Solution:

Given: A is nilpotent of index 2.

Definition: A matrix A is nilpotent of index 2 if

A^2=O,\qquad A\ne O.

We have to prove that, for every positive integer n,

A(I\pm nA)=A.

Calculation:

Using distributivity of matrix multiplication,

A(I+nA)=AI+nA^2.

Since AI=A and A^2=O,

A(I+nA)=A+nO=A.

Similarly,

A(I-nA)=AI-nA^2=A-nO=A.

Hence both signs give the same result.

Final Answer:

\boxed{A(I\pm nA)=A\quad\text{for every positive integer }n.}

Solution:

Given parametric curve:

x=6t^2,\qquad y=4t^3-3t.

For a parametrically represented curve,

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}, \qquad \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt},

provided dx/dt\ne0.

Step 1: First derivative

\frac{dx}{dt}=12t, \qquad \frac{dy}{dt}=12t^2-3.

Therefore, for t\ne0,

\frac{dy}{dx} = \frac{12t^2-3}{12t} = \frac{4t^2-1}{4t} = t-\frac{1}{4t}.

Step 2: Second derivative

Differentiate dy/dx with respect to t:

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(t-\frac{1}{4t}\right) = 1+\frac{1}{4t^2} = \frac{4t^2+1}{4t^2}.

Hence

\frac{d^2y}{dx^2} = \frac{\dfrac{4t^2+1}{4t^2}}{12t} = \frac{4t^2+1}{48t^3}.

Since 4t^2+1>0 for every real t, the sign of d^2y/dx^2 is determined entirely by the sign of t^3. Thus:

• if t>0, then d^2y/dx^2>0, so the branch is convex (concave upward);
• if t<0, then d^2y/dx^2<0, so the branch is concave (concave downward).

At t=0, dx/dt=0 and the above expression for the second derivative with respect to x is not finite. The corresponding point of the curve is (0,0).

Final Answer:

\boxed{\frac{d^2y}{dx^2}=\frac{4t^2+1}{48t^3}}

Hence the curve is convex/concave upward for \boxed{t>0} and concave downward for \boxed{t<0}, with t=0 corresponding to the exceptional point (0,0).

Solution:

Given differential equation:

\left|1+\left(\frac{dy}{dx}\right)^2\right|^{2/3} = \rho\,\frac{d^2y}{dx^2}.

Definitions:

• The order of a differential equation is the order of the highest derivative occurring in it.
• The degree is the power of the highest-order derivative after the differential equation has been made polynomial in all its derivatives.

Since

1+\left(\frac{dy}{dx}\right)^2>0,

the absolute-value sign does not affect the expression:

\left[1+\left(\frac{dy}{dx}\right)^2\right]^{2/3} = \rho\,\frac{d^2y}{dx^2}.

The highest derivative present is d^2y/dx^2, so the order is 2.

To find the degree, remove the fractional exponent. Cubing both sides gives

\left[1+\left(\frac{dy}{dx}\right)^2\right]^2 = \rho^3\left(\frac{d^2y}{dx^2}\right)^3.

This is polynomial in the derivatives, and the highest-order derivative d^2y/dx^2 occurs to power 3.

Final Answer:

\boxed{\text{Order}=2,\qquad \text{Degree}=3.}

Solution:

Definition: A vector field \vec F is called irrotational in a region if

\nabla\times\vec F=\vec0.

Here

\vec r=x\mathbf i+y\mathbf j+z\mathbf k.

Therefore

\nabla\times\vec r = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ x & y & z \end{vmatrix}.

Expanding the determinant,

\nabla\times\vec r = \left(\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z}\right)\mathbf i - \left(\frac{\partial z}{\partial x}-\frac{\partial x}{\partial z}\right)\mathbf j + \left(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y}\right)\mathbf k.

Now

\frac{\partial z}{\partial y}=0,\quad \frac{\partial y}{\partial z}=0,\quad \frac{\partial z}{\partial x}=0,\quad \frac{\partial x}{\partial z}=0,\quad \frac{\partial y}{\partial x}=0,\quad \frac{\partial x}{\partial y}=0.

Hence

\nabla\times\vec r=0\mathbf i+0\mathbf j+0\mathbf k=\vec0.

Final Answer:

\boxed{\nabla\times\vec r=\vec0,}

so \boxed{\vec r\text{ is irrotational}}.

Solution:

Standard result: If T:V\to V is linear and B=\{e_1,e_2,e_3\} is an ordered basis, then the columns of the matrix [T]_B are the coordinate vectors [T(e_1)]_B,[T(e_2)]_B,[T(e_3)]_B.

