Solution:

Given:

x\frac{dy}{dx}+(2-x)y=e^{3x},\qquad x>0.

Standard form and method:

A first-order linear differential equation is written as

\frac{dy}{dx}+P(x)y=Q(x),

and its integrating factor is

\mathrm{I.F.}=e^{\int P(x)\,dx}.

Since x>0, divide the given equation by x:

\frac{dy}{dx}+\left(\frac{2}{x}-1\right)y=\frac{e^{3x}}{x}.

Thus

P(x)=\frac{2}{x}-1,\qquad Q(x)=\frac{e^{3x}}{x}.

Integrating factor:

\begin{aligned} \mathrm{I.F.} &=e^{\int\left(\frac{2}{x}-1\right)dx}\\ &=e^{\,2\ln x-x}\\ &=x^2e^{-x}, \end{aligned}

where \ln x is valid because x>0.

Multiplying the standard form by x^2e^{-x},

x^2e^{-x}\frac{dy}{dx} +x^2e^{-x}\left(\frac{2}{x}-1\right)y =x^2e^{-x}\frac{e^{3x}}{x}.

The left-hand side is the derivative of x^2e^{-x}y. Indeed,

\begin{aligned} \frac{d}{dx}\left(x^2e^{-x}y\right) &=x^2e^{-x}\frac{dy}{dx} +y\frac{d}{dx}\left(x^2e^{-x}\right)\\ &=x^2e^{-x}\frac{dy}{dx} +y\left(2xe^{-x}-x^2e^{-x}\right)\\ &=x^2e^{-x}\frac{dy}{dx} +x^2e^{-x}\left(\frac{2}{x}-1\right)y. \end{aligned}

Therefore,

\frac{d}{dx}\left(x^2e^{-x}y\right)=xe^{2x}.

Integrating both sides,

x^2e^{-x}y=\int xe^{2x}\,dx+C.

Evaluate the integral by parts. Take

u=x,\qquad dv=e^{2x}dx,\qquad du=dx,\qquad v=\frac12e^{2x}.

Hence

\begin{aligned} \int xe^{2x}\,dx &=\frac{x}{2}e^{2x}-\frac12\int e^{2x}dx\\ &=\frac{x}{2}e^{2x}-\frac14e^{2x}\\ &=\frac{e^{2x}}{4}(2x-1). \end{aligned}

Thus

x^2e^{-x}y=\frac{e^{2x}}4(2x-1)+C.

Multiplying by e^x/x^2,

y=\frac{e^{3x}(2x-1)}{4x^2}+\frac{Ce^x}{x^2}.

Verification:

Since the solution was obtained from

\frac{d}{dx}(x^2e^{-x}y)=xe^{2x},

multiplying this identity by e^x/x gives back

x\frac{dy}{dx}+(2-x)y=e^{3x}.

Final Answer:

\boxed{\,y=\frac{e^{3x}(2x-1)}{4x^2}+\frac{Ce^x}{x^2},\qquad x>0\,}.

Solution:

Relevant result: If a linear transformation T:\mathbb{R}^n\to\mathbb{R}^m has standard matrix A, then

\operatorname{Im}(T)=\operatorname{Col}(A),

the column space of A. Therefore, it is enough to choose the columns of A so that they span precisely the plane x+y+z=0.

Step 1: Find a basis of the required plane.

Let (u,v,w) be any vector in the plane. Then

u+v+w=0\quad\Longrightarrow\quad w=-u-v.

Hence

\begin{pmatrix}u\\v\\w\end{pmatrix} = \begin{pmatrix}u\\v\\-u-v\end{pmatrix} = u\begin{pmatrix}1\\0\\-1\end{pmatrix} + v\begin{pmatrix}0\\1\\-1\end{pmatrix}.

Thus the plane is spanned by

v_1=\begin{pmatrix}1\\0\\-1\end{pmatrix}, \qquad v_2=\begin{pmatrix}0\\1\\-1\end{pmatrix}.

These vectors are linearly independent: if \alpha v_1+\beta v_2=0, then the first coordinate gives \alpha=0 and the second gives \beta=0. Therefore \{v_1,v_2\} is a basis of the plane.

Step 2: Construct a transformation.

Choose the standard matrix

A= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -1 & -1 & 0 \end{pmatrix}.

Its first two columns are v_1,v_2, while the third column is zero. Therefore its column space is exactly \operatorname{span}\{v_1,v_2\}.

For (x,y,z)\in\mathbb{R}^3, define

\begin{aligned} T(x,y,z) &= A\begin{pmatrix}x\\y\\z\end{pmatrix}\\ &= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -1 & -1 & 0 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix}\\ &= \begin{pmatrix}x\\y\\-x-y\end{pmatrix}. \end{aligned}

Verification that the image is contained in the plane:

If T(x,y,z)=(X,Y,Z), then

X=x,\qquad Y=y,\qquad Z=-x-y,

and therefore

X+Y+Z=x+y-x-y=0.

Thus every image vector lies in the required plane.

Verification that every vector of the plane occurs as an image:

Let (X,Y,Z) satisfy X+Y+Z=0. Then Z=-X-Y, and

T(X,Y,0)=(X,Y,-X-Y)=(X,Y,Z).

Hence every vector of the plane belongs to \operatorname{Im}(T). Therefore the image is exactly the plane.

Final Answer:

\boxed{T(x,y,z)=(x,\ y,\ -x-y)}

is one required linear transformation, and

\boxed{\operatorname{Im}(T)=\{(X,Y,Z)\in\mathbb{R}^3:X+Y+Z=0\}}.

Solution:

Relevant theorem: Intermediate Value Theorem.

