 2/6 at 9am

Problem Set Guidelines
Section 11.3: Motion in Space

Reading questions:

1. If r(t) is a vector-valued function representing the motion of an object at time t, what physical information does r'(t) give you? How about ||r'(t)||?

   NGP says:
   

   r'(t) tells you the object's velocity at time t, and the magnitude of r'(t) is the speed of the object.

2. Find the velocity and acceleration vectors, if the position of an object moving in space is given by r(t)=(5/sqrt(t))i+ln(t³)j-tan(3t)k.

   SR says:
   

   v(t)=r'(t)= <-5/2sqrt(t^3), 3/t, -3sec^2(3t)>
   a(t)=r''(t)=<15/4sqrt(t^5), - 3/t^2, -18sec^2(3t)tan(3t)>

Due Monday 2/4 at 9am

Section 11.2: The Calculus of Vector-Valued Functions

1. If r(t)=tcos(t)i+exp(t^2)j+ln(t)k, what is r'(t)?
   Remember exp(t^2) is Maple notation for e^t2.

   AJ says:
   

   r'(t) = <-t*sin(t) + cos(t), 2t*exp(t^2), 1/t> by the usual differentiation rules.

2. If r(t) is a vector-valued function, what geometric/graphical information does r'(a) give you?

   EP says:
   

   r'(a) gives the tangent vector to the curve c at the point corresponding to t=a for the vector-valued function r(t)

   XZ adds:
   

   If r(t) is on a interval I, the curve traced by the vector-valued function r(t) is smooth if r' is continuous on I and r'(t) is not equal 0, except possibly at any endpoints of I.

3. Let r(t)=cos(t)i+sin(t)j and s(t)=sin(5t)i+cos(5t)j.

   (a) What do the graphs of r(t) and s(t) look like?
   

   MK says:
   

   For r(t), the parametric equations are:
   x = cos(t), y = sin(t).

   x^2+y^2 = cos^2(t)+sin^2(t) = 1, so, x^2+y^2 = 1
   => A circle centered at the origin, with radius = 1

   For s(t), the parametric equations are:
   x = sin(5t), y = cos(5t).

   x^2+y^2 = sin^2(5t)+cos^2(5t) = 1, so x^2+y^2 = 1
   => A circle centered at the origin, with radius = 1

   Therefore, the graphs will look alike because they have the same equation x^2+y^2 = 1.

   (b) If the graphs of two vector-valued functions r(t) and s(t) are the same, must r'(0)=s'(0)? (Is this a new result, or was it also true for functions f(x) and g(x)?)

   MB says:
   

   Because the derivative of a vector valued function is a vector, and not a slope, there is the potential for similar-looking graphs of two functions to have different derivatives. This differs from functions such as f(x) and g(x).

Due Friday 2/1 at 9am

Section 11.1: Vector-Valued Functions

Reading questions:

1. The vector-valued function r(t)=cos(t)i+sin(t)j lies in the plane.

   (a) Using Example 1.2 as a guide, what will the graph of this function look like?
   

   XZ says:
   

   x = cos(t), y = sin(t)
   => x^2 + y^2 = cos^2(t) +sin^2(t)
   => x^2 + y^2 = 1

   It is a circle with radius = 1, centered at the origin.

   (b) Is it possible to rewrite r(t) as a function y=f(x)?

   CN says:
   

   It is not a function. It is possible to rewrite r(t) as two functions y=f(x), one positive, and one negative.

2. What are some advantages to using vector-valued functions?

   ZH and AJ say:
   

   Some advantages of using vector valued functions include:

   • Extraordinarily complex functions in Cartesian coordinates can be made much easier. This is especially true for 3D cases where the equations in xy can get scary.
   • With vector-valued functions you can create things that cannot be created with a function y = f(x).
   • Since vectors and parametric equations are so similar, the notation used for vectors provides a GREAT way of creating vector equations that is more compact.

Due Wednesday 1/30 at 9am

Section 9.1: Plane Curves and Parametric Equations Section 10.5: Lines and Planes in Space

Reading questions:

1. Find parametric equations for the line through the points (1,2,3) and (7,-1,0).

   KB says:
   

   x-1= - 6t y-2=3t z-3=3t

2. What information about a plane P do you need to determine an equation for the plane?

   BM says:
   

   To determine the equation of a plane you need either a point and a normal vector, three pts, or a point and a parallel plane.

3. Find an equation for the plane containing the point (1,2,3) with normal vector <3,0,-1>.

   EL says:
   

   3(x-1)+0(y-2)-1(z-3) = 0 -> 3x-z = 0

Reminder:

• Make sure you clear up any difficulties on PS 1 by coming to my office hours or taking advantage of the tutors at the Kollett Center.

Due Monday 1/28 at 9am

Problem Set Guidelines
Section 10.4: The Cross Product
Section 10.5: Lines and Planes in Space

Reading questions:

1. If a, b, and c are arbitrary vectors, should we expect that (a x b) x c=a x (b x c)?

   MK says:
   

   No, because the cross product is not associative.

2. How can we tell whether two vectors in V₃ are parallel? How can we tell whether two vectors are orthogonal?

   CG says:
   

   If the cross product is zero then they are parallel. If the dot product is zero they are orthogonal.

3. What information about a line L do you need to determine an equation for the line?

   JB says:
   

   In order to find an equation for a line L, we need to have either two points on the line, or we must know one point on the line and the direction of the line.

4. Find parametric equations for the line through the point (1,2,3) and parallel to the vector <3,0,-1>.

   NA says:
   

   Parametric equations are:
   x-1=3t
   y-2=0t
   z-3= -1t
   

Due Friday 1/25 at 9am

guidelines for submitting reading assignments
suggestions for reading a math text
course policies
syllabus
Section 10.1: Vectors in the Plane
Section 10.2: Vectors in Space
Section 10.3: The Dot Product
Section 10.4: The Cross Product

Reading questions:

1. Find a unit vector in the direction of <-2,3,-6>.

   LD says:
   

   ||<-2,3,-6>|| = sqrt[(-2)^2+(3)^2+(-6)^2] = sqrt(49) = 7

   u = (1/7)*<-2,3,-6> = <(-2/7),(3/7),(-6/7)>

2. Let a=<2,-1,3>, b=<4,10,-1>, and c=<3,-1,2> be vectors, and let x.y represent the dot product of any vectors x and y.
   

   (a) Find a.b.
   

   JE says:
   

   a.b.= 8+(-10)+(-3)= -5

   (b) Does (a.b).c make sense, and if so, what is it in this case?
   

   AJ says:
   

   No because you can't do a dot product with the scalar that results from a . b.

   (c) Are b and c orthogonal?
   

   ED says:
   

   <4,10,-1> * <3,-1,2> = 12-10-2 = 0. Therefore b and c are orthogonal, because b.c = 0

3. Briefly explain what, geometrically, proj_ba and comp_ba are.

   JH says:
   

   proj_ba is the projection of a onto b, meaning the vector that is parallel to b and has the same component as a along b. comp_ba is the magnitude of the segment formed when a perpendicular line is dropped from vector a onto the line of vector b.

4. How is a x b related to a and b geometrically?

   AM says:
   

   axb is a vector in V3 which is perpendicular to both a and b.

5. If a and b are vectors in V₂ (that is, in the plane), is a x b defined?

   EP says:
   

   a x b is not defined in V2. The cross product does not work for vectors in V2. If it was set up using a matrix, it would not be able to be set up and calculated out.