## Project 1

VectorPlot[{1, 2 x}, {x, -3, 3}, {y, -3, 3}] StreamPlot[{1, 2 x}, {x, -3, 3}, {y, -3, 3}] VectorPlot[{1, ((2 y)/x)}, {x, -3, 3}, {y, -3, 3}] StreamPlot[{1, ((2 y)/x)}, {x, -3, 3}, {y, -3, 3}] VectorPlot[{1, ((x – y)/(x + y))}, {x, -3, 3}, {y, -3, 3}] StreamPlot[{1, ((x – y)/(x + y))}, {x, -3, 3}, {y, -3, 3}] VectorPlot[{1, (x^2) + (2*y^2) }, {x, -3, 3}, {y, -3, 3}] StreamPlot[{1, (x^2) + (2*y^2) }, {x, -3, 3}, {y, -3, 3}] VectorPlot[{1, (cos (y)/(x*sin (y) – y^2)) }, {x, -3, 3}, {y, -3, 3}] VectorPlot[{1, (-2*x*y)/((x^2) + (y^2)) }, {x, -3, 3}, {y, -3, 3}] StreamPlot[{1, (-2*x*y)/((x^2) + (y^2)) }, {x, -3, 3}, {y, -3, 3}] Question 1:

Question 1:

In question one our group was required to generate a pair of random vectors in dimensions two, three, four, and five and determine in which dimension are the vectors more likely to be perpendicular. We used the following code to generate the pair of random vectors for each dimension.

This code randomly generated 10,000 pairs of vectors with a specified number of dimensions with components ranging from -1 to 1. Maple then finds the angle between two vectors. The map function is used to convert radians into degrees.  The last two lines generate a histogram based on the data.

Our group found that a pair of random vectors is most likely to be perpendicular to each other in the fifth dimension. Each of the four histograms represent the likely hood for the vectors to create a 90 degree angle. In each histogram the x-axis corresponds to the angle, and the y-angle corresponds to the frequency of the angles. Clearly, the fifth dimension has a higher percentage of 90 degree angles.

As the number dimensions increases, the number of different ways one can have a vector be perpendicular to another vector increases.  When expanding the number of dimensions from two to three, the number of possible perpendicular vectors for a given vector increase. This suggests that we will be more likely to find perpendicular vectors in dimension three than in dimension two. Looking at the histogram for two dimensions, we see that all the angles form at about the same frequency. When looking at the dimension third dimension graph, we can see that the graph peaks at about 90 degrees.

The same goes for the graphs in the fourth and fifth dimension. It is easy to see that as the dimensions increase the percentage of 90 degree angles increase.   The above histograms are featured in ascending order (2 Dimensions, 3 Dimensions, 4 Dimensions, 5 Dimensions) as you scroll down.  The independent axis corresponds to the angle, and the dependent axis corresponds to the frequency of the angles.  As the number of dimensions increases, the odds of one vector being perpendicular to another vector also increases, forming a direct relationship. After completing this process for each specified dimension, we concluded that we are more likely to find a pair of corresponding vectors to be perpendicular to each other in the fifth dimension.  As the number of dimensions increase, there is a greater probability of finding two vectors that are perpendicular to one another. When analyzing the first histogram, which depicts the angles formed by corresponding vectors in the second dimension, we can see there is a relatively modest distribution.  There is no specific trend that represents the greater probability of perpendicular angles forming. However; when we look at the histogram that illustrates the results of the corresponding angles in the third dimension, we can see there is a greater probability of perpendicular angles to form. Since a third axis (in this case, a z-axis) is added to the coordinate system, there are many more opportunities for any two random vectors to form 90 degree angles.

When comparing both the second and third dimensions, it is obvious that the third dimension represents the greatest possibility of vectors forming 90 degree angles. However, as we add more dimensions, the frequency of finding 90 degree angles increases dramatically. This is represented by analyzing the histograms for both the fourth and fifth dimensions. As the number of dimensions increases, we see a peak form around the 90 degree mark, representing a greater likelihood of finding perpendicular angles. Also, if we were to increase the number of random vectors generated, we would clearly see a more accurate representation of this distinction between dimensions. We can conclude that if a sixth dimension were added to this program, we would produce a histogram with an even steeper slope around 90 degrees.

