Vector Calculus 4th Edition Chapter 4 Maxima and Minima in Several Variables
Page 262 Problem 1 Answer
Given f(x) = e2x
To find the Taylor polynomials.Using the method of polynomial.
To find the Taylor polynomials.
Recall that the Taylor polynomial of the function f at the given point a of the kth order is calculated with:
Tk(x) = f(a)+f′(a)(x−y)+f′′(a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
So in this task we have f(x) = e2x,k=4 and a=0.
From there we calculate:
f(a) = f(0) ⇒ 1
f′(0) = 2
f′′(0) = 4
f′′′(0) = 8
f(iv) = 16
Finally the polynomial is given with:
T4(x) = 1+2x+2x2+4/3x3+2/3x4
The Taylor polynomials is T4 (x) = 1+2x+2x2+4/3x3+2/3x4
Page 262 Problem 2 Answer
Given f(x)= ln(1+x)
To find the Taylor polynomials.Using the method of polynomial.
To find the Taylor polynomials.
Recall that the Taylor polynomial of the function f at the given point a of the k th order is calculated with:
Tk(x) = f(a)+f′(a)(x−y)+f′′(a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
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So in this task we have f(x)=ln(1+x), k=3 and a=0.
From there we calculate:
f(a) = f(0) = 0
f′(x) = 1/1+x
f′(0) = 1
f′′(x) = −1/(1+x)2
f′′(0) = −1
f′′′(x) = 2/(1+x)3
f′′′(0) = 2
Finally the polynomial is given with:
T3(x) = x−x2/2+x3/3
The Taylor polynomials is T3 (x) = x−x2/2+x3/3
Page 262 Problem 3 Answer
Given f(x) = 1/x2
To find the Taylor polynomials.Using the method of polynomial.
To find the Taylor polynomials.
Recall that the Taylor polynomial of the function f at the given point a of the kth order is calculated with:
Tk(x) = f(a)+f′(a)(x−y)+f′′(a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
So in this task, we have f(x) = 1/x2, k = 4 and a=1.
From there we calculate:
f(a) = f(1) = 1
f′(x) = −2/x3
f′(1) = −2
f′′(x) = 6/x4
f′′(1) = 6
f′′′(x) = −24/x5
f′′′(1) = −24
f(iv)(x) = 120/x6
f{(iv)}(1) = 120
Finally the polynomial is given with:
T4(x) = 1 − 2(x−1) + 3(x−1)2 − 4(x−1)3 + 5(x−1)4
The Taylor polynomials is T4 (x) = 1 − 2(x−1) + 3(x−1)2 − 4 (x−1)3 + 5(x−1)4
Page 262 Problem 4 Answer
Given f(x)=√x
To find the Taylor polynomials.Using the method of polynomial.
To find the Taylor polynomials.
Recall that the Taylor polynomial of the function f at the given point a of the k th order is calculated with:
Tk (x)=f(a)+f′(a)(x−y)+f′′(a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
So in this task we have f(x)=√x, k=3 and a=1.
From there we calculate:
f(a) = 1
f′(x) = 1/2√x
f′(1) =1 /2
f′′(x) = −1/4×3/2
f′′(1) = −1/4
f′′′(x) = 3/8×5/2
f′′′(1) = 3/8
Finally the polynomial is given with:
T3(x) = 1 + 1/2(x−1 )−1/8 (x−1)2 + 1/16(x−1)3
The Taylor polynomials is T3
(x) = 1+ 1/2(x−1) − 1/8(x−1)2 + 1/16(x−1)3
Page 262 Problem 5 Answer
Given f(x)=√x
To find the Taylor polynomials.Using the method of polynomial.
To find the Taylor polynomials.
Recall that the Taylor polynomial of the function f at the given point a of the kth order is calculated with:
Tk(x) = f(a)+f′(a)(x−y)+f′′(a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
So in this task we have f(x)=√x, k=3 and a=9.
From there we calculate:
f(a)=f(9)=3
f′(x)=1/2√x
f′(1)=1/6
f′′(x)=−1/4x3/2
f′′(1)=−1/108
f′′′(x)=3/8x5/2
f′′′(1)=1/648
Finally the polynomial is given with:
T3 (x)=3+1/6(x−9)−1/216(x−9)2+1/3888(x−9)3
The Taylor polynomials is T3 (x)=3+1/6(x−9)−1/216(x−9)2+1/3888(x−9)3
Page 262 Problem 6 Answer
Given f(x)=sinx
To find the Taylor polynomials.Using the method of polynomial.