If a real-valued function g is continuous on [a,b] and a number L lies between g(a) and g(b), then there exists c\in[a,b] such that g(c)=L. In particular, if g(a)g(b)\le 0, then g(c)=0 for some c\in[a,b].

Define an auxiliary function on [0,3] by

g(x)=f(x)-f(x+3).

Since f is continuous on [0,6], both f(x) and f(x+3) are continuous for 0\le x\le3. Therefore g is continuous on [0,3].

At the two endpoints,

g(0)=f(0)-f(3),

whereas

\begin{aligned} g(3) &=f(3)-f(6)\\ &=f(3)-f(0)\qquad\text{because }f(6)=f(0)\\ &=-\{f(0)-f(3)\}\\ &=-g(0). \end{aligned}

There are now two possibilities.

1. If g(0)=0, then f(0)=f(3), and the required conclusion holds with x=0.
2. If g(0)\ne0, then g(3)=-g(0), so g(0) and g(3) have opposite signs. Hence g(0)g(3)<0. By the Intermediate Value Theorem there exists x_0\in(0,3) such that g(x_0)=0.

Thus, in either case, there exists x_0\in[0,3] for which

g(x_0)=f(x_0)-f(x_0+3)=0.

Therefore

f(x_0)=f(x_0+3).

Final Answer:

The required point exists; namely,

\boxed{\exists\,x\in[0,3]\text{ such that }f(x)=f(x+3).}

Hence the required result is established.

Solution:

The statement specifies only that the particle is constrained to the surface of a right circular cylinder of radius b. Therefore a unique numerical velocity cannot be obtained without an additional law of motion. The general velocity of a particle constrained to such a cylinder can, however, be found completely.

Take the axis of the cylinder to be the z-axis. Its equation is

x^2+y^2=b^2.

Introduce the angular coordinate \theta=\theta(t) and the axial coordinate z=z(t). Then a general point on the cylindrical surface is

\mathbf r(t)=b\cos\theta\,\mathbf i+b\sin\theta\,\mathbf j+z\,\mathbf k.

Differentiating with respect to time,

\begin{aligned} \mathbf v =\frac{d\mathbf r}{dt} &=-b\sin\theta\,\dot\theta\,\mathbf i +b\cos\theta\,\dot\theta\,\mathbf j +\dot z\,\mathbf k\\ &=b\dot\theta(-\sin\theta\,\mathbf i+\cos\theta\,\mathbf j) +\dot z\,\mathbf k. \end{aligned}

If \mathbf e_\theta=-\sin\theta\,\mathbf i+\cos\theta\,\mathbf j is the unit vector in the circumferential direction, then

\mathbf v=b\dot\theta\,\mathbf e_\theta+\dot z\,\mathbf k.

The two components are perpendicular. Hence the speed is

|\mathbf v| =\sqrt{(b\dot\theta)^2+\dot z^{\,2}}.

The same tangency condition follows directly from the cylinder equation. Differentiating x^2+y^2=b^2,

2x\dot x+2y\dot y=0\quad\Longrightarrow\quad x\dot x+y\dot y=0,

showing that the velocity has no component normal to the cylindrical surface.

Final Answer:

\boxed{\mathbf v=(-b\dot\theta\sin\theta,\ b\dot\theta\cos\theta,\ \dot z)}

or, equivalently,

\boxed{\mathbf v=b\dot\theta\,\mathbf e_\theta+\dot z\,\mathbf k,\qquad |\mathbf v|=\sqrt{b^2\dot\theta^{\,2}+\dot z^{\,2}}.}

Solution:

Definition of a basis:

A subset B=\{v_1,\ldots,v_r\} of a vector space V is called a basis of V if:

1. B is linearly independent, i.e. c_1v_1+\cdots+c_rv_r=0 implies c_1=\cdots=c_r=0; and
2. B spans V, i.e. every vector of V is a linear combination of v_1,\ldots,v_r.

Let the given vectors be the rows of the matrix

A= \begin{pmatrix} 1&1&0&-1\\ 2&4&6&0\\ -2&-3&-3&1\\ -1&-2&-2&2\\ 4&6&4&-6 \end{pmatrix}.

Relevant result: Elementary row operations replace rows by linear combinations of the old rows through an invertible operation. Hence row-equivalent matrices have the same row space. The nonzero rows of a row-echelon form are linearly independent and therefore form a basis of the row space.

Step 1: Eliminate the first column below R_1.

Apply

R_2\to R_2-2R_1,\quad R_3\to R_3+2R_1,\quad R_4\to R_4+R_1,\quad R_5\to R_5-4R_1.

Calculate each new row:

\begin{aligned} R_2&=(2,4,6,0)-2(1,1,0,-1)=(0,2,6,2),\\ R_3&=(-2,-3,-3,1)+2(1,1,0,-1)=(0,-1,-3,-1),\\ R_4&=(-1,-2,-2,2)+(1,1,0,-1)=(0,-1,-2,1),\\ R_5&=(4,6,4,-6)-4(1,1,0,-1)=(0,2,4,-2). \end{aligned}

Therefore

A\sim \begin{pmatrix} 1&1&0&-1\\ 0&2&6&2\\ 0&-1&-3&-1\\ 0&-1&-2&1\\ 0&2&4&-2 \end{pmatrix}.

Step 2: Make the second pivot equal to 1.

Use R_2\to\frac12R_2:

\sim \begin{pmatrix} 1&1&0&-1\\ 0&1&3&1\\ 0&-1&-3&-1\\ 0&-1&-2&1\\ 0&2&4&-2 \end{pmatrix}.