Question 2:

Question 2 asks to generate a heap of random three-dimensional vector pairs and for each pair of vectors plot the cross product versus the dot product, and discuss any apparent observations.  Using maple we plugged in the following  codes below: When doing this, it let us find the needed magnitude of the cross products and the dot products and divide these values by the products of the lengths of both vectors, and plot the results against each other

This code was used to generate 100 pairs of three dimensional vectors. (We originally used 10,000 but found using 100 made it easier to see the relationship. With using 10,000 a solid line was created which was each of the 10,000 points.)  It then establishes a formula for the magnitude of the cross product, divided by the product of both vector lengths, and for the dot product divided by the product of both vector lengths. When trying the code for our list we chose to write it differently but still resulting in the same answer. We made it so the list would generate an “a” and a “b”, each being a random vector. We then made it so the list would generate the cross product vs. the dot product as shown above.  These results were then plotted onto a graph.

The graph appears to that of a semicircle. This is because the magnitude of the cross product over the product of the two vector lengths equals , which reduces to .  Similarly, the dot product over the product of the two vector lengths is , which reduces to . We can see that the graph forms a semicircle that has a radius of 1.0 and forms over the first and fourth quadrants of the coordinate system.  The formulas of the dot product and the cross product would respectively reduce to cos θ and sin θ. A parametric equation, therefore, is formed when these results are plotted.

The unit circle’s appearance reinforces the link of dot products and cross products to trigonometry. When the magnitude of one is at its maximum, the magnitude of the other is always zero.  We can see this at the angles of 0 and 180 degrees, where the magnitude of the dot product is maximized, and at 90 degrees, where the magnitude of the cross product is maximized. Question 3:

The final question asks us to generate a list of triples of random vectors in 3 dimensions and find out how the volumes of parallelpipeds are distributed. The code is as followed:

Maple then records the results and generates a histogram which is shown grouping the volumes of the parallelpipeds produced by the scalar triple product. From the histogram we can coonclude that the greatest frequency of parallelpipeds is with smaller volumes. Parallelpipeds with larger volumes are much less frequent than the smaller. This is because  for a large parallelpiped to be generated, the vectors must have close to 90 degree angles in between them and the vectors themselves must be long.

If the histogram was divided into smaller bins, we would probably see that there was a frequency peak shortly to the right of the y-axis, with the frequency falling off rapidly at either side, forming a skewed distribution with the tail to the right.  The smallest volumes in the data set would be exceedingly unlikely, as one would require the combination of very small angles between the vectors and/or small vectors, preferably both.  Seeing that the data is being randomly generated, this extreme seems that this combination would be almost as unlikely as the largest of the parallelpipeds. (This would probably still be  more likely; a single vector of length zero, or and angle between the vectors of 0 degrees would yield a zero volume, while a single angle of 90 degrees or vector of maximum length would not guarantee a maximum volume. )

By examining the graph, we can see a trend in the volume with respect to frequency. As the volume of the parallelepiped decrease, the frequency increased and vice versa. This makes sense because in order to have larger volumes, the vectors forming the shape must be long and have large angles between them. Since we are randomly generating vectors, it is very unlikely to form a parallelepiped with three vectors with large magnitudes and large angles between them. Even though we are more likely to generate 90 degree angles in the third dimension, the distribution of angles over the entire shape would decrease that likelihood

## Project 6

Part A: Line Integrals using Plane Curves

For this first part, we were first asked to select two 2D vector fields and plot them in Maple. As we know, vector fields describe the magnitude and direction of vectors in space. Using two variables, we can see how our vector field acts in two dimensions. We were then asked to use two parametrize plane curves and evaluate the line integrals $\int_C \textbf{F}\cdot d\textbf{r}$ using our previously selected vector fields. This curve will act as the path a particle takes as it moves through a region of space; in this case, as it moves through a vector field. The line integral will describe the work done on the particle while it moves along its path.