The given basis is the standard basis

e_1=(1,0,0),\qquad e_2=(0,1,0),\qquad e_3=(0,0,1).

The transformation is

T(x,y,z)=(2y+z,\;x-4y,\;3x).

Step 1: Image of e_1

T(1,0,0) = (2\cdot0+0,\;1-4\cdot0,\;3\cdot1) = (0,1,3).

Thus

[T(e_1)]_B=\begin{pmatrix}0\\1\\3\end{pmatrix}.

Step 2: Image of e_2

T(0,1,0) = (2,\;-4,\;0),

so

[T(e_2)]_B=\begin{pmatrix}2\\-4\\0\end{pmatrix}.

Step 3: Image of e_3

T(0,0,1) = (1,\;0,\;0),

so

[T(e_3)]_B=\begin{pmatrix}1\\0\\0\end{pmatrix}.

Placing these coordinate vectors as columns gives

[T]_B = \begin{pmatrix} 0 & 2 & 1\\ 1 & -4 & 0\\ 3 & 0 & 0 \end{pmatrix}.

Verification:

\begin{pmatrix} 0 & 2 & 1\\ 1 & -4 & 0\\ 3 & 0 & 0 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2y+z\\x-4y\\3x\end{pmatrix},

which agrees with the given transformation.

Final Answer:

\boxed{ [T]_B= \begin{pmatrix} 0 & 2 & 1\\ 1 & -4 & 0\\ 3 & 0 & 0 \end{pmatrix}}

Solution:

Cauchy-Schwarz Inequality — Statement:

For any vectors u,v in an inner product space,

|\langle u,v\rangle| \le \|u\|\,\|v\|.

Equivalently,

|\langle u,v\rangle|^2 \le \langle u,u\rangle\,\langle v,v\rangle.

Equality holds if and only if u and v are linearly dependent.

Proof:

If v=0, then \langle u,v\rangle=0, and both sides are zero. Thus the inequality holds.

Now assume v\ne0. Put

\lambda=\frac{\langle u,v\rangle}{\langle v,v\rangle}

in the complex case with the usual convention that the inner product is linear in its first argument; the same computation reduces to the familiar real formula in a real inner product space.

By positive definiteness,

0\le \|u-\lambda v\|^2 = \langle u-\lambda v,\;u-\lambda v\rangle.

Expanding,

\|u-\lambda v\|^2 = \langle u,u\rangle -\overline{\lambda}\langle u,v\rangle -\lambda\langle v,u\rangle +|\lambda|^2\langle v,v\rangle.

Using \langle v,u\rangle=\overline{\langle u,v\rangle} and the chosen value of \lambda, we get

0 \le \|u\|^2 - \frac{|\langle u,v\rangle|^2}{\|v\|^2}.

Since \|v\|^2>0, multiplying by \|v\|^2 yields

|\langle u,v\rangle|^2 \le \|u\|^2\|v\|^2.

Taking non-negative square roots,

|\langle u,v\rangle| \le \|u\|\,\|v\|.

Equality condition: Equality holds exactly when

\|u-\lambda v\|^2=0,

that is, when u=\lambda v. Hence equality occurs if and only if the two vectors are linearly dependent.

Final Answer:

\boxed{|\langle u,v\rangle|\le\|u\|\,\|v\|}

with equality precisely for linearly dependent vectors.

Solution:

Let

F(x,y)=x^2y-xy^2+xy+y^2+x-y.

The curve F(x,y)=0 is cubic. We first determine its three asymptotes and then find the finite point in which each asymptote meets the curve.

Step 1: Directions of the asymptotes

The homogeneous part of highest degree is

F_3(x,y)=x^2y-xy^2=xy(x-y).

Thus the three asymptotic directions correspond to

x=0,\qquad y=0,\qquad x-y=0.

Step 2: Asymptote parallel to the y-axis

Write the equation as a polynomial in y:

(1-x)y^2+(x^2+x-1)y+x=0.

For a vertical asymptote x=a, the coefficient of the highest power y^2 must vanish as |y|\to\infty. Therefore

1-a=0\quad\Rightarrow\quad a=1.

Hence the vertical asymptote is

\boxed{x=1}.

Step 3: Asymptote parallel to the x-axis

Write the equation as a polynomial in x:

yx^2+(-y^2+y+1)x+(y^2-y)=0.

For a horizontal asymptote y=b, the coefficient of x^2 must vanish. Hence

b=0,

so the horizontal asymptote is

\boxed{y=0}.

Step 4: Oblique asymptote

Since its direction is x-y=0, let the asymptote be

y=x+b.

Substitute y=x+b into F(x,y). Direct expansion gives

F(x,x+b) = (2-b)x^2+(-b^2+3b)x+(b^2-b).

