Linear approximation of quotient

$\frac{(2.01{)}^{2}}{\sqrt{.95}}$

$\frac{(2.01{)}^{2}}{\sqrt{.95}}$

Riya Hansen
2022-07-12
Answered

Linear approximation of quotient

$\frac{(2.01{)}^{2}}{\sqrt{.95}}$

$\frac{(2.01{)}^{2}}{\sqrt{.95}}$

You can still ask an expert for help

Kaya Kemp

Answered 2022-07-13
Author has **18** answers

set $f(x)=\frac{(2+x{)}^{2}}{\sqrt{1-5x}}$ expand it then set $x=0.01$.

using linear approximation $f(x)=4+14x$ $f(0.01)=4.14$.

using linear approximation $f(x)=4+14x$ $f(0.01)=4.14$.

Savanah Boone

Answered 2022-07-14
Author has **5** answers

You may also do it using the linear approximation of $f(x,y)=\frac{{x}^{2}}{\sqrt{y}}$ at $({x}_{0},{y}_{0})=(2,1)$ with $\mathrm{\Delta}x=0.01$ and $\mathrm{\Delta}y=0.05$:

$f(x,y)\stackrel{linear}{\approx}f({x}_{0},{y}_{0})+\frac{\mathrm{\partial}f}{\mathrm{\partial}x}({x}_{0},{y}_{0})\mathrm{\Delta}x+\frac{\mathrm{\partial}f}{\mathrm{\partial}y}({x}_{0},{y}_{0})\mathrm{\Delta}y$

With

$\frac{\mathrm{\partial}f}{\mathrm{\partial}x}=\frac{2x}{\sqrt{y}}$

$\frac{\mathrm{\partial}f}{\mathrm{\partial}y}=-\frac{{x}^{2}}{2\sqrt{{y}^{3}}}$

you get

$\frac{(2.01{)}^{2}}{\sqrt{.95}}\approx 4+4\cdot 0.01-2\cdot 0.05=3.94$

$f(x,y)\stackrel{linear}{\approx}f({x}_{0},{y}_{0})+\frac{\mathrm{\partial}f}{\mathrm{\partial}x}({x}_{0},{y}_{0})\mathrm{\Delta}x+\frac{\mathrm{\partial}f}{\mathrm{\partial}y}({x}_{0},{y}_{0})\mathrm{\Delta}y$

With

$\frac{\mathrm{\partial}f}{\mathrm{\partial}x}=\frac{2x}{\sqrt{y}}$

$\frac{\mathrm{\partial}f}{\mathrm{\partial}y}=-\frac{{x}^{2}}{2\sqrt{{y}^{3}}}$

you get

$\frac{(2.01{)}^{2}}{\sqrt{.95}}\approx 4+4\cdot 0.01-2\cdot 0.05=3.94$

asked 2021-02-21

Find the Laplace transforms of the following time functions.

Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.

a)$f(t)=1+2t$
b)$f(t)=\mathrm{sin}\omega t\text{Hint: Use Euler\u2019s relationship,}\mathrm{sin}\omega t=\frac{{e}^{(}j\omega t)-{e}^{(}-j\omega t)}{2j}$

c)$f(t)=\mathrm{sin}(2t)+2\mathrm{cos}(2t)+{e}^{-t}\mathrm{sin}(2t)$

Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.

a)

c)

asked 2021-02-09

In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative
$\frac{dy}{dt}$ also appears. Consider the following initial value problem, defined for t > 0:

$\frac{dy}{dt}+4{\int}_{0}^{t}y(t-w){e}^{-4w}dw=3,y\left(0\right)=0$

a) Use convolution and Laplace transforms to find the Laplace transform of the solution.

$Y\left(s\right)=L\left\{y\left(t\right)\right)\}-?$

b) Obtain the solution y(t).

y(t) - ?

a) Use convolution and Laplace transforms to find the Laplace transform of the solution.

b) Obtain the solution y(t).

y(t) - ?

asked 2022-07-12

I solved the differential equation

$\frac{dy}{dx}=\frac{x}{{x}^{2}+1}$

to get the general solution

$y=\frac{ln|x+1|+c}{2}$

Im given the initial condition

$y{y}^{\prime}-2{e}^{x}=0,y(0)=3$

but I dont know what to do with it

$\frac{dy}{dx}=\frac{x}{{x}^{2}+1}$

to get the general solution

$y=\frac{ln|x+1|+c}{2}$

Im given the initial condition

$y{y}^{\prime}-2{e}^{x}=0,y(0)=3$

but I dont know what to do with it

asked 2020-11-14

Use the Laplace transform to solve the given initial-value problem

$y\prime \prime +2y\prime +y=\delta (t-4)$

$y(0)=0$

$y\prime (0)=0$

asked 2021-03-18

Determine ${L}^{-1}\left[\frac{(s-4){e}^{-3s}}{{s}^{2}-4s+5}\right]$

asked 2021-12-26

Solve the given differential equation. If an initial condition is given, also find the solution that satisfies it.

$\frac{dy}{dx}=3-6x+y-2xy$

asked 2022-06-12

Let the first order differential equation be

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$