Our first graph is as followed: As we can see, the vector field lies on the interval [0,1] in both the x and y directions. Our parametric curve itself follows a counterclockwise path along the time interval [0, $\frac{\pi}{2}$]. To solve the line integral, we simply substituted $x = cos(t)$ and $y = sin(t)$ into our vector field equation Above is the integral taken after finding the dot product of our original equation and the derivative of sin(t),cos(t). It was as followed * <-sin(t),cos(t)>

Once this was done we tool the integral of the dot product to find how much work was done when moving from 0 to Pi/2. We got -1/3. This tells us that the vector field will travel toward the positive x direction, opposing the path our the curve ( which we know travels counterclockwise). This opposition will give us a negative value for our line integral, indicating the particle has done negative work.  If we look at the graph itself. The path the curve takes moving in the y direction cannot cancel out the change in movement in the negative x direction, thus our value must be negative.

We then used the following maple code to plot our second graph:  Again  above is the integral taken after finding the dot product of our original equation and the derivative of sin(t),cos(t). It was as followed * <-sin(t),cos(t)>

ANd once again we got a negative answer of -9/40. which once again shows that negative work is being done This tells us that the vector field will travel toward the positive x direction, opposing the path our the curve ( which we know travels counterclockwise). This opposition will give us a negative value for our line integral, indicating the particle has done negative work.  If we look at the graph itself. The path the curve takes moving in the y direction cannot cancel out the change in movement in the negative x direction, thus our value must be negative.

Line Integrals using Space Curves

The second part of this project asked us this time to choose two 3D vector fields of the form $\textbf{F}(x,y,z)$ and plot them using Maple.  This will allow us to get a general picture of the field in 3d. We then were asked to choose two parametric space curves and evaluate the line integrals.  THe integral above is the dot product we calculated which was <3*cos(t)*sin(t),2sin(t)^2,(cos(t)-sin(t))*z>*<-sin(t),cos(t)>

Once again we get a negative work done as we move from 0 to Pi/2 (counter clockwise)  The integral above is the dot product of <-3cos(t)^2,cos(t)*sin(t),(sin(t)-cos(t))*z>*<-sin(t),cos(t)>

This time we have positive work done. WHat this means is that as we move from 0 to Pi/2 less work is required because the field vector is pointing along the same path, counterclock wise. It is moving with the curve thus less work is done giving us a positive answer.

## Project 5

In this project, we were asked to use double integrals to calculate the volume over a particular region using three different surfaces of our choosing. These three functions each contained two variables and were bounded by a set of coordinates.

Part 1: Double Integrals and Rectangular Bases

Surface 1 Above is the double integral over a set of points.  When evaluating the function from 0 to 5 we get a volume of zero which makes perfect sense because as you can see in the 3d graph as one side increases so does the other side which would cause the function to equal 0. More simply put  If we compare this result to our original 3D plot, we can see  this object is symmetrical and has an equal volume above and below the xy plane. Therefore, our total volume should equal zero when integrated.

Since our second volume is negative, our surface must be contained within the specified boundaries so that negatively calculated values of z out weigh the positively calculated values of z. As we can see in the 3D plot, this is the case; more of the volume is negative than positive.

Surface 2 Our total volume ends up being a positive value. Therefore more of our surface lies in a defined area where the z values are positive.

Surface 3 (ignore the second double integral for now)

Again our total volume ends up being a positive value. Therefore more of our surface lies in a defined area where the z values are positive.

Part 2: Double Integrals and Non-Rectangular Bases

For this part, we were asked to use the same functions for are previous surfaces; however, their corresponding bases had to be defined by a region of the form $R:=\{x,y\vert a\leq x\leq b, g_1(x)\leq y \leq g_2(x)$. Simply put instead of finding the volume over a rectangle it is between a defined plot. For the first two functions, we decided to use the boundaries of $y =2x^2$ and $y=x^2+1$ for our integral, as used in example 1 of the textbook.  The following is the maple codes we used foe each surface. In the first two cases which are surface 1 and 2 in the order above our total volume ended up being negative. As we increase our y value for either function, the elevation decreases and the function’s corresponding base has only positive y values.