To find the Taylor polynomials.
Recall that the Taylor polynomial of the function f at the given point a of the kth order is calculated with:
Tk(x)=f(a)+f′(a)(x−y)+f′′(a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
So in this task we have f(x)=sin(x),k=5 and a=0.
From there we calculate:
f(a)=f(0)=0
f′(x)=cos(x)
f′(0)=1
f′′(x)=−sin(x)
f′′(0)=0
f′′′(x)=−cos(x)
f′′′(0)=−1
Forf(iv) (x)=sin(x)
f{(iv)} (0)=0
f(5) (x)=cos(x)
f{(5)}(0)=1
Finally the polynomial is given with:
T5(x)=x−x3/6+x5/120
The Taylor polynomials is T5 (x)=x−x3/6+x5/120
Page 262 Problem 7 Answer
Given: f(x)=sinx
To find the Taylor polynomials pk of given order k at the indicated point a.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the Taylor polynomials pk of given order k at the indicated point a,
Tk (x)=f(a)+f′(a)(x−a)+f′′ (a)/2!(x−a)2+…+f(k)(a)/k!(x−a)k
f′(x)⇒cos(x)⇒f′(π2)=0
f′′(x)⇒−sin(x)⇒f′′(π2)=−1
f′′′(x)⇒−cos(x)⇒f′′′(π2)=0
f4(x)⇒sin(x)⇒f4(π2)=1
f(5)(x)⇒cos(x)⇒f(5)(π2)=0
T5(x)=1−(x−π2)2/2+(x−π2)4/24
The Taylor polynomial is T5
(x)=1−(x−π/2)2/2+(x−π/2)4/24.
Page 262 Problem 8 Answer
Given: f(x,y)=1/(x2+y2+1)
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(0,0)=1
∂f/∂x⇒−2x(x2+y2+1)2
⇒∂f/∂x(0,0)=0
∂f/∂y⇒−2y
(x2+y2+1)2⇒∂f/∂y(0,0)=0
∂2f/∂x2⇒6x2−2y2−2/(x2+y2+1)3⇒∂2f/∂x2
(0,0)=−2/∂2f/∂x∂y⇒8xy
(x2+y2+1)3⇒∂2f/∂x∂y(0,0)=0
T1(x)=1
T2(x)=1−x2−y2
The polynomial is given as T2(x)=1−x2−y2.
Page 262 Problem 9 Answer
Given: f(x,y)=1/(x2+y2+1)
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.The first-order Taylor Polynomial at a point (a,b)
is given by L(x)=f(a,b)+fx(a,b)⋅(x−a)+fy(a,b)⋅(y−b)
where fx,fy are the partial derivatives with respect to x and y, respectively.
The second-order Taylor Polynomial, given that the first-order polynomial has been calculated, is given by Q(x)=L(x)+fxx(a,b)2(x−a)2+fxy(a,b)2(x−a)(y−b)+fyy(a,b)2(y−b)2
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(1,−1)=1/3
∂f/∂x⇒−2x(x2+y2+1)2
⇒∂f/∂x(1,−1)=−2/9
∂f/∂y⇒−2y(x2+y2+1)2
⇒∂f/∂y(1,−1)=2/9
∂2f/∂y2⇒6y2−2x2−2(x2+y2+1)3
⇒∂2f∂y2(1,−1)=2/27
∂2f/∂x2⇒6x2−2y2−2/(x2+y2+1)3
⇒∂2f/∂x2(1,−1)=2/27
∂2f/∂x∂y⇒8xy(x2+y2+1)3
⇒∂2f/∂x∂y(1,−1)=−8/27
T1(x)=1/3−2(x−1)/9+2(y+1)/9
T2(x)=1/3−2(x−1)/9+2(y+1)/9+(x−1)2/27−8(x−1)(y+1)/27+(y+1)2/27
The polynomials are given as T1 (x)=1/3−2(x−1)/9+2(y+1) 9
and T2(x)=1/3−2(x−1)/9+2(y+1)/9+(x−1)2/27−8(x−1)(y+1)/27+(y+1)2/27.
