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# derivative of a constant raised to a variable

From Wikipedia

Constant function

In mathematics, a constant function is a function whose values do not vary and thus are constant. For example, if we have the function f(x) = 4, then f is constant since f maps any value to 4. More formally, a function f&nbsp;:&nbsp;Aâ†’ B is a constant function if f(x) = f(y) for all x and y in A.

Every empty function is constant, vacuously, since there are no x and y in A for which f(x) and f(y) are different when A is the empty set. Some find it more convenient, however, to define constant function so as to exclude empty functions.

In the context of polynomial functions, a non-zero constant function is called a polynomial of degree zero.

## Properties

Constant functions can be characterized with respect to function composition in two ways.

The following are equivalent:

1. f&nbsp;:&nbsp;Aâ†’ B is a constant function.
2. For all functions g, h&nbsp;:&nbsp;Câ†’ A, fog = foh, (where "o" denotes function composition).
3. The composition of f with any other function is also a constant function.

The first characterization of constant functions given above, is taken as the motivating and defining property for the more general notion of constant morphism in category theory.

In contexts where it is defined, the derivative of a function measures how that function varies with respect to the variation of some argument. It follows that, since a constant function does not vary, its derivative, where defined, will be zero. Thus for example:

For functions between preordered sets, constant functions are both order-preserving and order-reversing; conversely, if f is both order-preserving and order-reversing, and if the domain of f is a lattice, then f must be constant.

Other properties of constant functions include:

A function on a connected set is locally constant if and only if it is constant.

Variable star

A star is classified as variable if its apparent magnitude as seen from Earth changes over time, whether the changes are due to variations in the star's actual luminosity, or to variations in the amount of the star's light that is blocked from reaching Earth. Many, possibly most, stars have at least some variation in luminosity: the energy output of our Sun, for example, varies by about 0.1% over an 11 year solar cycle, equivalent to a change of one thousandth of a magnitude.

It is convenient to classify variable stars as belonging to one of two types:

• Intrinsic variables, whose luminosity actually changes; for example, because the star periodically swells and shrinks.
• Extrinsic variables, whose apparent changes in brightness are due to changes in the amount of their light that can reach Earth; for example, because the star has an orbiting companion that sometimes eclipses it.

## Discovery

The first variable star was identified in 1638 when Johannes Holwarda noticed that Omicron Ceti (later named Mira) pulsated in a cycle taking 11 months; the star had previously been described as a nova by David Fabricius in 1596. This discovery, combined with supernovae observed in 1572 and 1604, proved that the starry sky was not eternally invariable as Aristotle and other ancient philosophers had taught. In this way, the discovery of variable stars contributed to the astronomical revolution of the sixteenth and early seventeenth centuries.

The second variable star to be described was the eclipsing variable Algol, by Geminiano Montanari in 1669; John Goodricke gave the correct explanation of its variability in 1784. Chi Cygni was identified in 1686 by G. Kirch, then R Hydrae in 1704 by G. D. Maraldi. By 1786 ten variable stars were known. John Goodricke himself discovered Delta Cephei and Beta Lyrae. Since 1850 the number of known variable stars has increased rapidly, especially after 1890 when it became possible to identify variable stars by means of photography.

The latest edition of the General Catalogue of Variable Stars (2008) lists more than 46,000 variable stars in our own galaxy, as well as 10,000 in other galaxies, and over 10,000 'suspected' variables.

## Detecting variability

The most common kinds of variability involve changes in brightness, but other types of variability also occur, in particular changes in the spectrum. By combining light curve data with observed spectral changes, astronomers are often able to explain why a particular star is variable.

### Variable star observations

Variable stars are generally analysed using photometry, spectrophotometry and spectroscopy. Measurements of their changes in brightness can be plotted to produce light curves. For regular variables, the period of variation and its amplitude can be very well established; for many variable stars, though, these quantities may vary slowly over time, or even from one period to the next. Peak brightnesses in the light curve are known as maxima, while troughs are known as minima.

