tag:blogger.com,1999:blog-79828357075937106682019-01-09T01:43:50.676-08:00things ponderedThoughts on science, technology, math, engineering, and societyPhillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-7982835707593710668.post-88580296699260394832019-01-08T21:42:00.000-08:002019-01-08T21:42:59.600-08:00Understanding the physics of a pendulum <div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-yIsO2G1iEd8/Wkl7nllQnYI/AAAAAAAAAwI/0ndhlGurzaUVXoiQsxE4MYi0ynnt7xmOwCLcBGAs/s1600/SimplePendulum1.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="452" data-original-width="750" height="384" src="https://1.bp.blogspot.com/-yIsO2G1iEd8/Wkl7nllQnYI/AAAAAAAAAwI/0ndhlGurzaUVXoiQsxE4MYi0ynnt7xmOwCLcBGAs/s640/SimplePendulum1.png" width="640" /></a></div><br />You've probably seen a pendulum swing, but have you ever thought about why it is that a pendulum swings at all? Further, why is it that a pendulum only swings for some period of time becoming motionless? It turns out, a simple pendulum is a great means through which we can develop intuition about the conservation of energy and the relationship between gravitational potential energy and kinetic energy -- not to mention oscillations.<br /><br />Consider a pendulum of mass $m$, attached to a point of rotation by a massless string of length $L$. Physicists call this a simple pendulum because we are making a simplification by ignoring the mass of the string.<br /><br />The pendulum begins in an initial state:<br /><br /><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-Hd3MLkpcsV8/WkmKMTHbi8I/AAAAAAAAAxU/1-4IcvPAcwM8wxY5G5fVifqcrZrsY1pkgCLcBGAs/s1600/pend1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="213" data-original-width="341" height="249" src="https://3.bp.blogspot.com/-Hd3MLkpcsV8/WkmKMTHbi8I/AAAAAAAAAxU/1-4IcvPAcwM8wxY5G5fVifqcrZrsY1pkgCLcBGAs/s400/pend1.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><div class="separator" style="clear: both; text-align: center;"></div><br />And at some point, for the sake of this analysis, it reaches a state mid-swing, which we'll call the final state:<br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-gJomUiqHKZU/WkmKMY9kJ0I/AAAAAAAAAxY/_DjIFg9Wz2wenuPjIEBg1oAi8AUFtdIKACEwYBhgL/s1600/pend2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="217" data-original-width="451" src="https://2.bp.blogspot.com/-gJomUiqHKZU/WkmKMY9kJ0I/AAAAAAAAAxY/_DjIFg9Wz2wenuPjIEBg1oAi8AUFtdIKACEwYBhgL/s1600/pend2.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><br /><br />Do not be confused by the use of the word "final" here. This is not the final state of the pendulum's swinging motion. It's true final state is zero motion. Rather by final state, we mean the state of the pendulum after some time has elapsed from the initial state. We are free to choose any initial and final state we want. However, as you'll see, I've chosen these states for a very specific reason -- it will make our analysis easier, and in fact, give us more meaningful information. I encourage you to try this by defining different initial and final conditions, however.<br /><br />From the conservation of energy, we know it must be true that:<br />$$E_i = E_f $$<br />Where, $E_i$ is the initial total energy of the pendulum and $E_f$ is its final total energy.<br /><br />We can describe $E_i$ and $E_f$ more specifically, in terms of kinetic energy $K$ and potential energy $U$. Therefore, we can write:<br /><br />$$E_i = K_i + U_i $$<br />$$E_f = K_f + U_f $$<br /><br />Consider the definitions of $U$ and $K$.<br /><br />$$U = mgh $$<br />Where $m$ is mass, $g$ is the gravitational acceleration with value $9.8 \frac{m}{s^2}$, and $h$ is height.<br />$$K = \frac{1}{2}mv^2$$<br />Where $v$ is velocity.<br /><br />Now, here's the trick. When using the conservation of energy to analyze a phenomenon, it can be in our best interest to choose our initial and final conditions such that they make the math simpler. If we look carefully at the initial and final conditions, we'll see that $K_i =0$ because the initial velocity has been chosen such that it is at a point in time when the pendulum has zero velocity. We don't know the exact time when this is true, but we know from watching a real pendulum that it will swing up to some max height, at which point it will begin to swing back down. For some instantaneous time, its velocity must be zero. Meanwhile, we also know that $U_f=0$ because we've defined the final condition at a point in time when the pendulum has zero height. As a result, the conservation of energy says:<br /><br />$$U_i = K_f$$<br />$$mgh = \frac{1}{2}mv^2$$<br /><br />The next problem that must be solved is determining the height $h$ of the pendulum.