The third surface has a positive which means most of our surface lies in a defined area where the z values are positive.

Part 3: Double Integrals and Differently Bound Non-Rectangular Bases

Similar to part 2 of this project, part 3 required us to use the sam calculations as before; however, we had to select a differently bound base to find the volume over. We used a region of the form $R:=\{x,y\vert c\leq x\leq d, h_1(y)\leq y \leq h_2(y)$ where we had to select two new functions that are continuous and on the interval.  For this we decided to use the 2y and y^2. This is shown as the following maple code: With the exact same results as the part two, In the first two cases which are surface 1 and 2 in the order above our total volume ended up being negative. As we increase our y value for either function, the elevation decreases and the function’s corresponding base has only positive y values.

The third surface has a positive which means most of our surface lies in a defined area where the z values are positive

With a little more time i would have went back and changed what i was bounding this integral by. I could have changed it to x-1 or something along those lines which would have given us different answers. The different answers would be due to the new shape the bounded lines would form.

Part 4: Double Integrals and Polar Coordinates

In the final part of this project, we were asked to calculate the volume under each of our surfaces and inside of a cylinder using polar coordinates.  Polar coordinates are defined by a radial distance and an angle (r, $\theta$). Any point in the polar coordinate system has a radial distance from the origin to the specified location and an angle that is counterclockwise from the positive x axis. For our project, we used the functions $x=rcos(\theta)$ and $y = rsin(\theta)$ in order to calculate the total volume. Using the following formula, we are able to find the total volume of each surface in a polar coordinate system: $v = \Sigma f(r, \theta) rdrd\theta = \int^{\alpha}_{\beta}\int^a_b(r(f(r, \theta))drd\theta$

The following is the maple code we used: As you can see above we first found the integral from 0 to 1. We then took the double integral and evaluated it from 0 to 2Pi. (There is only two functions above. For some reason maple wasnt giving us an answer when we tried to evaluate both integrals at the same time. So as shown above we just separated them.)

For surface 1 (the first two integrals) the total volume for the surface with a circular base is  the same as the total volume for the surface with a rectangular base.  If we consider the structure of the 3D shape. Since our surface has a saddle shape, it is symmetrical in both the x and y directions. The volume in both the positive and negative regions should be the same and cancel each other out.

For surface 2 (The last two integrals)  using a circular base for surface #3 changed the total volume significantly due to the new bounds of the integral.

## Project 4

In this project, we were asked to look at the gradient vector field of a surface, whose function is composed of two variables.  The 3d graphs we used are as followed the gradient vector of a function is used to indicate the magnitude and direction of the slope on any given surface. Visually,we express this change in slope on a contour plot. In this plot we have both contours lines and arrows indicating the change in the steepest slopes on a surface.

The change is expressed as $\nabla(z) = \frac{\partial z}{\partial z} i + \frac{\partial z}{\partial y}j$

The gradient is a vector function, which contains i and j. We then took the partial derivatives, which gave us the rate of change of the surface. This is shown as the following maple code.

﻿

Using the gradient, we were able to find the directional derivatives of each function.  We can define the directional derivative of a function mathematically using the equation $D_u(z) = \nabla(z) \cdot u$

Where the variable u is the direction of the unit vector of the function f. We used this equation for both geometrical surfaces.

As you can see there is a gradual change in slope as we move along the surface. This surface has a lower change in slope which is due to large denominator of the function. This surface also has a parabola shape in its surface which gives its hill look.

As you can see there is a gradual change in slope as we move along the surface. This surface has a lower change in slope which is due to large denominator of the function. This surface also has a parabola shape in its surface which gives its hill look.

This is the contour plot of surface 1. The contour lines are shown as solid lines and  our gradient field vectors are represented by the small black arrows. Each arrow gives us the direction and the magnitude of each vector. YOu can determine this by looking at the arrow’s length and direction. As the length of the vector increases in a certain direction, the magnitude increases.