Amateur astronomers can do useful scientific study of variable stars by visually comparing the star with other stars within the same telescopic field of view of which the magnitudes are known and constant. By estimating the variable's magnitude and noting the time of observation a visual lightcurve can be constructed. The American Association of Variable Star Observers collects such observations from participants around the world and shares the data with the scientific community.

From the light curve the following data are derived:

• are the brightness variations periodical, semiperiodical, irregular, or unique?
• what is the period of the brightness fluctuations?
• what is the shape of the light curve (symmetrical or not, angular or smoothly varying, does each cycle have only one or more than one minima, etcetera)?

From the spectrum the following data are derived:

• what kind of star is it: what is its temperature, its luminosity class (dwarf star, giant star, supergiant, etc.)?
• is it a single star, or a binary? (the combined spectrum of a binary star may show elements from the spectra of each of the member stars)
• does the spectrum change with time? (for example, the star may turn hotter and cooler periodically)
• changes in brightness may depend strongly on the part of the spectrum that is observed (for example, large variations in visible light but hardly any changes in the infrared)
• if the wavelengths of spectral lines are shifted this points to movements (for example, a periodical swelling and shrinking of the star, or its rotation, or an expanding gas shell) (Doppler effect)
• strong magnetic fields on the star betray themselves in the spectrum
• abnormal emission or absorption lines may be indication of a hot stellar atmosphere, or gas clouds surrounding the star.

In very few cases it is possible to make pictures of a stellar disk. These may show darker spots on its surface.

### Interpretation of obs

Motion graphs and derivatives

In mechanics, the derivative of the position vs. timegraph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the y-axis and time on the x-axis, the slope of the curve is given by:

v = \frac{\Delta y}{\Delta x} = \frac{\Delta s}{\Delta t}.

Here s is the position of the object, and t is the time. Therefore, the slope of the curve gives the change in position (in metres) divided by the change in time (in seconds), which is the definition of the average velocity (in meters per second (\begin{matrix} \frac{m}{s} \end{matrix})) for that interval of time on the graph. If this interval is made to be infinitesimally small, such that {\Delta s} becomes {ds} and {\Delta t} becomes {dt}, the result is the instantaneous velocity at time t, or the derivative of the position with respect to time.

A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in y over change in x:

a = \frac{\Delta y}{\Delta x} = \frac{\Delta v}{\Delta t}.

Where v is the velocity, measured in \begin{matrix} \frac{m}{s} \end{matrix}, and t is the time measured in seconds. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives \begin{matrix} \frac{dv}{dt} \end{matrix}, the instantaneous acceleration at time t, or the derivative of the velocity with respect to time (or the second derivative of the position with respect to time). The units of this slope or derivative are in meters per second per second (\begin{matrix} \frac{m}{s^2} \end{matrix}, usually termed "meters per second-squared"), and so, therefore, is the acceleration.

Since the acceleration of the object is the second derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the seconds cancel out and only meters remain. \begin{matrix} \frac{m}{s} \end{matrix}s = m.)

The same multiplication rule holds true for acceleration vs. time graphs. When (\begin{matrix} \frac{m}{s^2} \end{matrix}) is multiplied by time (s), velocity is obtained. (\begin{matrix} \frac{m}{s^2} \end{matrix}s = \begin{matrix} \frac{m}{s} \end{matrix}).

## Variable rates of change

The expressions given above apply only when the rate of change is constant or when only the average (mean) rate of change is required. If the velocity or positions change non-linearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. Differentiation reduces the time-spans used above to be extremely small and gives a velocity or acceleration at each point on the graph rather than between a start and end point. The derivative forms of the above equations are

v = \frac{ds}{dt},
a = \frac{dv}{dt}.

Since acceleration differentiates the expression involving position, it can be rewritten as a second derivative with respect to position:

a = \frac{d^2 s}{dt^2}.

Since, for the purposes of mechanics such as this, integration is the opposite of differentiation, it is also possible to express position as a function of velocity and velocity as a function of acceleration. The process of determining the area under the curve, as described above, can give the displacement and change in velocity over particular time intervals by using definite integrals:

s(t_2)-s(t_1) = \int_{t_1}^{t_2}{v}\, dt,
v(t_2)-v(t_1) = \int_{t_1}^{t_2}{a}\, dt.