Step 3: Eliminate the second column below R_2.

Apply

R_3\to R_3+R_2,\qquad R_4\to R_4+R_2,\qquad R_5\to R_5-2R_2.

Then

\begin{aligned} R_3&=(0,-1,-3,-1)+(0,1,3,1)=(0,0,0,0),\\ R_4&=(0,-1,-2,1)+(0,1,3,1)=(0,0,1,2),\\ R_5&=(0,2,4,-2)-2(0,1,3,1)=(0,0,-2,-4). \end{aligned}

Thus

\sim \begin{pmatrix} 1&1&0&-1\\ 0&1&3&1\\ 0&0&0&0\\ 0&0&1&2\\ 0&0&-2&-4 \end{pmatrix}.

Interchange R_3 and R_4:

\sim \begin{pmatrix} 1&1&0&-1\\ 0&1&3&1\\ 0&0&1&2\\ 0&0&0&0\\ 0&0&-2&-4 \end{pmatrix}.

Finally, eliminate the third entry of R_5 by R_5\to R_5+2R_3:

\sim \begin{pmatrix} 1&1&0&-1\\ 0&1&3&1\\ 0&0&1&2\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}.

The three nonzero rows are linearly independent because each has a leading entry in a new column. Since all row operations preserve the row space, these rows span the same subspace as the original five vectors.

Final Answer:

A basis of the subspace generated by the given subset is

\boxed{ \left\{ (1,1,0,-1),\; (0,1,3,1),\; (0,0,1,2) \right\}}.

Hence the dimension of the generated subspace is

\boxed{3}.

Solution:

Important orientation of the diagonalization:

The usual eigenvector matrix S satisfies S^{-1}AS=D. Here the question specifically asks for a matrix P satisfying

PAP^{-1}=D.

Therefore, after constructing the usual eigenvector matrix S, we must take P=S^{-1} (or any nonzero scalar multiple of it).

Given

A=\begin{pmatrix}2&2\\1&3\end{pmatrix}.

Step 1: Find the eigenvalues.

The characteristic equation is

\det(A-\lambda I)=0.

Thus

\begin{aligned} 0 &=\det \begin{pmatrix} 2-\lambda&2\\ 1&3-\lambda \end{pmatrix}\\ &=(2-\lambda)(3-\lambda)-2\\ &=6-2\lambda-3\lambda+\lambda^2-2\\ &=\lambda^2-5\lambda+4\\ &=(\lambda-4)(\lambda-1). \end{aligned}

Hence

\lambda_1=4,\qquad \lambda_2=1.

Step 2: Find an eigenvector for \lambda_1=4.

We solve

(A-4I)\begin{pmatrix}x\\y\end{pmatrix}=0.

Now

A-4I= \begin{pmatrix} -2&2\\ 1&-1 \end{pmatrix},

so the equations are

-2x+2y=0,\qquad x-y=0.

Both give x=y. Taking x=y=1, one eigenvector is

v_1=\begin{pmatrix}1\\1\end{pmatrix}.

Step 3: Find an eigenvector for \lambda_2=1.

We solve

(A-I)\begin{pmatrix}x\\y\end{pmatrix}=0.

Since

A-I= \begin{pmatrix} 1&2\\ 1&2 \end{pmatrix},

we obtain

x+2y=0\quad\Longrightarrow\quad x=-2y.

Taking y=1,

v_2=\begin{pmatrix}-2\\1\end{pmatrix}.

Step 4: Form the usual eigenvector matrix.

Put the eigenvectors as columns:

S= \begin{pmatrix} 1&-2\\ 1&1 \end{pmatrix}.

Its determinant is

\det S=(1)(1)-(-2)(1)=3\ne0,

so S is invertible.

Its inverse is

\begin{aligned} S^{-1} &=\frac1{\det S} \begin{pmatrix} 1&2\\ -1&1 \end{pmatrix}\\ &=\frac13 \begin{pmatrix} 1&2\\ -1&1 \end{pmatrix}. \end{aligned}

Since the question asks for PAP^{-1}, choose P=S^{-1}. A common nonzero scalar factor cancels in PAP^{-1}, so for simpler arithmetic we may take

P= \begin{pmatrix} 1&2\\ -1&1 \end{pmatrix}.

Then

\det P=(1)(1)-2(-1)=3\ne0,

and

P^{-1} =\frac13 \begin{pmatrix} 1&-2\\ 1&1 \end{pmatrix}.

Step 5: Verify the exact product PAP^{-1}.

First,

\begin{aligned} PA &= \begin{pmatrix} 1&2\\ -1&1 \end{pmatrix} \begin{pmatrix} 2&2\\ 1&3 \end{pmatrix}\\ &= \begin{pmatrix} 1\cdot2+2\cdot1 & 1\cdot2+2\cdot3\\ -1\cdot2+1\cdot1 & -1\cdot2+1\cdot3 \end{pmatrix}\\ &= \begin{pmatrix} 4&8\\ -1&1 \end{pmatrix}. \end{aligned}

Therefore

\begin{aligned} PAP^{-1} &= \begin{pmatrix} 4&8\\ -1&1 \end{pmatrix} \frac13 \begin{pmatrix} 1&-2\\ 1&1 \end{pmatrix}\\ &=\frac13 \begin{pmatrix} 4+8 & -8+8\\ -1+1 & 2+1 \end{pmatrix}\\ &=\frac13 \begin{pmatrix} 12&0\\ 0&3 \end{pmatrix}\\ &= \begin{pmatrix} 4&0\\ 0&1 \end{pmatrix}. \end{aligned}

Final Answer:

One suitable invertible matrix is

\boxed{ P=\begin{pmatrix}1&2\\-1&1\end{pmatrix}}

for which

\boxed{ PAP^{-1}= \begin{pmatrix}4&0\\0&1\end{pmatrix}}.