<br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-h0aAsn8UNp4/WkmKMW4i9oI/AAAAAAAAAxc/D-6WIN338lAC_Cs7NY_t41hwBZaB0iMLgCEwYBhgL/s1600/endtest.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="210" data-original-width="405" src="https://2.bp.blogspot.com/-h0aAsn8UNp4/WkmKMW4i9oI/AAAAAAAAAxc/D-6WIN338lAC_Cs7NY_t41hwBZaB0iMLgCEwYBhgL/s1600/endtest.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><br /></div><br /><br />From this diagram, we know that $L = h + z$ and that $z = Lcos\theta$. We can write the height $h$ as:<br /><br />$$h = L(1-cos\theta)$$<br /><br />Referring back to our conservation of energy statement, it follows that:<br /><br />$$gL(1-cos\theta) = \frac{1}{2}v^2$$<br /><br />Consider the velocity $v$. This is a special type of velocity because of the semicircular motion the pendulum traces out as it swings.<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-JIkH0L8a5UQ/WkmKMg-izVI/AAAAAAAAAxg/As7YnGZ6FPw8wDyiiVNTFUicSwfoCAJqwCEwYBhgL/s1600/pend_4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="244" data-original-width="455" src="https://3.bp.blogspot.com/-JIkH0L8a5UQ/WkmKMg-izVI/AAAAAAAAAxg/As7YnGZ6FPw8wDyiiVNTFUicSwfoCAJqwCEwYBhgL/s1600/pend_4.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><br /><br />Indeed, this velocity is a tangential velocity $v_t$. And we know that $v_t = \omega R$, where $R$ is the radius of the circle and $\omega$ is the angular velocity. In this case, $R = L$.<br /><br />We can re-write our equation as:<br /><br />$$gL(1-cos\theta) = \frac{1}{2}\omega^2L^2$$<br /><br />Solve for $\omega$:<br /><br />$$\omega = \sqrt{\frac{2g}{L}(1-cos\theta)}$$<br /><br />If we remember that $\omega = 2\pi f$, then we can develop some intuition for what this equation means. It tells us that the motion of the pendulum has a frequency $f$ associated with it. This knowledge combined with our understanding of the relationship between kinetic energy and potential energy suggests that its motion has an oscillation-like nature. Remember: the total energy of a particle consists of kinetic and potential energy. When one increases, the other decreases, and vice versa.<br /><br />The pendulum swings back and forth, with each swing its gravitational potential energy converted to kinetic energy. In a perfect vacuum, this swinging motion would continue forever. However, on earth -- and in any room with air -- some of the kinetic energy is lost to the surrounding air, as the pendulum collides with air molecules and transfers some of its energy to those molecules. This explains why the pendulum gradually loses height after each period of swinging.<br /><br />Imagine our pendulum swinging back and forth. At the beginning of each fall, the pendulum has some amount of potential energy; it is converted entirely to kinetic energy and moves downward. But not all of that kinetic energy is returned to potential energy because it is transferred in those collisions with air molecules. As a result, at the end of its upswing, the pendulum reaches a lower height. Some of the motion of the pendulum has been given to the air molecules, so the pendulum is less energetic overall. This leaking of energy to the surrounding air continues with each period of swinging. Given enough time, the pendulum runs out of energy and becomes motionless.<br /><br />Without considering the surrounding air, it looks like the energy has disappeared and the conservation of energy has been violated. But it is not so: the energy has been conserved -- it has been transferred outside of the system of the swinging pendulum.<br /><br />Perhaps a question has entered your mind: how does the height that the pendulum is dropped from relate to its speed and kinetic energy?<br /><br />Using just the first two terms of the Taylor series expansion, we can approximate $cos\theta$ as:<br /><br />$$ cos\theta = 1 - \frac{\theta^2}{2} $$<br /><br />Remember, this is an approximation, as we are removing many of the terms from the Taylor series expansion. However, if we use this value and plug it into our equation, we find that:<br /><br />$$\omega = \theta \sqrt{\frac{g}{L}}$$<br /><br />The angle $\theta$ is directly related to the height $h$. The greater its value, the higher the pendulum starts its swinging at. This means the greater height from which we drop the pendulum, the greater its frequency of swinging. Further, because the angular velocity $v_t = \omega R$, the initial kinetic energy of the pendulum is greater since it depends on velocity.