For y=x+b to be an asymptote, the coefficient of the highest remaining power x^2 must vanish:

2-b=0 \quad\Rightarrow\quad b=2.

Hence the oblique asymptote is

\boxed{y=x+2}.

Step 5: Finite intersections of the curve with its asymptotes

For x=1,

F(1,y)=y+1.

Thus y=-1, giving

P_1=(1,-1).

For y=0,

F(x,0)=x,

so

P_2=(0,0).

For y=x+2,

F(x,x+2)=2(x+1).

Hence x=-1 and y=1, giving

P_3=(-1,1).

Now each of the three points satisfies

x+y=0.

Indeed,

1+(-1)=0,\qquad 0+0=0,\qquad -1+1=0.

Final Answer:

The three finite points where the cubic meets its asymptotes are

(1,-1),\quad(0,0),\quad(-1,1),

and they lie on the straight line

\boxed{x+y=0.}

Solution:

To show that boundedness alone does not imply Riemann integrability, it is enough to give one bounded function that is not Riemann integrable.

Consider the Dirichlet function on [0,1]:

f(x)= \begin{cases} 1,& x\in\mathbb Q,\\ 0,& x\notin\mathbb Q. \end{cases}

Step 1: Boundedness

For every x\in[0,1],

0\le f(x)\le1.

Hence f is bounded.

Step 2: Behaviour on every subinterval

Both the rational numbers and the irrational numbers are dense in \mathbb R. Therefore every non-degenerate subinterval of [0,1] contains both rational and irrational points.

Consequently, on every subinterval I of any partition,

\sup_{x\in I}f(x)=1, \qquad \inf_{x\in I}f(x)=0.

Let

P:\quad 0=x_0<x_1<\cdots<x_n=1

be any partition. Then its upper Darboux sum is

U(P,f) = \sum_{i=1}^{n}1\cdot(x_i-x_{i-1}) = 1,

whereas its lower Darboux sum is

L(P,f) = \sum_{i=1}^{n}0\cdot(x_i-x_{i-1}) = 0.

Thus for every partition P,

U(P,f)-L(P,f)=1.

In particular, the difference cannot be made arbitrarily small. Therefore the upper and lower Riemann integrals are

\overline{\int_0^1}f(x)\,dx=1, \qquad \underline{\int_0^1}f(x)\,dx=0,

which are unequal.

Final Answer:

The Dirichlet function is bounded on [0,1] but is not Riemann integrable. Hence

\boxed{\text{a bounded function need not be Riemann integrable}.}

Solution:

Given straight line in polar form:

\frac{l}{r}=A\cos\theta+B\sin\theta.

Multiplying by r and using x=r\cos\theta, y=r\sin\theta, we obtain

l=Ar\cos\theta+Br\sin\theta=Ax+By.

Thus the straight line is

Ax+By-l=0.

The circle is

r=2a\cos\theta.

Multiplying by r,

r^2=2ar\cos\theta,

so in Cartesian coordinates,

x^2+y^2=2ax.

Completing the square,

(x-a)^2+y^2=a^2.

Therefore the circle has centre (a,0) and radius a.

Condition for tangency: A line touches a circle if and only if the perpendicular distance from the centre of the circle to the line equals the radius.

The distance from (a,0) to Ax+By-l=0 is

\frac{|Aa-l|}{\sqrt{A^2+B^2}}.

Hence tangency requires

\frac{|Aa-l|}{\sqrt{A^2+B^2}}=a.

Squaring,

(Aa-l)^2=a^2(A^2+B^2).

Expanding,

a^2A^2-2aAl+l^2 = a^2A^2+a^2B^2.

After cancelling a^2A^2,

l^2-2aAl-a^2B^2=0.

Final Answer:

The required condition that the straight line touch the circle is

\boxed{l^2-2aAl-a^2B^2=0.}

Solution:

Let the centre of the required sphere be C(h,k,l) and its radius be R. The sphere passes through

P=(1,0,0),\qquad Q=(0,1,0),\qquad R_1=(0,0,1).

Therefore the squared distances from C to the three points are equal:

(h-1)^2+k^2+l^2 = h^2+(k-1)^2+l^2 = h^2+k^2+(l-1)^2.

Comparing the first two expressions,

h^2-2h+1+k^2 = h^2+k^2-2k+1,

which gives

h=k.

Comparing the second and third gives

k=l.

Hence all possible centres lie on the line

h=k=l=t.

For such a centre, the radius satisfies

R^2=(t-1)^2+t^2+t^2 = 3t^2-2t+1.