Solution:

Let the three mutually perpendicular tangent planes meet at the point

P=(X,Y,Z).

Write the conicoid as

ax^2+by^2+cz^2=1.

Tangent-plane formula:

At a point Q_i=(x_i,y_i,z_i) on the conicoid, the tangent plane is

ax_ix+by_iy+cz_iz=1.

Its normal vector is

\mathbf n_i=(ax_i,\ by_i,\ cz_i).

Because P lies on this tangent plane,

\mathbf n_i\cdot\mathbf P =ax_iX+by_iY+cz_iZ=1, \qquad i=1,2,3.

The three tangent planes are mutually perpendicular. Hence their normal vectors \mathbf n_1,\mathbf n_2,\mathbf n_3 are mutually perpendicular. Put

N_i=|\mathbf n_i|, \qquad \mathbf e_i=\frac{\mathbf n_i}{N_i}.

Then \mathbf e_1,\mathbf e_2,\mathbf e_3 form an orthonormal basis.

From \mathbf n_i\cdot\mathbf P=1,

\mathbf e_i\cdot\mathbf P=\frac1{N_i}.

Since \{\mathbf e_1,\mathbf e_2,\mathbf e_3\} is orthonormal, the squared length of \mathbf P is the sum of the squares of its components in this basis:

X^2+Y^2+Z^2 =|\mathbf P|^2 =\sum_{i=1}^3(\mathbf e_i\cdot\mathbf P)^2 =\sum_{i=1}^3\frac1{N_i^2}.

We now evaluate this sum by using the fact that each Q_i lies on the conicoid. Let

A=\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}.

Since \mathbf n_i=AQ_i, we have Q_i=A^{-1}\mathbf n_i. The condition Q_i^TAQ_i=1 therefore becomes

\mathbf n_i^TA^{-1}\mathbf n_i=1.

Substitute \mathbf n_i=N_i\mathbf e_i:

N_i^2\,\mathbf e_i^TA^{-1}\mathbf e_i=1.

Hence

\frac1{N_i^2}=\mathbf e_i^TA^{-1}\mathbf e_i.

Summing for i=1,2,3,

\sum_{i=1}^3\frac1{N_i^2} = \sum_{i=1}^3\mathbf e_i^TA^{-1}\mathbf e_i.

For any orthonormal basis, the sum on the right is the trace of the linear operator A^{-1}. Therefore

\sum_{i=1}^3\frac1{N_i^2} =\operatorname{tr}(A^{-1}) =\frac1a+\frac1b+\frac1c.

Consequently,

X^2+Y^2+Z^2=\frac1a+\frac1b+\frac1c.

Final Answer:

The locus of the common point of three mutually perpendicular tangent planes is the sphere

\boxed{x^2+y^2+z^2=\frac1a+\frac1b+\frac1c}.

Solution:

To prove that

\lim_{x\to0}\cos\frac1x

does not exist, it is sufficient to find two sequences tending to 0 along which the function has two different limiting values.

Sequential criterion for a limit:

If \lim_{x\to a}f(x)=L exists, then for every sequence x_n\to a with x_n\ne a, we must have f(x_n)\to L. Therefore, if two such sequences produce different limits, the original limit cannot exist.

Choose

x_n=\frac1{2\pi n},\qquad n=1,2,3,\ldots

Then x_n\to0, and

\cos\frac1{x_n} =\cos(2\pi n) =1.

Hence

\lim_{n\to\infty}\cos\frac1{x_n}=1.

Now choose a second sequence

y_n=\frac1{(2n+1)\pi},\qquad n=1,2,3,\ldots

Again y_n\to0, but

\cos\frac1{y_n} =\cos((2n+1)\pi) =-1.

Therefore

\lim_{n\to\infty}\cos\frac1{y_n}=-1.

The two sequences x_n and y_n both approach 0, whereas the corresponding function values approach 1 and -1, respectively. A single limit at 0 therefore cannot exist.

Final Answer:

\boxed{\lim_{x\to0}\cos\frac1x\text{ does not exist}.}

Hence the required result is established.

Solution:

(i) Evaluate \displaystyle\lim_{x\to\infty}(\sqrt{x^2+3x}-x).

The expression is of the indeterminate form \infty-\infty. Rationalize it:

\begin{aligned} \sqrt{x^2+3x}-x &= \frac{(\sqrt{x^2+3x}-x)(\sqrt{x^2+3x}+x)} {\sqrt{x^2+3x}+x}\\ &= \frac{x^2+3x-x^2}{\sqrt{x^2+3x}+x}\\ &= \frac{3x}{\sqrt{x^2+3x}+x}. \end{aligned}

For x\to+\infty, divide numerator and denominator by x:

\begin{aligned} \frac{3x}{\sqrt{x^2+3x}+x} &= \frac{3}{\sqrt{1+\frac3x}+1}. \end{aligned}

Therefore

\lim_{x\to\infty}(\sqrt{x^2+3x}-x) = \frac{3}{\sqrt{1+0}+1} =\frac32.

(ii) Evaluate \displaystyle\lim_{x\to0^+}(1-\sin x)^{1/x}.

This is of the indeterminate form 1^\infty. Put

L=\lim_{x\to0^+}(1-\sin x)^{1/x}.

Taking logarithms,

\ln L = \lim_{x\to0^+}\frac{\ln(1-\sin x)}{x}.