Surface 2:

We then did the same as above using our second surface.

The following is the partial derivative: Using the unit vector we were able to plug this value into the directional derivative formula and use the previously calculated partial derivatives.

THis then gave us the graph of the directional derivative.

In this graph our slope changes sharply as we move away from the origin. From the graph we can see that there is a greater change in slope in the y-direction . This drastic change in slope is due to the parabolic shape our equation provides.

Once again, the countour lines are shown as solid lines and  gradient field vectors are represented by the small black arrows. Each arrow gives us the direction and the magnitude of each vector. YOu can determine this by looking at the arrow’s length and direction. As the length of the vector increases in a certain direction, the magnitude increases. This graph shows us that the corners have the greatest slope and as you move towards the center the slope is 0. The arrows are also greater in length at the corners which tell us that the magnitude is greatest there. This is evident as you can see in the 3d graph the slope is greatest as you get further away from the center.

The gradient vectors always point uphill and are also perpendicular to the contour lines on the plot. This tells us where the surface of a function is steepest.

If we look at our second plot, we can see the gradient field vectors follow a similar path alongside the contour lines. By looking at the arrow’s length and direction we can see a  there is a sharp increase in slope. As stated before as you move away from the center the slope increases thus the gradient lines increase.

If we look at the second surface, our gradient vector arrows all point from the positive y direction toward the negative y direction. Therefore our slope is greatest as we approach the x axis. These lines are a visual aid to determine how the surface changes.  As we move further away from the x axis, the change in the slope decreases until it becomes level at a certain point. This gradual increase and decrease in the gradient are a result of the overall shape.

We then used this equation to generate our contour plot as followed:

## Project 3

For project 3 we had to work to gain an analytical understanding of how surfaces change as we travel on them in different directions.  We did this by creating and analyzing graphs of surfaces and particles as they move along the surface.

For our first surface we used the following Maple code to generate a 3d graph and its corresponding contour lines. The three dimensional plot and the contour plot are each shown above. We can see the correlation between the two plots with respect to the rapid change in elevation of the surface. As we move closer to the x-axis on our contour plot, our change in slope becomes larger. Which is shown as the contour lines become closer and closer to each other. . The radius of these circles tell us the maximum and minimum values of the 3D plot.

After creating these initial graphs we found the partial derivatives of the surface. The following is the code we used in Maple.  The partial derivative is labeled fx and fy. These partial derivatives show how fast the surface changes in different directions. We can see that both of these expressions contain an exponent of 2, which contributes to the vertical curvature of the 3D plot.  There is a greater exponential change in the partial derivative of y. This is very clear because when we look at the 3D plot, our change in y drastically changes the surface’s change in slope, while the change in x only slightly changes.

We then used the parametric equation above and applied the chain rule to give us

(fx*fxx)+(fy*fyy). We then plugged in our x and y values using our parametric equation,

cos(2*Pi*t),sin(2*Pi*t). Once this was plugged in we then generated the equation above which is shown as z=.  When this equation was generated we then plotted its graphing using the following Maple code: This graph represents the speed of a particle as it moves on the surface. The graph illustrates, that there is a gradual, change in elevation in this surface. It can be seen as a wavelength. This surface has a gradual change in slope. The magnitude of this curve is about -10..10.

We then used the following code to plot our second surface.  A contour plot has lines that connect points of the same elevation.  The space between two contour lines represents a constant change in elevation.  Hence, the closer the lines are to each other, the steeper the slope of the surface in that area. The contour plot here shows that the surface starts to curve as it gets further from 0. As you get further from 0 the slope increases as seen in the contour plot where the lines get closer together in the corners of the grid. The central part of the surface is flat, so there are no central contour lines.

Our surface has both a slope upward and downward at an exponential rate as our corresponding x and y values change. This makes sense if we consider the original formula.  We can  see, by examining the 3D plot, the rapid change in elevation of the surface, we constructed a contour plot to illustrate the drastic change in slope of the surface. As we increase our x and y values, our changes in elevation increases, as we can see in the plot. The closer the lines are together, the greater the change in elevation of the surface.