To minimize R, it is enough to minimize R^2. Differentiate:

\frac{d}{dt}(R^2)=6t-2.

The stationary value occurs when

6t-2=0 \quad\Rightarrow\quad t=\frac13.

Also,

\frac{d^2}{dt^2}(R^2)=6>0,

so this gives the minimum radius.

Thus the centre is

C=\left(\frac13,\frac13,\frac13\right),

and

R^2 = 3\left(\frac19\right) - 2\left(\frac13\right) + 1 = \frac13-\frac23+1 = \frac23.

Therefore

R=\sqrt{\frac23}.

The sphere is

\left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z-\frac13\right)^2 = \frac23.

Equivalently, after expansion,

x^2+y^2+z^2-\frac23(x+y+z)-\frac13=0.

Final Answer:

\boxed{ \left(x-\frac13\right)^2+ \left(y-\frac13\right)^2+ \left(z-\frac13\right)^2 = \frac23 }

with minimum radius \boxed{\sqrt{2/3}}.

Solution:

Given differential equation:

(1-x)y''+xy'-y=(1-x)^2.

For x\ne1, divide by 1-x to write it in standard linear form:

y'' + \frac{x}{1-x}y' - \frac{1}{1-x}y = 1-x.

Thus

R(x)=1-x.

Step 1: Solve the complementary equation

The associated homogeneous equation is

(1-x)y''+xy'-y=0.

We first observe that y_1=e^x is a solution, because

y_1'=e^x,\qquad y_1''=e^x,

and therefore

(1-x)e^x+xe^x-e^x=0.

Also y_2=x is a solution, since

y_2'=1,\qquad y_2''=0,

and

(1-x)\cdot0+x\cdot1-x=0.

The Wronskian is

W= \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} e^x & x\\ e^x & 1 \end{vmatrix} = e^x(1-x).

Hence W\ne0 on any interval not containing x=1, and y_1,y_2 form a fundamental pair there.

Therefore

y_c=C_1e^x+C_2x.

Step 2: Variation of parameters

For

y''+P(x)y'+Q(x)y=R(x),

a particular solution may be written as

y_p=u_1y_1+u_2y_2,

where

u_1'=-\frac{y_2R}{W}, \qquad u_2'=\frac{y_1R}{W}.

Here

u_1' = -\frac{x(1-x)}{e^x(1-x)} = -xe^{-x}.

Integrating,

u_1 = \int -xe^{-x}\,dx.

Using integration by parts, or differentiating the result directly,

\int -xe^{-x}\,dx = (x+1)e^{-x}.

Similarly,

u_2' = \frac{e^x(1-x)}{e^x(1-x)} = 1,

so

u_2=x.

Constants arising in u_1,u_2 may be omitted because they reproduce complementary-function terms.

Hence

y_p = u_1y_1+u_2y_2 = (x+1)e^{-x}e^x+x\cdot x = x+1+x^2.

Step 3: General solution

y=y_c+y_p = C_1e^x+C_2x+x^2+x+1.

Verification of the particular part:

For y_p=x^2+x+1,

y_p'=2x+1,\qquad y_p''=2.

Therefore

(1-x)2+x(2x+1)-(x^2+x+1) = x^2-2x+1 = (1-x)^2,

as required.

Final Answer:

\boxed{y=C_1e^x+C_2x+x^2+x+1.}

Solution:

To prove:

2nJ_n(x)=x\left[J_{n-1}(x)+J_{n+1}(x)\right].

We use the standard series representation of the Bessel function of the first kind:

J_n(x) = \sum_{r=0}^{\infty} \frac{(-1)^r}{r!\,\Gamma(n+r+1)} \left(\frac{x}{2}\right)^{n+2r}.

Step 1: Derive the first differential recurrence

Multiply by x^n:

x^nJ_n(x) = \sum_{r=0}^{\infty} \frac{(-1)^r x^{2n+2r}} {2^{n+2r}r!\,\Gamma(n+r+1)}.

Differentiate term by term:

\frac{d}{dx}\left(x^nJ_n(x)\right) = \sum_{r=0}^{\infty} \frac{(-1)^r(2n+2r)x^{2n+2r-1}} {2^{n+2r}r!\,\Gamma(n+r+1)}.

Since

\Gamma(n+r+1)=(n+r)\Gamma(n+r),

we have

\frac{2n+2r}{\Gamma(n+r+1)} = \frac{2(n+r)}{(n+r)\Gamma(n+r)} = \frac{2}{\Gamma(n+r)}.

Therefore

\frac{d}{dx}\left(x^nJ_n(x)\right) = x^nJ_{n-1}(x).

Expanding the derivative,

nx^{n-1}J_n+x^nJ_n' = x^nJ_{n-1}.