Multiply and divide by \sin x:

\begin{aligned} \ln L &= \lim_{x\to0^+} \left( \frac{\ln(1-\sin x)}{\sin x} \right) \left( \frac{\sin x}{x} \right). \end{aligned}

Use the standard limits

\lim_{u\to0}\frac{\ln(1-u)}u=-1, \qquad \lim_{x\to0}\frac{\sin x}{x}=1.

Since u=\sin x\to0,

\ln L=(-1)(1)=-1.

Exponentiating,

L=e^{-1}=\frac1e.

Final Answer:

\boxed{ \lim_{x\to\infty}(\sqrt{x^2+3x}-x)=\frac32, \qquad \lim_{x\to0^+}(1-\sin x)^{1/x}=\frac1e.}

Solution:

Sufficient conditions for a local maximum or minimum:

Let f be twice continuously differentiable near a critical point (a,b), so that

f_x(a,b)=0,\qquad f_y(a,b)=0.

Define

D=f_{xx}f_{yy}-(f_{xy})^2

at the critical point. Then:

1. If D>0 and f_{xx}>0, f has a strict local minimum.
2. If D>0 and f_{xx}<0, f has a strict local maximum.
3. If D<0, the point is a saddle point.
4. If D=0, the test is inconclusive.

Now consider

f(x,y)=x^3+y^3-9xy+1.

Step 1: Find the critical points.

f_x=3x^2-9y, \qquad f_y=3y^2-9x.

At a critical point,

3x^2-9y=0, \qquad 3y^2-9x=0.

Dividing by 3,

x^2=3y,\qquad y^2=3x.

From the first equation,

y=\frac{x^2}{3}.

Substitute this into y^2=3x:

\left(\frac{x^2}{3}\right)^2=3x.

Thus

\frac{x^4}{9}=3x \quad\Longrightarrow\quad x^4=27x \quad\Longrightarrow\quad x(x^3-27)=0.

Hence

x=0\quad\text{or}\quad x=3.

When x=0, y=x^2/3=0. When x=3, y=9/3=3. Thus the critical points are

(0,0),\qquad(3,3).

Step 2: Compute the second derivatives.

f_{xx}=6x,\qquad f_{yy}=6y,\qquad f_{xy}=-9.

Therefore

D=(6x)(6y)-(-9)^2=36xy-81.

At (0,0):

D(0,0)=36(0)(0)-81=-81<0.

Hence (0,0) is a saddle point. Also

f(0,0)=1.

At (3,3):

\begin{aligned} D(3,3) &=36(3)(3)-81\\ &=324-81\\ &=243>0, \end{aligned}

and

f_{xx}(3,3)=6(3)=18>0.

Therefore (3,3) is a strict local minimum. Its value is

\begin{aligned} f(3,3) &=3^3+3^3-9(3)(3)+1\\ &=27+27-81+1\\ &=-26. \end{aligned}

No critical point satisfies the sufficient conditions for a local maximum.

Final Answer:

\boxed{(0,0)\text{ is a saddle point},\qquad (3,3)\text{ is a strict local minimum with }f(3,3)=-26.}

There is no local maximum among the critical points.

Solution:

Let

f(x,y)=\bigl(f_1(x,y),f_2(x,y)\bigr) =\bigl(x^2+y^2,\ xy\bigr).

Relevant result:

For a differentiable mapping f:\mathbb{R}^2\to\mathbb{R}^2, the total derivative at (a,b) is the linear map represented by the Jacobian matrix

Df(a,b)= \begin{pmatrix} \dfrac{\partial f_1}{\partial x} & \dfrac{\partial f_1}{\partial y}\\ \dfrac{\partial f_2}{\partial x} & \dfrac{\partial f_2}{\partial y} \end{pmatrix}_{(a,b)}.

Compute the four partial derivatives:

\frac{\partial f_1}{\partial x}=2x,\qquad \frac{\partial f_1}{\partial y}=2y,

and

\frac{\partial f_2}{\partial x}=y,\qquad \frac{\partial f_2}{\partial y}=x.

Hence

Df(x,y)= \begin{pmatrix} 2x&2y\\ y&x \end{pmatrix}.

At (1,2),

Df(1,2)= \begin{pmatrix} 2(1)&2(2)\\ 2&1 \end{pmatrix} = \begin{pmatrix} 2&4\\ 2&1 \end{pmatrix}.

Therefore, for an increment (h,k), the total derivative acts as

\begin{aligned} Df(1,2)(h,k) &= \begin{pmatrix} 2&4\\ 2&1 \end{pmatrix} \begin{pmatrix}h\\k\end{pmatrix}\\ &= \begin{pmatrix} 2h+4k\\ 2h+k \end{pmatrix}. \end{aligned}

Verification from the increment:

\begin{aligned} f(1+h,2+k)-f(1,2) &= \bigl((1+h)^2+(2+k)^2-5,\ (1+h)(2+k)-2\bigr)\\ &= \bigl(2h+4k+h^2+k^2,\ 2h+k+hk\bigr). \end{aligned}

The linear part is exactly (2h+4k,\,2h+k), while the remaining terms (h^2+k^2,hk) are of higher order.

Final Answer:

\boxed{ Df(1,2)= \begin{pmatrix} 2&4\\ 2&1 \end{pmatrix}, \qquad Df(1,2)(h,k)=(2h+4k,\ 2h+k).}

Solution:

We have

3x^2+5y^2+3z^2+2yz+2zx+2xy-4x-8z+5=0.

Step 1: Write the quadratic part in matrix form.

