# Introduction to Partial Derivatives

In previous lessons, we learned how the derivative $$f'(x)$$ gives us the steepness at each point along a function $$f(x)$$. In this lesson, we'll discuss how using the concept of a partial derivative we can find the steepness at each point along a surface $$z=f(x,y)$$. To find the partial derivative we treat one of the variables as a constant and then take the ordinary derivative of $$f(x,y)$$. Using this concept, we can specify how steep a surface $$f(x,y)$$ is along the $$x$$ direction and along the $$y$$ direction at each point along the surface. In other words, for every point along the surface, there is a steepness of the surface associated with both the $$x$$ and the $$y$$ directions at that point.

# Quasars

In this article, we'll discuss the history behind the discovery of one of the most exotic and remarkable objects ever discovered in science—a quasar. Quasars consist of an accretion disk made of ultra-hot gas and dust surrounding a supermassive black hole; two immense strands of super-heated plasma extended in a direction perpendicular to the disk for millions of the light-years through the mostly empty void of intergalactic space. Quasars were ubiquitous in the early, young universe and were located at the center of most galaxies; but today, most quasars are gone because all of the matter comprising the accretion disk eventually got gobbled up by the super massive black hole. For example, our home galaxy—the Milky Way—now only has the left-over remnant of a quasar at its center—a super massive black hole.

# Volume of an Oblate Spheroid

In this lesson, we'll discuss how by using the concept of a definite integral one can calculate the volume of something called an oblate spheroid. An oblate spheroid is essentially just a sphere which is compressed or stretched along one of its dimensions while leaving its other two dimensions unchanged. For example, the Earth is technically not a sphere—it is an oblate spheroid. To find the volume of an oblate spheroid, we'll start out by finding the volume of a paraboloid . (If you cut an oblate spheroid in half, the two left over pieces would be paraboloids.) To do this, we'll draw an $$n$$ number of cylindrical shells inside of the paraboloid; by taking the Riemann sum of the volume of each cylindrical shell, we can obtain an estimate of the volume enclosed inside of the paraboloid. If we then take the limit of this sum as the number of cylindrical shells approaches infinity and their volumes approach zero, we'll obtain a definite integral which gives the exact volume inside of the paraboloid. After computing this definite integral, we'll multiply the result by two to get the volume of the oblate spheroid.

# Our Future as Cyborgs

Undoubtedly, if our wisdom and foresight rises to be commensurate with our science and technology, the future of humanity in the 21st century is Utopian. We will have re-engineered the surface of the Earth with cities, transportation and communication systems, and new energy infrastructure which are designed and constructed to have optimal efficiency according to known science. We will have also spread across much of the solar system and, perhaps, have sent robotic spacecraft off to the nearest star system, Alpha Centauri. Aside from re-engineering the Earth and other worlds in our solar system, we will also likely re-engineer ourselves as we merge with our technology and machines. This will be the subject of discussion in this article.

# Gravitational Force Exerted by a Disk

To find the gravitational force exerted by a disk on a particle a height $$h$$ above the center of the disk, we must use Newton's law of gravity and the concept of a definite integral.

A Shkadov thruster is a type of megastructure which involves constructing a gargantuan orbital mirror next to a star. In this lesson, we'll start off by discussing how such a stellar engine works. The orbital mirror is placed in a position next to a star where it acts as a statite: that is, the star's gravity acting on the mirror is canceled out by the star's radiation pressure acting on the mirror. This allows the mirror to stay in a position that is stationary relative to the star's surface. The mirror bounces some of the star's light back at itself; when that reflected light collides with the star, it exerts a thrust on the star which causes it to accelerate and move. This will bring us to the second main focus of this lesson: namely, what are the possible uses of a Shkadov thruster? As we'll discuss, since a Shkadov thruster can move the star and since all of the planets, moons, comets, and asteroids in the star system is gravitationally bound to the star, not only does the star move but the entire solar system moves away also. In the distance future, our Sun will eventually die. But we might be able to use a Shkadov thruster to move the Earth to another solar system, but this would take many millions of years.

# Derivation of Snell's Law

The law of reflection had been well known as early as the first century; but it took longer than another millennium to discover Snell's law, the law of refraction. The law of reflection was readily observable and could be easily determined by making measurements; this law states that if a light ray strikes a surface at an angle $$θ_i$$ relative to the normal and gets reflected off of the surface, it will be reflected at an angle $$θ_r$$ relative to the normal such that $$θ_i=θ_r$$. The law of refraction, however, is a little less obvious and it required calculus to prove. The mathematician Pierre de Fermat postulated the principle of least time: that light travels along the path which gets it from one place to another such that the time $$t$$ required to traverse that path is shorter than the time required to take any other path. In this lesson, we shall use this principle to derive Snell's law.

# Optimization Problem

If $$(x,y)$$ represents any point on the circle, if $$P$$ is a point fixed at the coordinate point $$(4,0)$$, and if $$d$$ represents the distance between those two points then, by using only calculus, we can find the point $$(x,y)$$ on the circle associated with the minimum distance $$d$$.

# Maximizing the Area of a Rectangle

Given that the perimeter $$2x+2y$$ of any arbitrary rectangle must be constant, we can use calculus to find that particular rectangle with the greatest area. The solution to this problem has practical applications. For example, suppose that someone had only 30 meters of fencing to enclose their backyard and they wanted to know what fencing layout would maximize the size and total area of their backyard. Using calculus, we can answer such questions.

# Finding the Minima and Maxima of a Function

Calculus—specifically, derivatives—can be used to find the values of $$x$$ at which the function $$f(x)$$ is at either a minimum value or a maximum value. For example, suppose that we let $$x$$ denote the horizontal distance away from the beginning of a hiking trail near a mountain and we let $$f(x)$$ denote the altitude of the mountainous terrain at each $$x$$ value. $$f(x)$$ reaches a minimum value when the function "flattens out"—that is, when $$f'(x)$$ becomes equal to zero. These particular values of $$x$$ are associated with the bottom and top of the mountain. The condition that $$f'(x)=0$$ only tells us that $$f(x)$$ is at either a minimum or a maximum. To determine whether or not $$f(x)$$ is at a minimum or a maximum, we must use the concept of the second derivative. This will be the topic of discussion in this lesson.