The cube roots of ##8## are ##2##, ##2omega## and ##2omega^2## where ##omega=-1/2+sqrt(3)/2 i## is the primitive Complex cube root of ##1##.

Here are the cube roots of ##8## plotted in the Complex plane on the circle of radius ##2##:

graph{(x^2+y^2-4)((x-2)^2+y^2-0.01)((x+1)^2+(y-sqrt(3))^2-0.01)((x+1)^2+(y+sqrt(3))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

They can be written as:

##2(cos(0)+i sin(0)) = 2##

##2(cos((2pi)/3) + i sin((2pi)/3)) = -1 + sqrt(3)i = 2omega##

##2(cos((4pi)/3) + i sin((4pi)/3)) = -1 – sqrt(3)i = 2omega^2##

One way of finding these cube roots of ##8## is to find all of the roots of ##x^3-8 = 0##.

##x^3-8 = (x-2)(x^2+2x+4)##

The quadratic factor can be solved using :

##x = (-b +-sqrt(b^2-4ac))/(2a)##

##= (-2+-sqrt(2^2-(4xx1xx4)))/(2*1)##

##=(-2+-sqrt(-12))/2##

##=-1+-sqrt(3)i##

 

"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

I’ll simplify the equation, so it is easier to read.

##y=1ln(1x-2)+0##There is not shifting or movement of anykind except for the ##x-2##, so the ##ln## graph will be moved 2 units right.

So it’s just the usual ##ln## graph with 2 units moved right.

Here is a graph for reference. graph{ln(x-2) [-4.75, 15.25, -4.68, 5.32]}

 

"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

By using the center-radius form.The center is ##C## at ##(h,k)## meaning ##h = -4## and ##k = 2##.The equation will be ##(x – h)^2 + (y – k)^2 = r^2##That translates to:##(x + 4)^2 + (y – 2)^2 = 25##Expanding, we get##x^2 + 8x + 16 + y^2 – 4y + 4 = 25##or##x^2 + 8x + y^2 – 4y – 5 = 0##

For the graphing part, locate the center ##(-4,2)##. Open your compass 5 units to any direction, then make a circle. It should pass the following points (along with infinitely many other points):##(-4 + 5, 2)## ##->## ##(1,2)####(-4 – 5,2)## ##->## ##(-9, 2)####(-4, 2 + 5)## ##->## ##(-4, 7)####(-4, 2 – 5)## ##->## ##(-4, -3)##

 

"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

You must apply ##x=5/costheta## and do the mathematics step by step.

This integral does not seem solvable by other techniques, e.g. integration by part, just integration by trigonometric substitution.

On integration by trigonometric substitution, you either apply trigonometry or trigonometric functions, such as the one herein. Unfortunately there is no general recipes, just training. The trick is reducing the integral to a manageable state; the only way is trying.

Replace: ##x=5/costheta##; how do I know that? unfortunately, there is no way to explain it, it is a “guess and try” game.

Now you should transform the infinitesimal variable ##dx##, why? remember that in math everything should be coherent, such as, if you change the limits of integration, you must apply a proper transformation.

##dx=5/cos^2theta*sintheta d theta##

In order to arrive to this, it was applied the substitution rule:

##u=costheta##, ##x=5/u##, remember the basic derivatives, ##dx/(d theta)=-5/u^2(du)/(d theta)##, if you are not familiar, you must go back one step forward in your calculus life, and train. It is called the substitution method, quite needed. Remember that the derivative of ##costheta## is ##-sintheta##.

We should simplify, remember ##x=5/costheta## :

##sqrt(x^2 – 25)/x= sqrt(25/cos^2theta – 25)/(5/costheta)##

Basic rules from mathematics:

##sqrt(25/cos^2theta – 25)costheta/5##

##sqrt(1/cos^2theta – 1)costheta##

##sqrt(1 – cos^2theta)=sintheta##; here just apply ##cos^2theta + sin^2theta=1##

Finally,

##sqrt(x^2 – 25)/x= sintheta##

So, remember that ##tantheta=sintheta/costheta##:

##intsqrt(x^2 – 25)/xdx=5intsin^2theta/cos^2thetad theta##

##intsqrt(x^2 – 25)/xdx=5inttan^2thetad theta##

By using a table of integrals:

##5inttan^2thetad theta=5*(tantheta -theta + C)##

Where, ##C## is a constant.

remember ##x=5/costheta##, ##costheta=5/x##, ##theta=arccos(5/x)##

Finally:

##intsqrt(x^2 – 25)/xdx=5*(tanarccos(5/x) -arccos(5/x) + C)##

See:

https://en.wikipedia.org/wiki/Lists_of_integralshttps://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions

Guessing the “magic”

Our problem falls into the third case of the scheme below, replace u by x and a by 5.

We get ##sintheta=sqrt(x^2 – 25)/x##

Further: ##sintheta=sqrt(x^2 – 25)/x##

##sin^2theta=1 – 25/x^2##

##25/x^2=1 – sin^2theta##

##25/x^2=cos^2theta##

##25/cos^2theta=x^2 ##

##x=5/costheta ##

The rest you already know.

Hand calculations

 

"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

How do you know when a reaction is spontaneous? Would melting/freezing be considered spontaneous?

Spontaneous reactions are those which take place by themselves, given enough time. They do not haveto be rapid; speed is not a factor in definition of spontaneity. Explosions and many other…

 

"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"