Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 24

Determine the $f'(x)$ of the function $\displaystyle f(x) = \sqrt[3]{(x^2 - 1)^2}$
We have $f(x) =(x^2 - 1)^{\frac{2}{3}}$. So, by using Chain Rule

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{d}{dx} \left[ (x^2 - 1)^{\frac{2}{3}} \right]\\
\\
&= \frac{2}{3} (x^2 - 1)^{\frac{2}{3}-1} \cdot \frac{d}{dx} (x^2 - 1)\\
\\
&= \frac{2}{3} (x^2 - 1)^{-\frac{1}{3}} (2x) \\
\\
&= \frac{4x}{3} (x^2 - 1)^{-\frac{1}{3}} \text{ or } \frac{4x}{3(x^2 - 1)^{\frac{1}{3}}}
\end{aligned}
\end{equation}
$


Then,

$
\begin{equation}
\begin{aligned}
f''(x) &= \frac{4}{3} \left[ x \cdot \frac{d}{dx} (x^2 - 1)^{-\frac{1}{3}} + (x^2 - 1)^{-\frac{1}{3}} \cdot \frac{d}{dx} (x) \right]\\
\\
f''(x) &= \frac{4}{3} \left[ x \cdot -\frac{1}{3} (x^2 - 1)^{-\frac{1}{3} - 1} \cdot \frac{d}{dx} (x^2 - 1) + (x^2 - 1)^{-\frac{1}{3}} (1)\right]\\
\\
f''(x) &= \frac{4}{3} \left[ x \cdot -\frac{1}{3} (x^2 - 1)^{-\frac{2}{3}} (2x) + (x^2 - 1)^{-\frac{1}{3}} \right]\\
\\
f''(x) &= \frac{4}{3} \left[ \frac{-2x}{3(x^2 - 1)^{\frac{2}{3}}} + \frac{1}{(x^2 - 1)^{\frac{1}{3}}} \right]
\end{aligned}
\end{equation}
$

College Algebra, Chapter 3, 3.5, Section 3.5, Problem 4

Match the graph with the function

a.) $y = |x + 1|$

b.) $y = |x - 1|$

c.) $y = |x| - 1$

d.) $y = - |x|$

I.







II.







III.







IV.








a.) $y = |x + 1|$

The graph of $y = |x|$ is shifted 1 unit to the left. The equation matches graph II.

b.) $y = |x - 1|$

The graph of $y = |x|$ is shifted 1 unit to the right. The equation matches graph IV.

c.) $y = |x| - 1$

The graph of $y = |x|$ is shifted 1 unit downward. The equation matches graph I.

d.) $y = - |x|$

The graph of $y = |x|$ is reflected about $x$-axis. The equation matches graph III.

Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 48

Arc length of a curve C described by the parametric equations x=f(t) and y=g(t), a<=t<=b where f' and g' are continuous on [a,b] and C is traversed exactly once as t increases from a to b, then the length of the curve is given by,
L=int_a^bsqrt((dx/dt)^2+(dy/dt)^2)dt
We are given:x=arcsin(t),y=ln(sqrt(1-t^2)), 0<=t<=1/2
x=arcsin(t)
dx/dt=1/sqrt(1-t^2)
y=lnsqrt(1-t^2)
dy/dt=1/sqrt(1-t^2)d/dt(sqrt(1-t^2))
dy/dt=1/sqrt(1-t^2)(1/2)(1-t^2)^(1/2-1)(-2t)
dy/dt=-t/(1-t^2)
Now let's evaluate arc length by using the stated formula,
L=int_0^(1/2)sqrt((1/sqrt(1-t^2))^2+(-t/(1-t^2))^2)dt
L=int_0^(1/2)sqrt(1/(1-t^2)+t^2/(1-t^2)^2)dt
L=int_0^(1/2)sqrt((1-t^2+t^2)/(1-t^2)^2)dt
L=int_0^(1/2)sqrt(1/(1-t^2)^2)dt
L=int_0^(1/2)1/(1-t^2)dt
L=int_0^(1/2)1/((1+t)(1-t))dt
Using partial fractions integrand can be written as :
L=int_0^(1/2)1/2(1/(1+t)+1/(1-t))dt
Take the constant out and use the standard integral:int1/xdx=ln|x|+C
L=1/2int_0^(1/2)(1/(1+t)+1/(1-t))dt
L=1/2[ln|1+t|+ln|1-t|]_0^(1/2)
L=1/2{[ln|1+1/2|+ln|1-1/2|]-[ln1+ln1]}
L=1/2[ln|3/2|+ln|1/2|]
L=1/2[0.4054651081+0.69314718056]
L=1/2[1.09861228867]
L=0.54930614433
Arc length of the curve on the given interval is ~~0.549

How was Pennywise created?