Dividing by x^n,

J_n'+\frac{n}{x}J_n=J_{n-1}. \tag{1}

Step 2: Derive the second differential recurrence

From the series for J_n,

x^{-n}J_n(x) = \sum_{r=0}^{\infty} \frac{(-1)^r x^{2r}} {2^{n+2r}r!\,\Gamma(n+r+1)}.

Differentiating term by term, the r=0 term vanishes and

\frac{d}{dx}\left(x^{-n}J_n(x)\right) = \sum_{r=1}^{\infty} \frac{(-1)^r(2r)x^{2r-1}} {2^{n+2r}r!\,\Gamma(n+r+1)}.

Put r=s+1. Then

\frac{d}{dx}\left(x^{-n}J_n(x)\right) = -\sum_{s=0}^{\infty} \frac{(-1)^s x^{2s+1}} {2^{n+2s+1}s!\,\Gamma(n+s+2)}.

But

x^{-n}J_{n+1}(x) = \sum_{s=0}^{\infty} \frac{(-1)^s x^{2s+1}} {2^{n+2s+1}s!\,\Gamma(n+s+2)}.

Therefore

\frac{d}{dx}\left(x^{-n}J_n(x)\right) = -x^{-n}J_{n+1}(x).

Expanding the derivative,

-nx^{-n-1}J_n+x^{-n}J_n' = -x^{-n}J_{n+1}.

Multiplying by x^n,

J_n'-\frac{n}{x}J_n=-J_{n+1}. \tag{2}

Step 3: Eliminate J_n'

Subtract (2) from (1):

\left(J_n'+\frac{n}{x}J_n\right) - \left(J_n'-\frac{n}{x}J_n\right) = J_{n-1}+J_{n+1}.

Thus

\frac{2n}{x}J_n = J_{n-1}+J_{n+1}.

Multiplying by x,

2nJ_n(x) = x\left[J_{n-1}(x)+J_{n+1}(x)\right].

Final Answer:

\boxed{2nJ_n(x)=x\left[J_{n-1}(x)+J_{n+1}(x)\right].}

Hence the required Bessel recurrence relation is established.

Solution:

Let a sufficiently smooth space curve be represented by the position vector \vec r=\vec r(s), where s is arc length.

Define the unit tangent, principal normal and binormal by

\mathbf T=\frac{d\vec r}{ds}, \qquad \mathbf N=\frac{1}{\kappa}\frac{d\mathbf T}{ds}, \qquad \mathbf B=\mathbf T\times\mathbf N,

where \kappa is the curvature. Since s is arc length,

\|\mathbf T\|=1.

First Serret-Frenet formula

From \mathbf T\cdot\mathbf T=1, differentiate with respect to s:

2\mathbf T\cdot\frac{d\mathbf T}{ds}=0.

Hence d\mathbf T/ds is perpendicular to \mathbf T. By definition of the principal normal,

\frac{d\mathbf T}{ds}=\kappa\mathbf N. \tag{1}

If \rho=1/\kappa is the radius of curvature, this becomes

\frac{d\mathbf T}{ds}=\frac{\mathbf N}{\rho}.

Derivative of \mathbf N

Since \mathbf N\cdot\mathbf N=1,

\mathbf N\cdot\frac{d\mathbf N}{ds}=0,

so d\mathbf N/ds has no component along \mathbf N. Therefore it can be written as

\frac{d\mathbf N}{ds}=a\mathbf T+b\mathbf B.

Differentiate \mathbf T\cdot\mathbf N=0:

\frac{d\mathbf T}{ds}\cdot\mathbf N + \mathbf T\cdot\frac{d\mathbf N}{ds} = 0.

Using (1),

\kappa+a=0 \quad\Rightarrow\quad a=-\kappa.

The remaining coefficient is defined as the torsion \tau:

b=\tau.

Therefore

\frac{d\mathbf N}{ds} = -\kappa\mathbf T+\tau\mathbf B. \tag{2}

Derivative of \mathbf B

Since

\mathbf B=\mathbf T\times\mathbf N,

differentiate:

\frac{d\mathbf B}{ds} = \frac{d\mathbf T}{ds}\times\mathbf N + \mathbf T\times\frac{d\mathbf N}{ds}.

Using (1) and (2),

\frac{d\mathbf B}{ds} = \kappa\mathbf N\times\mathbf N + \mathbf T\times(-\kappa\mathbf T+\tau\mathbf B).

Now \mathbf N\times\mathbf N=0, \mathbf T\times\mathbf T=0, and for the right-handed triad (\mathbf T,\mathbf N,\mathbf B),

\mathbf T\times\mathbf B=-\mathbf N.