Let

\mathbf x= \begin{pmatrix}x\\y\\z\end{pmatrix}, \qquad A= \begin{pmatrix} 3&1&1\\ 1&5&1\\ 1&1&3 \end{pmatrix}, \qquad d= \begin{pmatrix}-2\\0\\-4\end{pmatrix}.

Then the equation is

\mathbf x^TA\mathbf x+2d^T\mathbf x+5=0.

Step 2: Find the centre.

For a central conicoid, the centre \mathbf c=(h,k,l)^T satisfies

A\mathbf c+d=0.

Thus

\begin{pmatrix} 3&1&1\\ 1&5&1\\ 1&1&3 \end{pmatrix} \begin{pmatrix}h\\k\\l\end{pmatrix} = \begin{pmatrix}2\\0\\4\end{pmatrix}.

This gives

3h+k+l=2,\qquad h+5k+l=0,\qquad h+k+3l=4.

Subtract the second equation from the first:

2h-4k=2\quad\Longrightarrow\quad h-2k=1.

Subtract the first equation from the third:

-2h+2l=2\quad\Longrightarrow\quad l=h+1.

From h-2k=1,

h=1+2k.

Substitute l=h+1 in h+5k+l=0:

h+5k+h+1=0 \quad\Longrightarrow\quad 2h+5k=-1.

Now substitute h=1+2k:

2(1+2k)+5k=-1 \quad\Longrightarrow\quad 2+9k=-1 \quad\Longrightarrow\quad k=-\frac13.

Therefore

h=1+2\left(-\frac13\right)=\frac13, \qquad l=h+1=\frac43.

Hence the centre is

\boxed{\left(\frac13,-\frac13,\frac43\right)}.

Step 3: Translate the origin to the centre.

Put

x=\frac13+X,\qquad y=-\frac13+Y,\qquad z=\frac43+Z.

Because the translation is to the centre, all linear terms vanish. The constant after translation is obtained by substituting the centre in the original expression:

\begin{aligned} F\left(\frac13,-\frac13,\frac43\right) &= 3\left(\frac19\right) +5\left(\frac19\right) +3\left(\frac{16}{9}\right) +2\left(-\frac49\right) +2\left(\frac49\right) +2\left(-\frac19\right)\\ &\quad -4\left(\frac13\right) -8\left(\frac43\right)+5\\ &=-1. \end{aligned}

Thus the translated equation is

3X^2+5Y^2+3Z^2+2YZ+2ZX+2XY=1.

Step 4: Find the principal directions.

The eigenvalues are obtained from

\det(A-\lambda I)=0.

Now

\det \begin{pmatrix} 3-\lambda&1&1\\ 1&5-\lambda&1\\ 1&1&3-\lambda \end{pmatrix}=0.

Expanding along the first row,

\begin{aligned} 0 &=(3-\lambda)\{(5-\lambda)(3-\lambda)-1\}\\ &\quad-\{(3-\lambda)-1\} +\{1-(5-\lambda)\}\\ &=(3-\lambda)\{(5-\lambda)(3-\lambda)-1\} -(2-\lambda)+(\lambda-4). \end{aligned}

Since

(5-\lambda)(3-\lambda)-1 =15-8\lambda+\lambda^2-1 =\lambda^2-8\lambda+14,

we get

\begin{aligned} 0 &=(3-\lambda)(\lambda^2-8\lambda+14)+2\lambda-6\\ &=-\lambda^3+11\lambda^2-36\lambda+36\\ &=-(\lambda-2)(\lambda-3)(\lambda-6). \end{aligned}

Hence the eigenvalues are

\lambda_1=2,\qquad\lambda_2=3,\qquad\lambda_3=6.

For \lambda=2:

(A-2I)= \begin{pmatrix} 1&1&1\\ 1&3&1\\ 1&1&1 \end{pmatrix}.

The equations give y=0 and x+z=0. Take

v_1=(-1,0,1),\qquad e_1=\frac1{\sqrt2}(-1,0,1).

For \lambda=3:

(A-3I)= \begin{pmatrix} 0&1&1\\ 1&2&1\\ 1&1&0 \end{pmatrix}.

The equations give y+z=0 and x+y=0. Taking y=-1 gives

v_2=(1,-1,1),\qquad e_2=\frac1{\sqrt3}(1,-1,1).

For \lambda=6:

(A-6I)= \begin{pmatrix} -3&1&1\\ 1&-1&1\\ 1&1&-3 \end{pmatrix}.

A solution is (1,2,1), since

\begin{pmatrix} -3&1&1\\ 1&-1&1\\ 1&1&-3 \end{pmatrix} \begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}.

Hence

v_3=(1,2,1),\qquad e_3=\frac1{\sqrt6}(1,2,1).

The three eigenvectors are mutually perpendicular:

v_1\cdot v_2=0,\qquad v_1\cdot v_3=0,\qquad v_2\cdot v_3=0.

Step 5: Rotate to principal axes.

Define the orthonormal principal coordinates

\xi=\frac{-X+Z}{\sqrt2},\qquad \eta=\frac{X-Y+Z}{\sqrt3},\qquad \zeta=\frac{X+2Y+Z}{\sqrt6}.

Because these are coordinates along orthonormal eigenvectors, the quadratic form becomes

2\xi^2+3\eta^2+6\zeta^2=1.

Equivalently,

\frac{\xi^2}{1/2}+\frac{\eta^2}{1/3}+\frac{\zeta^2}{1/6}=1.

All three coefficients are positive, so the conicoid is a real ellipsoid.