Page 262 Problem 10 Answer
Given: f(x,y)=e2x+y
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(0,0)=1
∂f/∂x⇒2e2x+y⇒∂f/∂x(0,0)=2
∂f/∂y⇒e2x+y⇒∂f/∂y(0,0)=1
∂2f/∂x2⇒4e2x+y⇒∂2f/∂x2(0,0)=4
∂2f/∂x∂y⇒2e2x+y ⇒∂2f/∂x∂y(0,0)=2
T1(x)=1+2x+y
T2(x)=1+2x+y+2x2+2xy+y2/2
The polynomial is given as T2(x)=1+2x+y+2x2+2xy+y2/2.
Page 262 Problem 11 Answer
Given: f(x,y)=e2xcos3y
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(0,π)=−1
∂f/∂x⇒2e2xcos3y
⇒∂f/∂x(0,π)=−2
∂f/∂y⇒−3e2x sin3y⇒∂f/∂y(0,π)=0
∂2f/∂y2⇒−9e2x cos3y⇒∂2f/∂y2(0,π)=9
∂2f/∂x∂y⇒−6e2xsin3y⇒∂2f/∂x∂y(0,π)=0
T1(x)=−1−2x
T2(x)=−1−2x−2x2+9/2(y−π)2
The polynomial is given as T2(x)=−1−2x−2x2+9/2(y−π)2.
Page 262 Problem 12 Answer
Given: f(x,y,z)=ye3x+ze2y
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(0,0,2)=2
∂f/∂x⇒3ye3x ⇒∂f/∂x(0,0,2)=0
∂f/∂z⇒e2y ⇒∂f/∂z(0,0,2)=1
∂2f/∂x∂y⇒3e3x ⇒∂2f/∂x∂y(0,0,6i)=3
Simplify,
∂2f/∂y∂z⇒2e2y⇒∂2f/∂y∂z(0,0,6)=2
∂2f/∂x∂z⇒0⇒∂2f/∂x∂z(0,0,6)=0
T1(x)=5y+z
T2 (x)=y+z+3xy+4y2+2yz
The polynomial is T2(x)=y+z+3xy+4y2+2yz.
Page 262 Problem 13 Answer
Given: f(x,y,z)=xy−3y2+2xz,a=(2,−1,1)
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(2,−1,1)=−1
∂f/∂x⇒y+2z⇒∂f/∂x(2,−1,1)=1
∂f/∂z⇒2x⇒∂f/∂z(2,−1,1)=4
T1(x)=1+x+8y+4z
T2=f
The polynomial is T2=f
Page 262 Problem 14 Answer
Given: f(x,y,z)=1/(x2+y2+z2+1)
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(0,0,0)=1
∂f/∂x⇒−2x
(x2+y2+z2+1)2
⇒∂f/∂x(0,0,0)=0
∂2f/∂x2⇒6x2−2y2−2z2−2(x2+y2+z2+1)3
⇒∂2f/∂x2(0,0,0)=−2
∂2f/∂z2⇒6z2−2x2−2y2−2(x2+y2+z2+1)3
⇒∂2f/∂z2(0,0,0)=−2
∂2f/∂x∂z⇒8xz(x2+y2+z2+1)3
⇒∂2f/∂x∂z(0,0,0)=0
T1(x)=1
T2(x)=1−x2−y2−z2
The polynomial is given as T1(x)=1
T2(x)=1−x2−y2−z2.
Page 262 Problem 15 Answer
Given: f(x,y,z)=sinxyz,a=(0,0,0)
To find the first- and second-order Taylor polynomials.
Using the method of maxima function method.
Critical points of a function of couple variables denote the points at which both partial derivatives of the function remain zero.
To find the first- and second-order Taylor polynomials,
f(0,0,0)=0
∂f/∂x⇒yzcosxyz
⇒∂f/∂x(0,0,0)=0
∂2f/∂x2⇒−y2z2sinxyz
⇒∂2f/∂x2(0,0,0)=0
∂2f/∂x∂y⇒zcosxyz−xyz2/sinxyz
⇒∂2f/∂x∂y(0,0,0)=0
∂2f/∂x∂z=cosxyz−xy2/zsinxyz
⇒∂2f/∂x∂z(0,0,0)=0
T1(x)=0
T2(x)=0
The polynomial is given as T1(x)=0.