Help to Integral Calculus Solutions & Examples

Evaluate the integrals by changing the order of integration in an appropriate way.
${\displaystyle {\int }_0^4{\int }_0^1{\int }_{2y}^2\frac{4\cos \left(x^2\right)}{2\sqrt{z}}dxdydz}$

Use polar coordinates to find the limit. [If are (r, theta) polar coordinates of the point (x,y) with r>=0, note that as (x,y) tends to (0,0).] ${\displaystyle lim(x,y)}$ tends to (0,0)${\displaystyle (x^2+y^2)\ln (x^2+y^2)}$

Find parametric equations for the tangent line to the curve with the given parametric equation s at the specified point.
${\displaystyle x=t^2+1,y=4\sqrt{t},z=e^{t^2-t};(2,4,1)}$

The image of polygon MNOP after a similarity transformation is polygon WXYZ. Each side of MNOP is 2 times as long as the corresponding side of WXYZ. What is the scale factor of the dilation in the similarity transformation?

Intervals of a derivative
I'm trying to find the intervals of the derivative of a function that I found on my book but I'm having some troubles understanding it, so I thought of trying to find some help here.
The function is $f(x)=\arcsin <!--\; \; -->(|x^2+3x+3|-<!--\; -\; -->1)$.
So I need to find the derivative of this function, study it and find the intervals (where it's increasing and where it's decreasing)
This is the solution: $-<!--\; -\; -->\frac{(2x+3)}{\sqrt{1-<!--\; -\; -->(x^2+3x+3{)}^2}}(-<!--\; -\; -->2,-<!--\; -\; -->1)$ and $\frac{(2x+3)}{\sqrt{1-<!--\; -\; -->(x^2+3x+1{)}^2}}(-<!--\; -\; -->3,-<!--\; -\; -->2)(-<!--\; -\; -->1,0)$.
So the function is decreasing in $[-<!--\; -\; -->3,-<!--\; -\; -->2]$ and $[-<!--\; -\; -->\frac{3}{2},-<!--\; -\; -->1]$ while increasing in $[-<!--\; -\; -->2,-<!--\; -\; -->\frac{3}{2}]$ and $[-<!--\; -\; -->1,0]$.
Here is where I am having some troubles:
- How do I find those intervals? I know that I need to solve the equations, but I 'm stuck there
- How can I find where the function is decreasing and increasing?

Why can some asymptotes be crossed?
I was taught that an asymptote cannot be crossed. My teacher then went and made my life a lot harder by countering what I've learned. Why can some asymptotes be crossed?

Consider this differential equation,
$dy/dx=x+\sin <!--\; \; -->(y)$
with initial condition $y=0.5$ when $x=1.2$:
1. Write down the recurrence relation for Euler's numerical method applied above.
2. With step size h = 0.1, calculate the approximations to y(1.3) and y(1.4).

1. This is my answer:
$Y_i+1=Y_i+0.1(X_i+\sin <!--\; \; -->(Y_i))$
2. I have problem solving this one...
From my txtbk reference, I sub $i=1$ into the eqn thus giving me
$Y_1=Y_0+0.1(1.2+\sin <!--\; \; -->(0.5))=0.1679$
Then, $i=2$
$Y_2=Y_1+0.1(1.3+\sin <!--\; \; -->(0.6)$
Im sure my method in part 2 is wrong and I don't really understand how to solve it via my txtbk.

Maximmum and minimum values of function in interval
How to find the intervals ( Increasing and decreasing) of the function and its Maximum and minimum value, where the function: $f(x)=a{x}^2+bx+c$.

Answer true or false to each of the following statement and explain your answer.
In using the method of transformations, a transformation of the predictor variable will change the conditional distribution of the response variable.

Use a double integral to find the volume of the indicated solid. z = 1-xy, y=x, y=1

Find the region E for which the triple integral
${\displaystyle \int \int \int E(1-x^2-2y^2-3z^2)dV}$
Is a maximum.

Solve the initial value problem related to $H(x,y)=x^2+y^2-<!--\; -\; -->1=0$ and $(x_0,y_0)=(1,0)$, using the Euler method.
The problems I've seen on the internet always give the $y^{\prime }$ in terms of $y$ and $x$ so what I did first was tro try to find this $y^{\prime }$ since I don't know it yet.
By implicit differentiation I have that $y^{\prime }=-<!--\; -\; -->x/y$
Now, If I set my $h=0.1$ for example, you can see on the first iteration I'll have an undefined value:
$x_0=1,y_0=0$
$x_1=x_0+h=1.01,y_1=y_0+y^{\prime }(x_0,y_0)=0+-<!--\; -\; -->1/0\to <!--\; \to \; -->UNDEFINED$
But I don't think this makes any sense, perhaps I differentiated incorrectly or this is just not the right way to go. I'd like someone to help me understand the very first step which is obtaining an ODE for a given function.