<br /><br />As you can see, the pendulum is a fantastic model for understanding the conservation of energy, the relationship between gravitational potential energy and kinetic energy, and oscillation behavior.<br /><br /><span style="font-size: x-small;">Image courtesy <a href="http://shareefjackson.com/mathlooksgood/sampleproblems/2016/12/20/physics-motion-of-a-simple-pendulum" target="_blank">Shareef Jackson</a></span>Phillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.com0tag:blogger.com,1999:blog-7982835707593710668.post-25794736735999711322018-01-01T14:38:00.000-08:002018-01-06T13:03:18.870-08:00How high up is outer space?<style> .post-timestamp { display: none; } </style> <br /><div class="MsoNormal"><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-A68u--2sC1s/Wkq6i8o7yoI/AAAAAAAAAzc/J7Eio9v6qWY2FN7TbDiXirHceKCM7GDSgCLcBGAs/s1600/strato_1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="168" data-original-width="645" src="https://4.bp.blogspot.com/-A68u--2sC1s/Wkq6i8o7yoI/AAAAAAAAAzc/J7Eio9v6qWY2FN7TbDiXirHceKCM7GDSgCLcBGAs/s1600/strato_1.jpg" /></a></div><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";"><br /></span><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";"><br /></span><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">The <a href="https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line" target="_blank">Kármán line</a> is the altitude of the boundary between earth’s atmosphere and outer space. This $100$ km or $328, 084$ ft. The value comes from Fédération Aéronautique Internationale, and it’s the same value that NASA uses to define the boundary between our planet’s atmosphere and outer space. <o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">If you’re like me, the highest you've ever been from sea level is around $30,000$ ft to $40,000$ ft, which is the range of altitudes at which most commercial airliners cruise at. <o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">For some context, the tallest mountain on earth is Mt. Everest, with a peak at $29,029$ ft, measured with respect to sea level.<o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">The boundary between earth and outer space, at $328,084$ ft, is roughly 11 times higher than Mt. Everest, as well as the highest up you’ve probably ever been. Try to picture that for a <span style="mso-spacerun: yes;"> </span>moment. <o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><br /></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="MsoNormal"><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-UdocWBImdmU/Wkq5MgUHYNI/AAAAAAAAAzU/_ODAt5eNccsboIp6DZowNAQ_1J6HqvnXACLcBGAs/s1600/320px-Atmosphere_layers-en.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="273" src="https://2.bp.blogspot.com/-UdocWBImdmU/Wkq5MgUHYNI/AAAAAAAAAzU/_ODAt5eNccsboIp6DZowNAQ_1J6HqvnXACLcBGAs/s1600/320px-Atmosphere_layers-en.png" /></a></div><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";"><br /></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="font-family: , serif;">If this picture doesn’t give you some sense of awe, perhaps another way of thinking about it will. Let's think about how much kinetic energy is gained due to the force of gravity at this altitude. After all, if you were to jump from such a height, the gravitational field of earth would impart onto you energy in the form of motion. Gravity would accelerate you to some speed. So, a natural question that arises is, if you were to fall back to earth from the Karman line, how fast would you be moving when you hit the ground? We'll calculate an upper bound on this speed. That is, what is the fastest you would be moving when you hit the ground if you experienced zero air resistance. </span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">The acceleration of gravity on earth is $g = 9.8 \frac{meters}{seconds^2}$ or $ 21.9\frac{miles}{hour^2}$. <o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">Using one of the kinematic equations, it</span><span style="color: black; mso-ascii-font-family: -webkit-standard; mso-bidi-font-family: "Times New Roman"; mso-hansi-font-family: -webkit-standard;">’</span><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">s possible to determine the speed at which you will hit the ground.<o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">$$d = vt + \frac{1}{2}at^2$$<o:p></o:p></span></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";"><br /></span></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">In case you want to work it out yourself, $d$ is distance, $t$ time, $a$ acceleration (in this case $g$).