Linear approximation of quotient ( 2.01 ) 2

Riya Hansen 2022-07-12 Answered
Linear approximation of quotient
( 2.01 ) 2 .95
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Kaya Kemp
Answered 2022-07-13 Author has 18 answers
set f ( x ) = ( 2 + x ) 2 1 5 x expand it then set x = 0.01.
using linear approximation f ( x ) = 4 + 14 x f ( 0.01 ) = 4.14.
Not exactly what you’re looking for?
Ask My Question
Savanah Boone
Answered 2022-07-14 Author has 5 answers
You may also do it using the linear approximation of f ( x , y ) = x 2 y at ( x 0 , y 0 ) = ( 2 , 1 ) with Δ x = 0.01 and Δ y = 0.05:
f ( x , y ) l i n e a r f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) Δ x + f y ( x 0 , y 0 ) Δ y
With
f x = 2 x y
f y = x 2 2 y 3
you get
( 2.01 ) 2 .95 4 + 4 0.01 2 0.05 = 3.94
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

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)=sinωtHint: Use Euler’s relationship, sinωt=e(jωt)e(jωt)2j
c)f(t)=sin(2t)+2cos(2t)+etsin(2t)
asked 2021-02-09
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative dydt also appears. Consider the following initial value problem, defined for t > 0:
dydt+40ty(tw)e4wdw=3,y(0)=0
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
Y(s)=L{y(t))}?
b) Obtain the solution y(t).
y(t) - ?
asked 2022-07-12
I solved the differential equation
d y d x = x x 2 + 1
to get the general solution
y = l n | x + 1 | + c 2
Im given the initial condition
y y 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+2y+y=δ(t4)
y(0)=0
y(0)=0
asked 2021-03-18
Determine L1[(s4)e3ss24s+5]
asked 2021-12-26
Solve the given differential equation. If an initial condition is given, also find the solution that satisfies it.
dydx=36x+y2xy
asked 2022-06-12
Let the first order differential equation be
d y / d x + P ( x ) y = Q ( x )
To solve this we multiply it by function say v(x) then it becomes
v ( x ) d y / d x + P ( x ) v ( x ) y = Q ( x ) v ( x )
d ( v ( x ) y ) / d x = Q ( x ) v ( x )
where d ( v ( x ) ) / d x = 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 l n | v ( x ) | = P ( x ) d x.
Now my question is why we neglect absolute value of v(x) and obtain integrating factor as e P ( x ) d x = v ( x )