Hence

\frac{d\mathbf B}{ds} = -\tau\mathbf N. \tag{3}

Serret-Frenet formulas:

\boxed{ \frac{d\mathbf T}{ds}=\kappa\mathbf N,\qquad \frac{d\mathbf N}{ds}=-\kappa\mathbf T+\tau\mathbf B,\qquad \frac{d\mathbf B}{ds}=-\tau\mathbf N. }

If \rho=1/\kappa is the radius of curvature and \sigma=1/\tau is the radius of torsion, the same formulas are

\boxed{ \frac{d\mathbf T}{ds}=\frac{\mathbf N}{\rho},\qquad \frac{d\mathbf N}{ds}=-\frac{\mathbf T}{\rho}+\frac{\mathbf B}{\sigma},\qquad \frac{d\mathbf B}{ds}=-\frac{\mathbf N}{\sigma}. }

Final Answer:

The Serret-Frenet system is

\boxed{ \frac{d\mathbf T}{ds}=\kappa\mathbf N,\qquad \frac{d\mathbf N}{ds}=-\kappa\mathbf T+\tau\mathbf B,\qquad \frac{d\mathbf B}{ds}=-\tau\mathbf N. }

Solution:

Stokes’ theorem: If S is an oriented smooth surface bounded by a positively oriented closed curve C, then

\oint_C\vec F\cdot d\vec r = \iint_S(\nabla\times\vec F)\cdot\hat n\,dS.

Here

\vec F=z\mathbf i+x\mathbf j+y\mathbf k.

The surface is the upper unit hemisphere

z=\sqrt{1-x^2-y^2},

and C is its boundary x^2+y^2=1,\;z=0. We choose the outward normal, so the positive orientation of C is counter-clockwise when viewed from the positive z-axis.

Left-hand side: line integral around C

Parametrize the unit circle by

x=\cos t,\qquad y=\sin t,\qquad z=0,\qquad 0\le t\le2\pi.

Then

dx=-\sin t\,dt,\qquad dy=\cos t\,dt,\qquad dz=0.

Since

\vec F\cdot d\vec r = z\,dx+x\,dy+y\,dz,

we have on C

\vec F\cdot d\vec r = 0\cdot dx+\cos t(\cos t\,dt)+\sin t\cdot0 = \cos^2t\,dt.

Therefore

\oint_C\vec F\cdot d\vec r = \int_0^{2\pi}\cos^2t\,dt.

Using \cos^2t=(1+\cos2t)/2,

\int_0^{2\pi}\cos^2t\,dt = \left[\frac t2+\frac{\sin2t}{4}\right]_0^{2\pi} = \pi. \tag{1}

Right-hand side: surface integral of the curl

Compute the curl:

\nabla\times\vec F = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ \partial/\partial x & \partial/\partial y & \partial/\partial z\\ z & x & y \end{vmatrix}.

Hence

\nabla\times\vec F = (1-0)\mathbf i+(1-0)\mathbf j+(1-0)\mathbf k = \mathbf i+\mathbf j+\mathbf k.

For the upper hemisphere regarded as z=z(x,y),

z_x=-\frac{x}{z}, \qquad z_y=-\frac{y}{z}.

Thus the upward/outward vector surface element is

\hat n\,dS = (-z_x,-z_y,1)\,dx\,dy = \left(\frac{x}{z},\frac{y}{z},1\right)dx\,dy.

Therefore

(\nabla\times\vec F)\cdot\hat n\,dS = \left(\frac{x}{z}+\frac{y}{z}+1\right)dx\,dy.

The projection is the unit disk D:x^2+y^2\le1. By symmetry, the integrals of x/z and y/z over D are zero because these functions are odd in x and y, respectively. Hence

\iint_S(\nabla\times\vec F)\cdot\hat n\,dS = \iint_D1\,dx\,dy = \text{area}(D) = \pi. \tag{2}

From (1) and (2),

\oint_C\vec F\cdot d\vec r = \iint_S(\nabla\times\vec F)\cdot\hat n\,dS = \pi.

Final Answer:

\boxed{\text{LHS}=\text{RHS}=\pi.}

Hence Stokes’ theorem is verified.

Solution:

Let the common length of each of the five rods be a. The rods are the four sides AB,BC,CD,DA of the rhombus together with its diagonal BD. Since all five rods have equal length,

AB=AD=BD=a \qquad\text{and}\qquad BC=CD=BD=a.

Therefore both triangles ABD and BCD are equilateral.

By symmetry, when the system is suspended from A and carries the weight W at C, the line AC is vertical and the diagonal BD is horizontal. Each of the rods BC and CD makes an angle 30^\circ with the vertical.