</span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">First, you must solve for the time it takes to fall a distance of $328,084$ ft. It would take you $143$ seconds or about $2.5$ minutes. For each hour, you</span><span style="color: black; mso-ascii-font-family: -webkit-standard; mso-bidi-font-family: "Times New Roman"; mso-hansi-font-family: -webkit-standard;">’</span><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">d accelerate by $21.9\frac{miles}{hour}$ because of gravity. <o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">By the time you hit the ground, you</span><span style="color: black; mso-ascii-font-family: -webkit-standard; mso-bidi-font-family: "Times New Roman"; mso-hansi-font-family: -webkit-standard;">’</span><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">d be moving at a speed of $3,131 \frac{miles}{hour}$. <o:p></o:p></span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">Again, keep in mind this back-of-the-envelope calculation ignores the effect of drag -- air resistance -- experienced by a falling person. So, you will actually be moving much slower than this, but this gives you some sense of the amount of energy gained from gravity at this altitude. </span></div><div class="MsoNormal"><br /></div><div class="MsoNormal"><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">So, how high up is outer space? High enough up that if you jumped from there, and didn't experience air resistance, you</span><span style="color: black; mso-ascii-font-family: -webkit-standard; mso-bidi-font-family: "Times New Roman"; mso-hansi-font-family: -webkit-standard;">’</span><span style="color: black; font-family: "-webkit-standard" , "serif"; mso-bidi-font-family: "Times New Roman";">d reach a speed of $3,131 \frac{miles}{hour}$. That</span><span style="color: black; mso-ascii-font-family: -webkit-standard; mso-bidi-font-family: "Times New Roman"; mso-hansi-font-family: -webkit-standard;">’</span><span style="font-family: , serif;">s faster than the speed of sound ($767 \frac{miles}{hour}$). If you kept moving at that speed, you could travel from Los Angeles to New York in less than an hour. </span></div><div class="MsoNormal"><span style="font-family: , serif;"><br /></span></div><div class="MsoNormal"><span style="font-family: , serif;">Of course, once you factor in air resistance, you'll find that you reach a terminal velocity. This is a calculation best left for another essay, but it's around half the speed predicted by the model of no air resistance. For some context, competitive skydiver </span><a href="https://en.wikipedia.org/wiki/Felix_Baumgartner" target="_blank">Felix Baumgartner</a> <span style="font-family: , serif;">holds the record for terminal velocity reached via skydiving. He reached a velocity of $834 \frac{miles}{hour}$ by jumping from </span>128,100 ft, about 40% of the altitude of the Karman line.</div>Phillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.com0tag:blogger.com,1999:blog-7982835707593710668.post-45030206352326613792017-12-28T13:38:00.001-08:002018-01-01T18:09:33.826-08:00Human Beings and the Next Revolution: Automation<br /><figure class="graf graf--figure" name="1503"><img class="graf-image" data-height="2709" data-image-id="1*OP6HocKn6PSdZDIs1bdWxg.jpeg" data-width="2500" src="https://cdn-images-1.medium.com/max/800/1*OP6HocKn6PSdZDIs1bdWxg.jpeg" /></figure><br /><div class="MsoNormal" style="line-height: normal; mso-margin-bottom-alt: auto; mso-margin-top-alt: auto;">“Can you believe people actually used to work?”<br /><br />Some day in the not-too-distance future, people will say words like these.<br /><br />There is a moment in humanity’s future that is approaching. And in this moment in time, people will no longer have careers or work full-time jobs.<br /><br />A common thread of argument is that “there will always be work for people to do.” To some degree, this may be true. But such optimism seems misguided if one can envision the potential for intelligent computers to surpass the abilities of mankind in all but few domains. Sure, a niche of humanity will work. But make no mistake, the majority will be unemployed, replaced by machines more efficient than they could ever be.<br /><br />The problem is that it’s hard to predict what will happen with artificial intelligence, and thus hard to predict how long this moment will last. There are at least two extreme outcomes to consider, and an infinite number of possibilities in between.