Equations of the axes:

The axes pass through the centre and have directions v_1,v_2,v_3. Thus they may be written parametrically as

\boxed{ \begin{aligned} \ell_1:\;&(x,y,z)=\left(\frac13,-\frac13,\frac43\right)+t(-1,0,1),\\ \ell_2:\;&(x,y,z)=\left(\frac13,-\frac13,\frac43\right)+t(1,-1,1),\\ \ell_3:\;&(x,y,z)=\left(\frac13,-\frac13,\frac43\right)+t(1,2,1). \end{aligned}}

Final Answer:

\boxed{2\xi^2+3\eta^2+6\zeta^2=1}

is the standard form. The conicoid is an ellipsoid with centre

\boxed{\left(\frac13,-\frac13,\frac43\right)}

and principal-axis directions proportional to

\boxed{(-1,0,1),\quad(1,-1,1),\quad(1,2,1)}.

Solution:

We have to find the volume common to

x^2+y^2+z^2\le4

and

(x-1)^2+y^2\le1.

Step 1: Describe the projection in the xy-plane.

Use cylindrical polar coordinates

x=r\cos\theta,\qquad y=r\sin\theta,\qquad dA=r\,dr\,d\theta.

The cylinder becomes

(r\cos\theta-1)^2+r^2\sin^2\theta\le1.

Expanding,

r^2\cos^2\theta-2r\cos\theta+1+r^2\sin^2\theta\le1.

Using \cos^2\theta+\sin^2\theta=1,

r^2-2r\cos\theta\le0.

Since r\ge0,

0\le r\le2\cos\theta.

This requires \cos\theta\ge0, so

-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}.

Step 2: Obtain the vertical limits from the sphere.

The sphere gives

r^2+z^2\le4 \quad\Longrightarrow\quad -\sqrt{4-r^2}\le z\le\sqrt{4-r^2}.

Therefore the required volume is

V= \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r\,dz\,dr\,d\theta.

Step 3: Integrate with respect to z.

\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}r\,dz = r\left[\,z\,\right]_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} = 2r\sqrt{4-r^2}.

Thus

V= \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} 2r\sqrt{4-r^2}\,dr\,d\theta.

Step 4: Integrate with respect to r.

Put u=4-r^2. Then du=-2r\,dr. Hence

\int2r\sqrt{4-r^2}\,dr =-\int u^{1/2}\,du =-\frac23u^{3/2} =-\frac23(4-r^2)^{3/2}.

Therefore

\begin{aligned} \int_0^{2\cos\theta}2r\sqrt{4-r^2}\,dr &= -\frac23 \left[(4-r^2)^{3/2}\right]_{0}^{2\cos\theta}\\ &= \frac23\left[ 4^{3/2}- (4-4\cos^2\theta)^{3/2} \right]\\ &= \frac23\left[ 8-(4\sin^2\theta)^{3/2} \right]. \end{aligned}

On [-\pi/2,\pi/2],

(4\sin^2\theta)^{3/2} =8|\sin\theta|^3.

Hence

V=\frac{16}{3} \int_{-\pi/2}^{\pi/2} \left(1-|\sin\theta|^3\right)d\theta.

The integrand is even, so

V=\frac{32}{3} \int_0^{\pi/2}(1-\sin^3\theta)\,d\theta.

Now

\int_0^{\pi/2}1\,d\theta=\frac{\pi}{2}.

Also,

\begin{aligned} \int_0^{\pi/2}\sin^3\theta\,d\theta &=\int_0^{\pi/2}\sin\theta(1-\cos^2\theta)\,d\theta. \end{aligned}

Put u=\cos\theta, du=-\sin\theta\,d\theta:

\begin{aligned} \int_0^{\pi/2}\sin^3\theta\,d\theta &= \int_1^0-(1-u^2)\,du\\ &=\int_0^1(1-u^2)\,du\\ &=\left[u-\frac{u^3}{3}\right]_0^1\\ &=1-\frac13=\frac23. \end{aligned}

Therefore

\begin{aligned} V &=\frac{32}{3}\left(\frac{\pi}{2}-\frac23\right)\\ &=\frac{16\pi}{3}-\frac{64}{9}\\ &=\frac{16}{9}(3\pi-4). \end{aligned}

Final Answer:

\boxed{V=\frac{16}{9}(3\pi-4)}

cubic units.

Solution:

Let r and \theta be polar coordinates with the origin at the centre of force, and put

u=\frac1r.

Binet’s formula for a central orbit:

If h=r^2\dot\theta is the angular momentum per unit mass and P(r) denotes the acceleration directed towards the origin, then

h^2u^2\left(\frac{d^2u}{d\theta^2}+u\right)=P\left(\frac1u\right).

Here, in the standard central-orbit notation consistent with the given initial speed, the stated force law is used as the acceleration (force per unit mass):

P(r)=\mu(r^5-c^4r).

If the printed expression is interpreted as the literal force on the particle of mass m, then P(r)=\mu(r^5-c^4r)/m and the parameter \mu in the following working is replaced by \mu/m. The supplied velocity \sqrt{2\mu/3}\,c^3 corresponds to the force-per-unit-mass convention used below.

Therefore

h^2u^2(u''+u) = \mu\left(u^{-5}-c^4u^{-1}\right),

where a prime now denotes differentiation with respect to \theta. Dividing by h^2u^2,

u''+u = \frac{\mu}{h^2} \left(u^{-7}-c^4u^{-3}\right).

Step 1: Determine h from the initial conditions.

The particle is projected from an apse at r=c. At an apse the radial velocity is zero, so the initial velocity is transverse. Its magnitude is

v_0=\sqrt{\frac{2\mu}{3}}\,c^3.

Hence

h=rv_\theta =c\sqrt{\frac{2\mu}{3}}\,c^3 =\sqrt{\frac{2\mu}{3}}\,c^4.