Step 1: Equilibrium of joint C

Let the force in each of the two rods BC and CD be T. By symmetry these two forces are equal.

At joint C, the horizontal components cancel. The sum of the upward vertical components must balance the downward weight W:

2T\cos30^\circ=W.

Since \cos30^\circ=\sqrt3/2,

2T\left(\frac{\sqrt3}{2}\right)=W,

so

T=\frac{W}{\sqrt3}. \tag{1}

Thus rods BC and CD carry equal tension W/\sqrt3.

Step 2: Forces in the upper rods

The whole framework is symmetric about the vertical line AC. The support reaction at A is W upward. If the common force in AB and AD is T_1, vertical equilibrium at A gives

2T_1\cos30^\circ=W.

Hence

T_1=\frac{W}{\sqrt3}=T. \tag{2}

Step 3: Equilibrium of joint B

At joint B, rods BA and BC each exert a force T=W/\sqrt3. Their vertical components are equal and opposite, so they cancel.

Each has horizontal component

T\sin30^\circ=\frac{T}{2}.

Both horizontal components act in the same direction, and their resultant is

\frac{T}{2}+\frac{T}{2}=T.

For equilibrium, rod BD must exert an equal force in the opposite direction at joint B. Such a force pushes the joint away from the other end of BD; hence BD is in compression, i.e. it carries a thrust.

Using (1), its magnitude is

T=\frac{W}{\sqrt3}.

Final Answer:

\boxed{\text{Thrust in }BD=\frac{W}{\sqrt3}.}

Solution:

Take the first force to have magnitude P and act along the x-axis y=0,z=0. Let the second force have magnitude Q and act parallel to the y-axis along x=0,z=c.

Thus the two force vectors may be written

\vec P=P\mathbf i, \qquad \vec Q=Q\mathbf j.

Step 1: Resultant force

\vec R = P\mathbf i+Q\mathbf j = (P,Q,0).

Hence

R^2=P^2+Q^2.

Step 2: Resultant moment about the origin

The force \vec P acts through the origin, so its moment about the origin is zero.

A point on the line of action of \vec Q is (0,0,c). Therefore its position vector is

\vec a=c\mathbf k.

The moment of \vec Q about the origin is

\vec M = \vec a\times\vec Q = c\mathbf k\times Q\mathbf j = -cQ\mathbf i.

Thus

\vec M=(-cQ,0,0).

Step 3: A point on the central axis

For a force system having resultant \vec R and resultant moment \vec M about the origin, a point \vec r_0 on the central axis is

\vec r_0 = \frac{\vec R\times\vec M}{R^2}.

Compute the cross product:

\vec R\times\vec M = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ P & Q & 0\\ -cQ & 0 & 0 \end{vmatrix} = cQ^2\mathbf k.

Therefore

\vec r_0 = \left(0,0,\frac{cQ^2}{P^2+Q^2}\right).

The central axis is parallel to \vec R=(P,Q,0). Hence its parametric equations are

x=\lambda P,\qquad y=\lambda Q,\qquad z=\frac{cQ^2}{P^2+Q^2}.

From the first two equations, for a non-zero resultant,

\frac{y}{x}=\frac{Q}{P}.

Also

\frac{Q^2}{P^2+Q^2} = \frac{(Q/P)^2}{1+(Q/P)^2} = \frac{(y/x)^2}{1+(y/x)^2} = \frac{y^2}{x^2+y^2}.

Hence

z = c\frac{y^2}{x^2+y^2}.

Multiplying by x^2+y^2,

(x^2+y^2)z=cy^2.

As P and Q vary, the corresponding central axes therefore generate this surface.

Final Answer:

\boxed{(x^2+y^2)z=cy^2.}

Solution:

Let v be the speed of the particle, s the arc length, \psi the angle made by the tangent with a fixed axis, and \rho the radius of curvature.

The tangential and normal accelerations are respectively

a_t=\frac{dv}{dt}, \qquad a_n=\frac{v^2}{\rho}.

Also, since

\frac{d\psi}{ds}=\frac1\rho \quad\text{and}\quad \frac{ds}{dt}=v,

the angular velocity of the tangent is

\frac{d\psi}{dt} = \frac{d\psi}{ds}\frac{ds}{dt} = \frac{v}{\rho}.

It is given that this angular velocity is constant. Write

\frac{d\psi}{dt}=\omega,

where \omega is constant. Therefore

\frac{v}{\rho}=\omega \quad\Rightarrow\quad \frac{v^2}{\rho}=v\omega.

The tangential and normal accelerations are equal, so

\frac{dv}{dt}=v\omega.