<br /><br />We do not know how rapidly artificial intelligence will progress. On one hand, there’s a chance that the time between the creation of the first machine with human-level general intelligence and the singularity will be infinitesimally small. That is, is the rate of advancement in artificial intelligence will be so great that shortly after developing such an AI, that very same AI will begin to make its own advancements with impeccable speed and success. Such an AI might modify itself, leading to a superintelligent being that wrestles control from and enslaves mankind. Clearly, such an outcome is more concerning than the unemployment rate.<br /><br />Or perhaps we just merge with machines and live happily ever after.<br /><br />But another real possibility is that we don’t reach such a level in artificial intelligence or biotechnology for a long time. And during that lengthy period of time, we continue to improve current methods of machine learning, computer vision, robotics, and computer reasoning to a point where computers and robots are a cheaper and superior option for most work-related tasks. In some ways, although this scenario is less horror film-esque than the one described above, it’s just as disconcerting because it raises a question: what value will human beings have in such a world?<br /><br />Simply put, we will need to radically change the way we perceive human value, or we could find ourselves living in a very dystopian reality.<br /><br />Part of this is dealing with the social challenges that await us. The most paramount of these will be establishing systems that can provide resources to the masses who can no longer reasonably be expected to support themselves. Fortunately, companies like YC are experimenting with approaches to basic income. There is a lot of fear about communism and socialism, but devout capitalists should try to understand that mass poverty and homelessness are the likely alternative. Fortunately, advances in technology also make it more likely that we can provide food, housing, energy, and transportation to people. It is unlikely that technology will be the bottleneck. Rather, poorly designed policies and electing the wrong types of people to positions of power will cause the most friction here.<br /><br />This coming era can be a high point for humanity. There is a real opportunity to reduce human suffering and free people from something they’ve been chained to: work. Freed from the slavery of having to earn an income, we’d be living in a truly new age — potentially a utopian one. The results of such a thing seem impossible to predict completely, but I look forward to seeing what people can do without such a burden.<br /><br />I just hope we remain aware of the path we are taking. Now is not the time to misstep.</div>Phillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.com0tag:blogger.com,1999:blog-7982835707593710668.post-36466662892499178842017-09-09T15:07:00.002-07:002017-12-28T13:42:45.903-08:00What's the difference between simple and elementary regions?<img src="https://qph.ec.quoracdn.net/main-qimg-e0b799d2a6e7e12e692aa78ebe83cf11-c" /><br />Simple regions are a topic that comes up in vector calculus.<br /><br />First, let's understand the difference between x-simple and y-simple regions. Visually, the Type 1 region seen here is an example of a y-simple region and the Type 2 region is an example of an x-simple region. See any textbook on vector calculus for a formal definition.<br /><br />But intuitively, a y-simple region is any region of space that can be bounded between two functions y=g1(x) and y=g2(x). If the region requires more than two such functions to bound it, then it’s not y-simple. The same idea applies to x-simple regions, but the two functions are x=g1(y) and x=g2(y).<br /><br />The idea, really, is a region is simple if you only need two functions to define its boundary.<br />Now, here's the easy part.<br /><br />A region is said to be <b>“simple”</b> if it is both y-simple <b>and</b> x-simple.<br /><br />A region is said to be <b>“elementary” </b>if it is y-simple <b>or</b> x-simple.<br /><br />Bonus idea: Do you see why this is useful for integration? Integration is defined from point A to point B. If a region is simple it means you can write limits of integration easily; the functions g1 and g2 are the limits of integration.<br /><br /><a href="http://mathonline.wikidot.com/evaluating-double-integrals-over-general-domains" target="_blank">Source of image used</a>Phillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.com1tag:blogger.com,1999:blog-7982835707593710668.post-41938341737668500362017-09-01T10:58:00.004-07:002017-09-11T10:59:54.080-07:00Complex numbers: the relationship between rectangular form, exponential form, angular notation<div class="separator" style="clear: both; text-align: left;">Learning complex analysis can be a bit daunting. Not only is this topic filled with abstract ideas, there are a number of equivalent ways of talking about the same idea. For example, there are at least three different forms one can use to represent a complex number. In this video, I explain these different forms, the relationships, and when to use each.