CALCU-Taylor polynomials for solutions to constant

Taylor polynomials for solutions to constant coefficient ordinary differential equations

Solution by Steps

step 1

To find the Taylor polynomial for solutions to constant coefficient ordinary differential equations, we start by considering the general form of a constant coefficient ODE

step 2

Suppose we have a differential equation of the form

a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0

step 3

The solution to this differential equation can be expressed as a Taylor series around a point

x = 0

:

y(x) = \sum_{k=0}^{\infty} \frac{y^{(k)}(0)}{k!} x^k

step 4

To find the Taylor polynomial of degree

n

, we truncate the series at the

n

-th term:

P_n(x) = \sum_{k=0}^{n} \frac{y^{(k)}(0)}{k!} x^k

step 5

The coefficients

y^{(k)}(0)

are determined by the initial conditions and the differential equation itself

step 6

For example, if the initial conditions are

y(0) = y_0

,

y'(0) = y_1

, etc., we can compute the higher derivatives using the differential equation

step 7

Substitute these values into the Taylor series to get the polynomial

P_n(x)

Answer

The Taylor polynomial for the solution to a constant coefficient ordinary differential equation is obtained by truncating the Taylor series of the solution at the desired degree and using the initial conditions to determine the coefficients.

Key Concept

Taylor polynomial for ODE solutions

Explanation

The Taylor polynomial for solutions to constant coefficient ordinary differential equations is derived by truncating the Taylor series of the solution and using initial conditions to find the coefficients.

Taylor polynomials for solutions to constant coefficient ordinary differential equations

Solution by Steps

step 1

To find the Taylor polynomial for solutions to constant coefficient ordinary differential equations, we start by considering the general form of a constant coefficient ODE

step 2

Suppose we have a differential equation of the form

a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0

step 3

The solution to this differential equation can be expressed as a Taylor series around a point

x = x_0

step 4

The Taylor series for

y(x)

around

x = x_0

is given by

y(x) = y(x_0) + y'(x_0)(x - x_0) + \frac{y''(x_0)}{2!}(x - x_0)^2 + \cdots + \frac{y^{(n)}(x_0)}{n!}(x - x_0)^n

step 5

To find the coefficients

y(x_0), y'(x_0), y''(x_0), \ldots, y^{(n)}(x_0)

, we use the initial conditions provided for the differential equation

step 6

By substituting the initial conditions into the Taylor series, we can determine the specific values of the coefficients

step 7

The resulting polynomial is the Taylor polynomial for the solution to the given constant coefficient ordinary differential equation

Answer

The Taylor polynomial for the solution to a constant coefficient ordinary differential equation is determined by expressing the solution as a Taylor series around a point and using the initial conditions to find the coefficients.

Key Concept

Taylor polynomial for ODE solutions

Explanation

The Taylor polynomial for solutions to constant coefficient ordinary differential equations is found by expressing the solution as a Taylor series and using initial conditions to determine the coefficients.

Solution by Steps

step 1

Identify the bounded domain

V

in

\mathbb{R}^3

surrounded by the plane

P := \{z = 0\}

and the surface

\Sigma := \{x^2 + y^2 + 1 = 0\}

step 2

Notice that the equation

x^2 + y^2 + 1 = 0

has no real solutions because

x^2 + y^2 \geq 0

for all real

x

and

y

, and adding 1 makes it strictly positive

step 3

Since there are no real solutions to the surface equation, the bounded domain

V

does not exist in the real number space

step 4

Therefore, the volume of the bounded domain

V

is

0

Answer

The volume of the bounded domain

V

is

0

.

Key Concept

Non-existence of real solutions

Explanation

The equation

x^2 + y^2 + 1 = 0

has no real solutions, implying that the bounded domain

V

does not exist in the real number space, resulting in a volume of

0

.

Solution by Steps

step 1

Identify the inequalities that define the domain

D

. The conditions are

x \geq 0

,

y \geq 0

, and

2x + y \leq 2

step 2

Plot the lines

x = 0

and

y = 0

on the coordinate plane. These lines represent the boundaries where

x

and

y

are non-negative

step 3

Plot the line

2x + y = 2

. This line intersects the

x

-axis at

x = 1

and the

y

-axis at

y = 2

step 4

Shade the region that satisfies all the inequalities. This region is bounded by the

x

-axis,

y

-axis, and the line

2x + y = 2

Question 2: Compute the integral

I:=\int_{D} f(x, y) \, dx \, dy

of the function defined as

f(x, y):=2-2x-y

.

step 1

Set up the double integral over the domain

D

. The limits for

x

are from

0

to

1

, and for

y

from

0

to

2 - 2x

step 2

Write the integral as

I = \int_{0}^{1} \int_{0}^{2-2x} (2 - 2x - y) \, dy \, dx

step 3

Integrate with respect to

y

:

\int_{0}^{2-2x} (2 - 2x - y) \, dy = \left[ (2 - 2x)y - \frac{y^2}{2} \right]_{0}^{2-2x}

step 4

Evaluate the inner integral:

\left[ (2 - 2x)(2 - 2x) - \frac{(2 - 2x)^2}{2} \right] - \left[ 0 \right] = (2 - 2x)^2 - \frac{(2 - 2x)^2}{2} = \frac{(2 - 2x)^2}{2}

step 5

Simplify and integrate with respect to

x

:

I = \int_{0}^{1} \frac{(2 - 2x)^2}{2} \, dx = \int_{0}^{1} 2(1 - x)^2 \, dx

step 6

Evaluate the outer integral:

2 \int_{0}^{1} (1 - 2x + x^2) \, dx = 2 \left[ x - x^2 + \frac{x^3}{3} \right]_{0}^{1} = 2 \left( 1 - 1 + \frac{1}{3} \right) = \frac{2}{3}

Question 3: Compute the volume of

E:=\left\{(x, y, z) \in \mathbb{R}^{3}: x, y, z \geq 0, 2x + y + z \leq 2\right\}

.

step 1

Set up the triple integral over the region

E

. The limits for

x

are from

0

to

1

, for

y

from

0

to

2 - 2x

, and for

z

from

0

to

2 - 2x - y

step 2

Write the integral as

V = \int_{0}^{1} \int_{0}^{2-2x} \int_{0}^{2-2x-y} dz \, dy \, dx

step 3

Integrate with respect to

z

:

\int_{0}^{2-2x-y} dz = (2 - 2x - y)

step 4

Integrate with respect to

y

:

\int_{0}^{2-2x} (2 - 2x - y) \, dy = \left[ (2 - 2x)y - \frac{y^2}{2} \right]_{0}^{2-2x} = \frac{(2 - 2x)^2}{2}

step 5

Integrate with respect to

x

:

V = \int_{0}^{1} \frac{(2 - 2x)^2}{2} \, dx = \int_{0}^{1} 2(1 - x)^2 \, dx

step 6

Evaluate the integral:

2 \int_{0}^{1} (1 - 2x + x^2) \, dx = 2 \left[ x - x^2 + \frac{x^3}{3} \right]_{0}^{1} = 2 \left( 1 - 1 + \frac{1}{3} \right) = \frac{2}{3}

Answer

The domain

D

is the region bounded by

x \geq 0

,

y \geq 0

, and

2x + y \leq 2

. The integral

I

over this domain is

\frac{2}{3}

. The volume

V

of the region

E

is also

\frac{2}{3}

.

Key Concept

Integration over a defined region

Explanation

To solve these problems, we set up and evaluate integrals over the specified regions, using the given boundaries and conditions.

Solution by Steps

step 1

Given the parametric equation of the curve

r(t) = (t, t^2, t)

, we first find the derivative

r'(t)

to get the tangent vector

step 2

Compute

r'(t) = (1, 2t, 1)

step 3

Evaluate

r'(t)

at

t = 0

:

r'(0) = (1, 0, 1)

step 4

Normalize

r'(0)

to get the unit tangent vector

T(p_0)

. The magnitude of

r'(0)

is

\sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}

step 5

Therefore,

T(p_0) = \frac{1}{\sqrt{2}}(1, 0, 1) = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)

Answer 1

T(p_0) = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)

step 1

To find the unit normal vector

N(p_0)

, we need the second derivative

r''(t)

step 2

Compute

r''(t) = (0, 2, 0)

step 3

Evaluate

r''(t)

at

t = 0

:

r''(0) = (0, 2, 0)

step 4

Normalize

r''(0)

to get the unit normal vector

N(p_0)

. The magnitude of

r''(0)

is

\sqrt{0^2 + 2^2 + 0^2} = 2

step 5

Therefore,

N(p_0) = \frac{1}{2}(0, 2, 0) = (0, 1, 0)

Answer 2

N(p_0) = (0, 1, 0)

step 1

The osculating plane

P(p_0)

is defined by the point

p_0

and the normal vector to the plane, which is given by the cross product of

T(p_0)

and

N(p_0)

step 2

Compute

T(p_0) \times N(p_0) = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right) \times (0, 1, 0)

step 3

The cross product is

\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)

step 4

The equation of the plane is

-\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}z = 0

, or equivalently

x = z

Answer 3

The osculating plane

P(p_0)

is

x = z

.

step 1

The center of the osculating circle

c(p_0)

can be found using the formula

c(p_0) = p_0 + \frac{1}{\kappa} N(p_0)

, where

\kappa

is the curvature

step 2

The curvature

\kappa

at

t = 0

is given by

\kappa = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3}

step 3

Compute

r'(0) \times r''(0) = (1, 0, 1) \times (0, 2, 0) = (-2, 0, 2)

step 4

The magnitude

|r'(0) \times r''(0)| = \sqrt{(-2)^2 + 0^2 + 2^2} = 2\sqrt{2}

step 5

The magnitude

|r'(0)| = \sqrt{2}

, so

|r'(0)|^3 = (\sqrt{2})^3 = 2\sqrt{2}

step 6

Therefore,

\kappa = \frac{2\sqrt{2}}{2\sqrt{2}} = 1

step 7

The center

c(p_0) = (0, 0, 0) + \frac{1}{1}(0, 1, 0) = (0, 1, 0)

Answer 4

The center of the osculating circle

c(p_0)

is

(0, 1, 0)

.

Key Concept

Unit Tangent Vector

Explanation

The unit tangent vector is the normalized first derivative of the parametric equation of the curve.

Key Concept

Unit Normal Vector

Explanation

The unit normal vector is the normalized second derivative of the parametric equation of the curve.

Key Concept

Osculating Plane

Explanation

The osculating plane is defined by the point on the curve and the normal vector to the plane, which is the cross product of the unit tangent and normal vectors.

Key Concept

Center of Osculating Circle

Explanation

The center of the osculating circle is found using the curvature and the unit normal vector at the given point on the curve.

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