Because d\psi/dt=\omega,

\frac{dv}{dt} = \frac{dv}{d\psi}\frac{d\psi}{dt} = \omega\frac{dv}{d\psi}.

Hence

\omega\frac{dv}{d\psi}=v\omega.

For \omega\ne0,

\frac{dv}{d\psi}=v.

Separating variables,

\frac{dv}{v}=d\psi.

Integrating,

\log v=\psi+\log C,

so

v=Ce^\psi.

From v/\rho=\omega,

\rho=\frac{C}{\omega}e^\psi.

Put K=C/\omega. Since ds/d\psi=\rho,

\frac{ds}{d\psi}=Ke^\psi.

Now

\frac{dx}{ds}=\cos\psi, \qquad \frac{dy}{ds}=\sin\psi.

Therefore

\frac{dx}{d\psi} = \frac{dx}{ds}\frac{ds}{d\psi} = Ke^\psi\cos\psi,

and

\frac{dy}{d\psi} = Ke^\psi\sin\psi.

Integrating,

x-x_0 = \frac K2 e^\psi(\cos\psi+\sin\psi), y-y_0 = \frac K2 e^\psi(\sin\psi-\cos\psi).

Choose the translated origin (x_0,y_0) as pole. Then

r^2 = x^2+y^2 = \frac{K^2}{2}e^{2\psi},

so

r=\frac{K}{\sqrt2}e^\psi.

Moreover, the above parametric expressions show that the polar angle \theta satisfies

\theta=\psi-\frac{\pi}{4}.

Thus \psi=\theta+\pi/4, and hence

r = \frac{K}{\sqrt2}e^{\theta+\pi/4} = Ae^\theta,

where A is a positive constant.

Final Answer:

The path is a logarithmic (equiangular) spiral:

\boxed{r=Ae^\theta}

up to a rotation of the polar axis and a translation of the origin.

Solution:

Let \mu/r^2 be the attractive acceleration towards the focus, where r is the instantaneous distance from the centre of force. The particle moves in an ellipse with the centre of force at a focus.

Let a be the semi-major axis of the ellipse.

Step 1: Determine a from the initial speed and distance

For inverse-square attraction, the specific mechanical energy is

E=\frac{v^2}{2}-\frac{\mu}{r}.

For an elliptic orbit of semi-major axis a, the same constant energy is

E=-\frac{\mu}{2a}.

At the point of projection the speed is V and the distance from the focus is r. Therefore

\frac{V^2}{2}-\frac{\mu}{r} = -\frac{\mu}{2a}.

Multiply by 2:

V^2-\frac{2\mu}{r} = -\frac{\mu}{a}.

Hence

\frac{\mu}{a} = \frac{2\mu}{r}-V^2.

Dividing by \mu,

\frac1a = \frac2r-\frac{V^2}{\mu}.

Thus

a = \left(\frac2r-\frac{V^2}{\mu}\right)^{-1}. \tag{1}

For an ellipse the bracket is positive.

Step 2: Obtain the period of an elliptic inverse-square orbit

Let h be the angular momentum per unit mass. Conservation of angular momentum gives the constant areal velocity

\frac{dA}{dt}=\frac h2.

For an ellipse of semi-major axis a, semi-minor axis b, and eccentricity e,

b=a\sqrt{1-e^2}.

The area of the ellipse is

\pi ab = \pi a^2\sqrt{1-e^2}.

For an inverse-square orbit, the semi-latus rectum p satisfies

p=\frac{h^2}{\mu}.

For an ellipse,

p=a(1-e^2).

Therefore

h^2=\mu a(1-e^2),

so

h=\sqrt{\mu a(1-e^2)}.

The periodic time T equals total area divided by areal velocity:

T = \frac{\pi a^2\sqrt{1-e^2}}{h/2} = \frac{2\pi a^2\sqrt{1-e^2}} {\sqrt{\mu a(1-e^2)}}.

Cancel \sqrt{1-e^2} and simplify a^2/\sqrt a=a^{3/2}:

T=\frac{2\pi}{\sqrt\mu}a^{3/2}. \tag{2}

Step 3: Substitute the value of a

From (1),

a^{3/2} = \left(\frac2r-\frac{V^2}{\mu}\right)^{-3/2}.

Substituting into (2),

T = \frac{2\pi}{\sqrt\mu} \left[ \frac2r-\frac{V^2}{\mu} \right]^{-3/2}.

Final Answer:

\boxed{ T= \frac{2\pi}{\sqrt{\mu}} \left[ \frac{2}{r}-\frac{V^2}{\mu} \right]^{-3/2} }.

Hence the required expression for the periodic time is proved.