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen="" class="YOUTUBE-iframe-video" data-thumbnail-src="https://i.ytimg.com/vi/i_Aob-0HOcs/0.jpg" frameborder="0" height="266" src="https://www.youtube.com/embed/i_Aob-0HOcs?feature=player_embedded" width="320"></iframe></div><br />Phillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.com2tag:blogger.com,1999:blog-7982835707593710668.post-75377080220223769302017-07-03T23:40:00.001-07:002017-09-11T11:00:00.782-07:00Calculus explained in a way that anyone can understand<div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-bG1kLfPIo2o/WVwsCD3NYrI/AAAAAAAAAqQ/AxcpLVkAPN4dwwHQtgfCa-ylc29-_7AhQCLcBGAs/s1600/pic_res_header.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="198" data-original-width="614" src="https://2.bp.blogspot.com/-bG1kLfPIo2o/WVwsCD3NYrI/AAAAAAAAAqQ/AxcpLVkAPN4dwwHQtgfCa-ylc29-_7AhQCLcBGAs/s1600/pic_res_header.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><a name='more'></a><span style="font-family: "times" , "times new roman" , serif;">Calculus is an essential mathematical tool used by scientists in many fields -- from physics to artificial intelligence. So, what is it?</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">Calculus is the study of change.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">If the field can be reduced to its two most fundamental ideas, these are the concepts of the derivative and integral.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><b><span style="font-family: "times" , "times new roman" , serif;">Differentiation: The Derivative</span></b><br /><span style="font-family: "times" , "times new roman" , serif;">Differentiation is the process of computing the derivative of a function. </span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">The derivative of a function refers to its slope. It is the rate of change of the function.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><br /><div style="text-align: center;"><span style="font-family: "times" , "times new roman" , serif;"><img id="equationview" name="equationview" src="https://latex.codecogs.com/svg.latex?Derivative%20%3D%20%5Cfrac%7BRise%7D%7BRun%7D" style="font-family: Arial, Helvetica, sans-serif; font-size: 12px; margin: 10px; text-align: center;" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></span></div><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">Further study of calculus involves learning techniques for computing the derivative of different functions, such as y = <span style="font-size: 18px;">x² </span>or y = sin ø.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">Mathematicians typically use the following notation to mean "take the derivative of y with respect to x":</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><br /><div style="text-align: center;"><span style="font-family: "times" , "times new roman" , serif;"><img id="equationview" name="equationview" src="https://latex.codecogs.com/svg.latex?%5Cinline%20%5CLARGE%20%5Cfrac%7B%5Cmathrm%7Bdy%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D" style="font-family: Arial, Helvetica, sans-serif; font-size: 12px; margin: 10px; text-align: center;" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></span></div><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">One can think of "dy" and "dx" standing for "difference in y" and "difference in x," and so this notation isn't all that confusing -- it means determine the slope of the function y!</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">The derivative of a function tells you something about the function at that point. For example, a large value means the slope of the function at that point is steep -- the function is changing quickly -- while a lower value suggests it is less steep -- it is changing less quickly.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-zw4gf5moMMA/WVwXz6YfFtI/AAAAAAAAApo/4NfXXeFXaQsD6Tf3Mee1B-SWooemvkNkgCLcBGAs/s1600/steepness.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="304" data-original-width="446" src="https://1.bp.blogspot.com/-zw4gf5moMMA/WVwXz6YfFtI/AAAAAAAAApo/4NfXXeFXaQsD6Tf3Mee1B-SWooemvkNkgCLcBGAs/s1600/steepness.png" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><span style="font-family: "times" , "times new roman" , serif;">The absolute value of the derivative at the position of the blue dot will be larger than the derivative at the red dot because the function is steeper at the blue dot</span></td></tr></tbody></table><div style="text-align: center;"><br /></div><span style="font-family: "times" , "times new roman" , serif;">Similarly, the sign of the derivative tells you whether the function at that point is increasing or decreasing. Positive means it's increasing, and negative means it's decreasing.