Some questons about recurrence sequences (using a problem).
A quick doubt, lets study the recurrence sequence:
$A_{n+1}=(4A_n+2)/A_n+3$
$A_0<-<!--\; -\; -->3$
First of all i do:
$A_{n+1}-<!--\; -\; -->A_n<0$
If this is true i can say that $A_n$ decrease. This is true for those $A_n$ values:
$-<!--\; -\; -->3>A_n>-<!--\; -\; -->1$
$A_n>2$
And false (so $A_n$ increase) for those $A_n$ values:
$A_n<-<!--\; -\; -->3$
$-<!--\; -\; -->1<A_n<2$
the case $A_n<-<!--\; -\; -->3$ interests me.
The limit L can be -1 or -2 but i cannot say it exists for sure because $A_n$ is not limited and monotone for all the $A_n$ possible values. For example, the sequence can go from $A_n>2$ then decrease and go in $A_n<-<!--\; -\; -->3$ then again increase and fall in $A_n>2$ etc...
Another doubt comes from this fact:
It's ok to remove -1 from the possible values of L because in this case the sequence still growing?
Anyways: It happens so many times that i know the sequence increase or decrease in an interval but i don't know if doing it it will fall in another interval where it starts decreasing or increasing and in this scenario i don't know how to demonstrate if it goes on some limit or just starts to "ping-pong" on different intervals.
Or in other words i don't know how many it decrease/increase so i cannot say if it will go out from the interval i'm considering.

Trigonometric equation with two functions and a parameter
The problem: For which values of a the equation $a.sinx.cosx=sinx-<!--\; -\; -->cosx$ has exactly 2 different roots in the interval $[0;\mathrm{\pi <!--\; \pi \; -->}]$.
So clearly $0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2},\mathrm{\pi <!--\; \pi \; -->}$ are not an answer. I divide by sinx.cosx and get $f(x)=\frac{1}{cosx}-<!--\; -\; -->\frac{1}{sinx}=a$. I look at the interval $(0;\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ cos x is decreasing so $\frac{1}{cosx}$ is increasing same for sin x is increasing and $\frac{-<!--\; -\; -->1}{sinx}$ is also increasing.
${\displaystyle \underset{x\to <!--\; \to \; -->0}{lim}f(x)=+\mathrm{\infty <!--\; \infty \; -->}}$ and ${\displaystyle \underset{x\to <!--\; \to \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}{lim}f(x)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}$ and f(x) is continuous so we have exactly 1 root in the interval $(0;\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ for every a.
My question: I guess I have to find for what value of a the function has 1 root in the interval $(\frac{\mathrm{\pi <!--\; \pi \; -->}}{2};\mathrm{\pi <!--\; \pi \; -->})$ how do I do that and is there an easier way to solve the whole problem?

What is the critical point of this function?
The problem reads: $f(x)=7\frac{e^{2x}}{x}+4.$.
I am unsure of how to approach this problem to find the derivative. If someone could break down the steps that would be greatly appreciated.
Also, the question asks for intervals of the increasing and decreasing parts of the function. How would I figure this out? I'm thinking I'd use a sign chart. But if you have any other useful methods, I am all ears, or rather eyes.

Show the function is decreasing in an interval
Let $s>1,s\in <!--\; \in \; -->\mathbb{R}$, and let f be a function defined by
$f(x)=\frac{ln(x)}{x^s},x>0$
Determine the monotone intervals of f.
I note that f(x)'s domain is $(0,\mathrm{\infty <!--\; \infty \; -->})$. I then find the derivative of f(x).
$f^{\prime }(x)=x^{-<!--\; -\; -->s-<!--\; -\; -->1}(1-<!--\; -\; -->s\cdot <!--\; \cdot \; -->ln(x))$
$f^{\prime }(x)=0⟹<!--\; ⟹\; -->x=e^{\frac{1}{s}}$
So $e^{\frac{1}{s}}$ is a critical point for f.
I'm uncertain how to evaluate the function around the critical point. I'm thinking to evaluate at the points $e^{\frac{0}{s}}$ and $e^{\frac{2}{s}}$?
So $f^{\prime }(e^{\frac{0}{s}})=e^{{\frac{0}{s}}^{-<!--\; -\; -->s-<!--\; -\; -->1}}(1-<!--\; -\; -->s\cdot <!--\; \cdot \; -->ln(e^{\frac{0}{s}}))$.
I then need to figure out if the expression above is greater, less or equal to 0. And likewise with the other critical point.
I'm not sure if I've done this correctly and I'm not sure how to make the expression for the evaluated critical point any nicer.

Asymptotes. Calculus.
I came across a question in a book asking for the asymptotes of $y=\sqrt{1+x^2}$. The answer says it as 2, but I am pretty convinced that there is no asymptote. I somehow think the book is wrong. Any help would be much appreciated.

use euler's method with step size $h=0.15$ to compute the approximate value of $y(0.6)$,correct upto five decimal places from initial value problem
$y^{\prime }=x(y+x)-<!--\; -\; -->2,y(0)=2$
Actually, I have found the answer for $y(0.6)$ with $h=0.15$ but i doesn't know wether the answer is correct upto 5 decimal places or not. How can i check the error and get the answer correct upto 5 decimal places?

How can I show that the Euler method fails to approximate the exact solution
$y(x)=(2x/3{)}^{3/2}$
to the IVP
$y^{\prime }=y^{\frac{1}{3}}$
$y(0)=0$
Here we have $f(t,y)=y^{\frac{1}{3}}$, $y_0=0$ and so $f(t_0,y_0)=f(0,0)=0$ and
$y_{n+1}=y_n+hf(t_n,y_n)$
Thus
$$\begin{gathered} y_1=0 \\ {y}_2=0 \\ ⋮<!--\; ⋮\; --> \\ {y}_n=0 \end{gathered}$$
So, I can't understand why it fails. Could you help me?