We then used the same parametric equations as before. The chain rule then gave us the following.  Again this graph shows the speed of a particle as it moves along the surface. This graph tells us that z(t)  frequently increases and decreases as our values change, which shows the changes in the elevation of the surface. It looks as though the change in magnitude along the curve repeats itself every cycle, as a sine or cosine wave would. Which makes sense since our parametric equation dealt with sine and cosine. We can attribute the drastic changes in slope to the magnitude of the curve along the y-axis, as it looks as though the maximum and minimum points on the curve lie at about y=20 and y=-20.

## Project 2

Project 2 asks to plot two curves, a plane curve and a space curve. We are asked to examine various properties of plane and space curves and demonstrate our findings through Maple. Once these curves are generated we were required to find the velocity, acceleration, arc length, and curvature.

For our plane curve we chose the serpentine curve. The following is the code we used to generate this curve. Once this curve was generated with maple we then were asked to find the velocity of the curve. This is simply finding the first derivative of the x coordinate and the y coordinate. Velocity as well as acceleration is shown as the following. To reiterate, the velocity is the first derivative of the x and y coordinate which is

v=csc^2(t),cos^2(t)-sin^2(t)

The following is a graph for the serpentine curve’s velocity.

The subsequent graph for the velocity curve is as followed: As we can see in the graph, the velocity curve is continuous until it reaches the coordinate value of (-1,-1). Since we are dealing with a curve that has the functions cosecant, sine and cosine within it, which are very responsive to value changes in the variable. Unlike the position vector, the velocity vector does not contain a cotangent function within its equation which contributes to the different shape.

Next we found the acceleration of the curve which is the second derivative.

For the acceleration, we  took the second derivative of both the x and y component functions. This is represented in the equation: $a=r''(t)=csc^2(t)*cot(t), cos(t)*-sin(t)]-[sin(t)*cos(t)]$

The graph for the acceleration curve is as followed: It is very evident that the graph closely resembles the original position graph This graph represents the curve $a = r''(t) = <4*csc^2(t)*cot(t), [4*cos(t)*-sin(t)]-[4*sin(t)*cos(t)]>$ Since i did not have time to plot my graph i used Andrew Grota’s serpentine graph. The main difference that his equation has from mine is that he uses coefficients. These coefficients led to the increase in the curves height making the max and min (2,-2). If i used my equation the graph would have a max and min of (1,-1).
The acceleration graph and position graph most likely look similar because of the cotangent which the velocity curve lacks.

Arc Length
Once the velocity and acceleration was derived from the original we were then able to find the arc length. The arc length is an the magnitude of the resultant velocity vector integrated over the time interval. If a curve $\gamma :[a,b]\to \mathbb{R}^2$ is defined on a closed interval $[a,b]$ given as $\gamma(t)=(x(t),y(t))$ the arc length of the curve is: $s:=\int_a^b\sqrt{\frac{dx(t)}{dt}^2+\frac{dy(t)}{dt}^2}dt$ (quoted from the project 2 blog)

Simply this is square root of the velocity in the x direction squared plus the velocity in the y direction squared. The following is the code used to plot the arc length. The arc length of the curve on the interval [1,u]. We then made u equal to 3 which then gave us an arc length of 7.98362349. In class with the help of Dr. Davis we first used Pi as our u value this gave us an arc length of infinity. This answer was due to the fact that we have a cosecant in our equation.  When increasing the value for u we  see a rapid increasing value for arc length. As it moved closer to Pi the value of the arc length continue to increase going to infinity.

We then found the curvature of the curve by using the following Maple code.
Curvature is $k(t):=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{3/2}}$ This is simply the:
[x-velocity *y-acceleration]*[x-acceleration*y-velocity] all divided by [x-velocity squared plus y-velocity squared raised to the 3/2 power.
After plugging in our velocities and accelerations into the equation we evaluated k using K(1) this gave us a curvature of 1.04121234. We then evaluated k(2) and got -0.983016151. After this we plotted the curvature. It is as followed:
From this graph we can see that at -1 and 1 the curve has its greatest curvature. When looking back at our original position graph it is very evident that the curve has it greatest curve at -1,1. From this graph we can also conclude that it works in a period restarting at what seems to by Pi.