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://4.bp.blogspot.com/-pek7s71tgTM/WVwYpcLVPDI/AAAAAAAAAps/dSmUoXvVZ4s7cedksuWpV_gnWh2Xwm6mwCLcBGAs/s1600/derivativesign.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="304" data-original-width="446" src="https://4.bp.blogspot.com/-pek7s71tgTM/WVwYpcLVPDI/AAAAAAAAAps/dSmUoXvVZ4s7cedksuWpV_gnWh2Xwm6mwCLcBGAs/s1600/derivativesign.png" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The sign of the derivative at the red dot will be negative because the function is decreasing, while the sign of the derivative at the blue dot will be positive because the function is increasing there</td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><i style="margin-left: 1em; margin-right: 1em;"></i></div><div class="separator" style="clear: both; text-align: center;"><br /></div><b><span style="font-family: "times" , "times new roman" , serif;">Integration: The Integral</span></b><br /><span style="font-family: "times" , "times new roman" , serif;">Integration is the process of computing the integral of a function.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">The integral is the area under the curve described by a function.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">The symbol and notation for an integral looks like this:</span><br /><br /><div style="text-align: center;"><span style="font-family: "times" , "times new roman" , serif;"><br /></span></div><div style="text-align: center;"><img id="equationview" name="equationview" src="https://latex.codecogs.com/svg.latex?%5Cint_%7Ba%7D%5E%7Bb%7Dsinx%20%5C%20dx" style="font-family: Arial, Helvetica, sans-serif; font-size: 12px; margin: 10px; text-align: center;" title="This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program." /></div><br />The "S" shaped symbol stands for sum. Remember, you can think of "dx" as being some piece of the x axis. And so, we can think of this notation as saying, "take the sum of all pieces of the function sin x from the point a to the point b on the x axis."<br /><br />Integration is simply the summing up of all the area under a curve over a given range, thus it represents the area under the curve.<br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-Sop4D-fqssY/WVwbCBbY5mI/AAAAAAAAApw/Wl3k2cTwYgE5vSFNlrh1WTpeVhjez3HbACLcBGAs/s1600/areaunder.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="304" data-original-width="446" src="https://1.bp.blogspot.com/-Sop4D-fqssY/WVwbCBbY5mI/AAAAAAAAApw/Wl3k2cTwYgE5vSFNlrh1WTpeVhjez3HbACLcBGAs/s1600/areaunder.png" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The red area under the sin wave curve from point a to b can be calculated using integration<br /><br /><br /></td></tr></tbody></table><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">Compared to differentiation, integration is often a more challenging process, sometimes requiring quite a bit of creativity. </span><br /><b><span style="font-family: "times" , "times new roman" , serif;"><br /></span></b><b><span style="font-family: "times" , "times new roman" , serif;">The Relationship Between Derivatives and Integrals</span></b><br /><span style="font-family: "times" , "times new roman" , serif;">Perhaps the most magical thing about calculus is the relationship between derivatives and integrals.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">Turns out, these two mathematical concepts are related in that they are inverses of one another. Indeed, just like one can use division to "undo" multiplication, one uses integration to "undo" differentiation of a function.</span><br /><span style="font-family: "times" , "times new roman" , serif;"><br /></span><span style="font-family: "times" , "times new roman" , serif;">The tools of differentiation (rate of change) and integration (area under the curve) enable us to quantitively describe phenomenon in science, engineering, economics, and more. These ideas were invented/discovered by </span>Isaac Newton<span style="font-family: "times" , "times new roman" , serif;"> and </span>Gottfried Leibniz, who we can thank for enabling much of the progress of these fields, which would be shadows of themselves without the ideas of calculus.Phillip Levinhttp://www.blogger.com/profile/14443277218388624680noreply@blogger.com0