Question:Think about it. How can you define a number to a single moment in time, if that time has passed so instantaneously, that it no longer exists. Break down a second into a nanosecond, and even smaller, as low as it can go, but you can keep going, because time does not stop, it is continuous. Surely it becomes a non-existent value, which means time does not actually exist, except in a meaningless form? Your thoughts please.

Answers:this is simply solved. CALCULUS! the mathematics of infinitesimal intervals! when you integrate you are literally taking each value at its instantaneous position (in time possibly) and summing its value (say velocity or something...). all physical equations are derived from calculus of constant functions. you can derive position by integrating velocity. you are correct, time cannot be broken into discrete moments, but it can however be summed over an interval through infinitesimally small segments. This was the revelation of Newton in his invention of Calculus. its not about the moment, its about the CHANGE in moment :) time is somewhat arbitrary expecially when taken from only a single moment. its like saying a line is made up of points, but how can you ever break it into every point which makes it up? you cant! there are too many! infinitely many! but that doesnt mean you cant measure the total length of the line and determine its slope :) good thinking though, you get a star. this whole time thing really hasnt been entirely figured out yet. weve got a long way to go to understand the nature of a single moment

Question:1/(x^3-3x^2+2x)=A/(x-a)+B/(x-b)+C/(x-c) is the only given equation. I am asked to find A,a,B,b,C,and c. How do I go about to solve this? Any tips are appreciated, thanks

Answers:Isn't a equation. Must become an identity and true for any x x^3-3x^2+2x = x(x-1)(x-2) a = 0 b =1 c =2 1/(x^3-3x^2+2x)=A/x+B/(x-1)+C/(x-2) for x = 3 1/6 = A/3+B/2+C x = -1 and another value for x and find 3 equations and find A,B,C Exists more way to find A,B,C like identification of coefficients using derivative of 1/(x^3-3x^2+2x)=A/x+B/(x-1)+C/(x-2) to reduce to only one undefined

Question:You fucccking fucking asssshole. That isnt what I was asking and you fucking know it if you had half a fuccking brain. I am not asking about graphing derivatives. Im asking about the traits of the graph itself.

Answers:it depends on your initial equation. like if it was a quadratic equation third would be a constant line and anything else higher is 0. I'm sorry the question is sort of vague do you mean that first derivative is the slope of the graph and the second is the concavity and the thirdis about the jerks of the original graph?

Question:Although the November 1949 Kilauea Iki eruption on the island of Hawaii bean with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater's floor, which at one point shot lava 1900 ft straight into the air (a world record). What was the lava's exit velocity in feet per second? in miles per hour? (Hint: If v0 is the exit velocity of a particle of lava, its height t seconds later will be s = (v0)(t) - 16t^2 feet. Begin by finding the time at which ds/dt = 0. Neglect air resistance.) PS: please use derivatives and show step by step! thanks! i can't decide whether v0 should be a constant or a variable. i think it should be a constant, so i treated it as such and came up with "s' = v0 - 32t" for the velocity function. but now i can't figure out how to solve for either variable.

Answers:It's funny, I'm in the Big Island of Hawaii for the summer and I just hiked along the Kilauea Iki (now dormant) a few days ago. v0 is supposed to be a constant. It is the exit velocity you are looking for, and you can assume it stays constant if the height of the fountain doesnot change. The derivative is ds/dt = v0 - 32 * t, as you found out yourself. Now let's go back to the physics : the lava reaches the apex of the fountain when its velocity is null. That gives you a relation between v0 and ta, time after which the lava reaches the apex : ds/dt = 0 = v0 - 32*ta (1) And you also have data about the ultimate height of the fountain, given by : ha = v0*ta - 16*ta^2 = 1900 ft (2) With these two equations, you can solve for v0 (and ta) : (1) => v0 = 32*ta => ta = v0/32 (2) => v0*ta - 16*ta*ta = 1900 By replacing ta you find : 1/32*v0^2 - 1/64*v0^2 = 1900 v0^2 = 1900 * 64 So v0 = 348 ft per sec is your answer.