Pennywise the Dancing Clown is the favored form of a greater entity, IT, the namesake of Steven King's novel. IT is a being of unknown age, though in the novel, the creature claims to be as old as time itself. Not only does IT lack a definite age, but it lacks a gender as well. Though the creature is referred to as male for the majority of the novel, IT does take the form of a female spider repeatedly. IT is, as best as can be assumed, a type of light and shadow being, made of chaotic orange light that exists in what the book refers to as a Macroverse. This place holds our own universe as well as the one beyond, where powerful entities such as IT originate. The novel never says conclusively if the orange light being is IT's true form, but it drives viewers to insanity and thus can only be assumed to be the monster's true form.
IT uses the form of Pennywise when on Earth in order to get close enough to lure children. IT prefers his meals to be filled with fright (for it's like salting the meat), and children are especially susceptible to their fear. Thus, Pennywise was created to satisfy specific cravings that the beast has. Every twenty-seven to thirty years, when a crime of viciousness and hate occurs, IT is summoned to Earth to feed. Until an act of equal horror as the first happens, IT continues its reign of terror as Pennywise the Dancing Clown.

In the Old English epic poem Beowulf, why did Hrothgar build Herot?

Beowulf, an Old English epic poem, was written somewhere around 600-900 A.D. It is the oldest known major work in Old English. It's author is unknown.
Much of the action takes place in a mead hall called Herot that was built by theDanish king, Hrothgar. The poem spends a great deal of time describing Herot and how it was central to the lives of the warriors. Hrothgar's motivation for building Herot is given in the first chapter (of the Burton Raffel version) of the poem:

Then Hrothgar . . . resolved
To build a hall that would hold his mighty
Band and reach higher toward Heaven than anything
That had ever been known to the sons of men.

Hrothgar wants Herot to serve several purposes. He wants to create an atmosphere of fellowship for his warriors. Kings in those days treated their key warriors respectfully, often rewarding them with gold for their efforts. Herot was actually made with some gold on the inside, so it helped foster a sense of loyalty in his men.
As the excerpt states, there is also a religious component at work here. Hrothgar wanted Herot to “reach higher toward Heaven.” Herot is a way for Hrothgar and his men to feel that they are more connected to God than they would be otherwise.
 

Why have the King and Queen invited Rosencrantz and Guildenstern to Elsinore, and what do they want them to do?

The king and queen have invited Hamlet's friends from Wittenberg University, Rosencrantz and Guildenstern, to Elsinore because they want them to keep an eye on Hamlet and try to figure out what has him acting so strangely. They also, of course, want those friends to inform on Hamlet to them. Claudius's motives are especially suspect, though Gertrude likely just wants to help her son feel better. Claudius tells the young men,

To draw [Hamlet] on to pleasures, and to gatherSo much as from occasion you may glean,Whether aught to us unknown afflicts him thusThat, opened, lies within our remedy (2.2.15–18).

In other words, the king tells Rosencrantz and Guildenstern to help Hamlet have a little fun, and then, when it will not seem out of place, they should ask Hamlet what afflicts, or saddens, him. Then, they should report on those conversations to the king and queen so that they can remedy, or help to fix, whatever is ailing Hamlet. The queen even says that they will reward Rosencrantz and Guildenstern for this service.

I assume that this question is referring to Rosencratz and Guildenstern in Act II, Scene 2.
At the beginning of the scene, Claudius, speaking to Rosencratz and Guildenstern, asks them to stay in Elsinore for a time and attempt to "glean" from Hamlet whether anything "afflicts him" of which the King and Queen are unaware. The Queen notes that they are close friends of Hamlet and might be better placed than others to help him.
When he runs into them, Hamlet assumes that they "were sent for," and demands to know the reason. Initially, Rosencratz says that they have come to see Hamlet, but Hamlet is eventually able to pry out of them the truth: that the King and Queen have, in fact, asked them to come to Elsinore. However, while Hamlet thinks his friends are here to betray him, the audience knows that they are actually trying to help him, whatever their promises to Gertrude.

xy + y' = 100x Solve the differential equation

The problem: xy+y'=100x is as first order differential equation that we can evaluate by applying variable separable differential equation:
N(y)y'=M(x)
N(y)(dy)/(dx)=M(x)
N(y) dy=M(x) dx
Apply direct integration: intN(y) dy= int M(x) dx to solve for the
 general solution of a differential equation.
Applying variable separable differential equation, we get:
xy+y'=100x
y' =100x-xy
y'=x(100-y)
(y')/(100-y)= x
Let y' =(dy)/(dx) :
((dy)/(dx))/(100-y)= x
(dy)/(100-y)= x dx
Apply direct integration on both sides:
int(dy)/(100-y)= int x dx
For the left side, we consider u-substitution by letting:
u= 100-y then du = -dy or -du=dy.
The integral becomes:
int(dy)/(100-y)=int(-du)/(u)
Applying basic integration formula for logarithm:
int(-du)/(u)= -ln|u|
Plug-in u = 100-y on "-ln|u| " , we get:
int(dy)/(100-y)=-ln|100-y|
 
For the right side, we apply the Power Rule of integration: int x^n dx = x^(n+1)/(n+1)+C
 
 int x* dx= x^(1+1)/(1+1)+C
 
                = x^2/2+C
 
Combing the results from both sides, we get the general solution of the differential equation as:
-ln|100-y|= x^2/2+C
or 
y =100- e^(-x^2/2-C)
 y = 100-Ce^(-x^2/2)