Space Curve
For our space curve we chose Vivian’s curve. the following is the code we used to plot the graph.  As defined by Gray (1997, p. 201), Viviani’s curve, sometimes also called Viviani’s window, is the space curve giving the intersection of the cylinder of radius and center the center of this curve is (0,0)

1908-1915, pp. 311-320; Struik 1988, pp. 10-11; Gray 1997, p. 201).
When rotated the curve will look as the following. (http://mathworld.wolfram.com/VivianisCurve.html)

To find the velocity of this curve did the same as above and found the first derivative. First we defined the curve then found its velocity using Maple: Using the velocity equation we then plotted it’s curve. We then used this equation to plot the speed be did this by:

What this graph tells us is that as the speed approaches Pi it nearly reaches zero. We can clearly see that the velocity is at its lowest at Pi.

The acceleration was determined the same way as it was before we found the second derivative using maple: It shown as the following: Once we found the acceleration and plotted the graph we then used the acceleration equation and used it to plot the speed of the acceleration, as shown above. This graph is basically the opposite of the velocity graph. The speed graph of acceleration has its greatest value at Pi. This makes sence because as Velocity is approaching Pi it is slowing down rapidly which makes its acceleration greater because acceleration is the change in velocity per unit of time. As it gets further away from Pi the velocity is increasing but not rapidly therefore the change in velocity isn’t as great making acceleration less.

Finding the Are length was the same as above. The coding for maple is as followed: Once the equation was defined we inserted s(1) which came out to be zero. And when we inserted s(3) we got 2.2928096

As you get further away from the origin the curvature should increase up to a point. At a certain point it should hit a max and return to zero. One can visualize this by looking at the original graph and see how it makes a figure eight. Being a figure eight makes it move in a cyclical motion.

Curvature

The following is the code used to determine the curvature of the space curve.

With more time i would have been able to plot this graph. But i can predict that its curvature would have been greatest around where the arrows below point. These points are most likely to have the greatest curvature because the curve looks the greatest here. As seen in the serpentine curve the curvature is greatest where the curve is the most drastic and noticeable.

## Project 1

Question 1:

Question 1:

In question one our group was required to generate a pair of random vectors in dimensions two, three, four, and five and determine in which dimension are the vectors more likely to be perpendicular. We used the following code to generate the pair of random vectors for each dimension.

This code randomly generated 10,000 pairs of vectors with a specified number of dimensions with components ranging from -1 to 1. Maple then finds the angle between two vectors. The map function is used to convert radians into degrees.  The last two lines generate a histogram based on the data.

Our group found that a pair of random vectors is most likely to be perpendicular to each other in the fifth dimension. Each of the four histograms represent the likely hood for the vectors to create a 90 degree angle. In each histogram the x-axis corresponds to the angle, and the y-angle corresponds to the frequency of the angles. Clearly, the fifth dimension has a higher percentage of 90 degree angles.

As the number dimensions increases, the number of different ways one can have a vector be perpendicular to another vector increases.  When expanding the number of dimensions from two to three, the number of possible perpendicular vectors for a given vector increase. This suggests that we will be more likely to find perpendicular vectors in dimension three than in dimension two. Looking at the histogram for two dimensions, we see that all the angles form at about the same frequency. When looking at the dimension third dimension graph, we can see that the graph peaks at about 90 degrees.

The same goes for the graphs in the fourth and fifth dimension. It is easy to see that as the dimensions increase the percentage of 90 degree angles increase.   As Eric Card put in his blog, “The above histograms are featured in ascending order (2 Dimensions, 3 Dimensions, 4 Dimensions, 5 Dimensions) as you scroll down.  The independent axis corresponds to the angle, and the dependent axis corresponds to the frequency of the angles.  As the number of dimensions increases, the odds of one vector being perpendicular to another vector also increases, forming a direct relationship.  Considering the two dimensional simulation in space, there is a limited number of ways to form a perpendicular angle between the two vectors by tail to tail visualization.  As the number of dimensions increases to 3, there is a drastic increase in the number of pairs of vectors that are perpendicular.  As Peter’s blog points out, when visualizing the vector pairs in a three dimensional space “there should now be an infinite number of unit vectors and multiples of these unit vectors that together form a circle with the tail of the first vector at the center, perpendicular to the plane of this circle.”  Thus, suggesting it is much more likely to find a pair of perpendicular vectors in three dimensions than when limited to only two dimensions.”

Question 2:

Question 2 asks to generate a heap of random three-dimensional vector pairs and for each pair of vectors plot the cross product versus the dot product, and discuss any apparent observations.  Using maple we plugged in the following  codes below: When doing this, it let us find the needed magnitude of the cross products and the dot products and divide these values by the products of the lengths of both vectors, and plot the results against each other

This code was used to generate 100 pairs of three dimensional vectors. (We originally used 10,000 but found using 100 made it easier to see the relationship. With using 10,000 a solid line was created which was each of the 10,000 points.)  It then establishes a formula for the magnitude of the cross product, divided by the product of both vector lengths, and for the dot product divided by the product of both vector lengths. When trying the code for our list we chose to write it differently but still resulting in the same answer. We made it so the list would generate an “a” and a “b”, each being a random vector. We then made it so the list would generate the cross product vs. the dot product as shown above.  These results were then plotted onto a graph.

The graph appears to that of a semicircle/parabola. This is because the magnitude of the cross product over the product of the two vector lengths equals , which reduces to .  Similarly, the dot product over the product of the two vector lengths is , which reduces to .

The unit circle’s appearance reinforces the link of dot products and cross products to trigonometry. When the magnitude of one is at its maximum, the magnitude of the other is always zero.  We can see this at the angles of 0 and 180 degrees, where the magnitude of the dot product is maximized, and at 90 degrees, where the magnitude of the cross product is maximized.

It is also important to take into account what Peter points out in his blog when he says ”the magnitude of the cross product over the the product of the two vector lengths equals , which reduces to .  Similarly, the dot product over the product of the two vector lengths is , which reduces to .”  Thus realizing that  is the independent variable and the  is the dependent variable, the parametric equation and  can be applied.  With this equation being equivalent for that of a circle, except only having positive values on the independent axis.  This could be related to the fact that we found the magnitude of the cross product by the vector norm function; the semicircle (being 1/2 of a unit circle with radius equal to 1) remains strictly in quadrants I and II.

Question 3:

The final question asks us to generate a list of triples of random vectors in 3 dimensions and find out how the volumes of parallelpipeds are distributed. The code is as followed: Maple then records the results and generates a histogram which is shown grouping the volumes of the parallelpipeds produced by the scalar triple product. From the histogram we can coonclude that the greatest frequency of parallelpipeds is with smaller volumes. Parallelpipeds with larger volumes are much less frequent than the smaller. This is because  for a large parallelpiped to be generated, the vectors must have close to 90 degree angles in between them and the vectors themselves must be long.

It is also important to take into account what Peter points out in his blog when he says “If the histogram was divided into smaller bins, we would probably see that there was a frequency peak shortly to the right of the y-axis, with the frequency falling off rapidly at either side, forming a skewed distribution with the tail to the right.  The smallest volumes in the data set would be exceedingly unlikely, as one would require the combination of very small angles between the vectors and/or small vectors, preferably both.  Seeing that the data is being randomly generated, this extreme seems that this combination would be almost as unlikely as the largest of the parallelpipeds. (This would probably still be  more likely; a single vector of length zero, or and angle between the vectors of 0 degrees would yield a zero volume, while a single angle of 90 degrees or vector of maximum length would not guarantee a maximum volume. )”

After concluding the project and analyzing all of the produced data you can clearly see that there is a connection between mathematical formulas and the makeup of vectors in space that they produce.