Thus

h^2=\frac{2\mu}{3}c^8, \qquad \frac{\mu}{h^2}=\frac{3}{2c^8}.

The orbit equation becomes

u''+u = \frac{3}{2c^8} \left(u^{-7}-c^4u^{-3}\right).

Step 2: Make the equation dimensionless.

Put

q=cu,\qquad\text{so that}\qquad u=\frac qc.

Then u''=q''/c. Substitution gives

\frac1c(q''+q) = \frac{3}{2c^8} \left(c^7q^{-7}-c^7q^{-3}\right).

Multiplying by c,

q''+q=\frac32(q^{-7}-q^{-3}).

Choose \theta=0 at the initial apse. Since r=c,

q(0)=cu(0)=1.

Also dr/d\theta=0 at an apse, and therefore du/d\theta=0; hence

q'(0)=0.

Step 3: Obtain a first integral.

From

q''=\frac32q^{-7}-\frac32q^{-3}-q,

multiply by 2q':

2q'q'' =3q^{-7}q'-3q^{-3}q'-2qq'.

Integrating with respect to \theta,

q'^2 = -\frac12q^{-6} +\frac32q^{-2} -q^2+C.

Apply q=1,q'=0 at \theta=0:

0=-\frac12+\frac32-1+C=C.

Thus C=0, and

q'^2=-q^2+\frac32q^{-2}-\frac12q^{-6}.

Multiply by q^6:

q^6q'^2 =-q^8+\frac32q^4-\frac12.

Factor the right-hand side:

\begin{aligned} -q^8+\frac32q^4-\frac12 &=-\frac12(2q^8-3q^4+1)\\ &=-\frac12(2q^4-1)(q^4-1)\\ &=\frac12(1-q^4)(2q^4-1). \end{aligned}

Hence

q^6q'^2=\frac12(1-q^4)(2q^4-1).

Step 4: Integrate the angular equation.

Put

s=q^4.

Then

s'=4q^3q',\qquad s'^2=16q^6q'^2.

Therefore

s'^2=8(1-s)(2s-1).

Now put

w=4s-3,\qquad s=\frac{w+3}{4},\qquad s'=\frac{w'}4.

Also

1-s=\frac{1-w}{4}, \qquad 2s-1=\frac{w+1}{2}.

Substitution gives

\frac{w'^2}{16} = 8\left(\frac{1-w}{4}\right)\left(\frac{w+1}{2}\right) =1-w^2.

Hence

w'^2=16(1-w^2).

At \theta=0, q=1, so s=1 and w=1. The solution satisfying this apse condition is

w=\cos4\theta.

Therefore

4s-3=\cos4\theta,

so

q^4=s=\frac{3+\cos4\theta}{4}.

Since q=cu=c/r,

\frac{c^4}{r^4}=\frac{3+\cos4\theta}{4}.

Thus

r^4\frac{3+\cos4\theta}{4}=c^4.

Use the identity

\cos^4\theta+\sin^4\theta =1-2\sin^2\theta\cos^2\theta =\frac{3+\cos4\theta}{4}.

Hence

r^4(\cos^4\theta+\sin^4\theta)=c^4.

Finally, x=r\cos\theta and y=r\sin\theta, so

x^4+y^4 =r^4\cos^4\theta+r^4\sin^4\theta =c^4.

Final Answer:

\boxed{x^4+y^4=c^4}.

Hence the required central orbit is established.

Solution:

Stokes’ theorem:

Let S be an oriented smooth surface bounded by a positively oriented simple closed curve C, and let \mathbf F have continuous first partial derivatives in a region containing S. Then

\boxed{ \oint_C\mathbf F\cdot d\mathbf r = \iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS}.

Here

\mathbf F=(x+y)\mathbf i+(2x-z)\mathbf j+(y+z)\mathbf k.

Let

A=(2,0,0),\qquad B=(0,3,0),\qquad C=(0,0,6).

We take the boundary orientation A\to B\to C\to A. Since

(B-A)\times(C-A) = (-2,3,0)\times(-2,0,6) = (18,12,6) = 6(3,2,1),

the corresponding surface normal points in the direction (3,2,1).

Part I: Evaluate the line integral.

Along AB:

Use

\mathbf r_1(t)=(2-2t,\ 3t,\ 0),\qquad 0\le t\le1.

Then

dx=-2dt,\qquad dy=3dt,\qquad dz=0.

Also

x+y=2-2t+3t=2+t,

and

2x-z=2(2-2t)=4-4t.

Therefore

\begin{aligned} I_{AB} &=\int_0^1\left[(2+t)(-2)+(4-4t)(3)\right]dt\\ &=\int_0^1(-4-2t+12-12t)\,dt\\ &=\int_0^1(8-14t)\,dt\\ &=\left[8t-7t^2\right]_0^1\\ &=1. \end{aligned}

Along BC:

Use

\mathbf r_2(t)=(0,\ 3-3t,\ 6t),\qquad0\le t\le1.

Then

dx=0,\qquad dy=-3dt,\qquad dz=6dt.

Further,

2x-z=-6t, \qquad y+z=3-3t+6t=3+3t.

Hence

\begin{aligned} I_{BC} &=\int_0^1\left[(-6t)(-3)+(3+3t)(6)\right]dt\\ &=\int_0^1(18t+18+18t)\,dt\\ &=\int_0^1(18+36t)\,dt\\ &=\left[18t+18t^2\right]_0^1\\ &=36. \end{aligned}

Along CA:

Use

\mathbf r_3(t)=(2t,\ 0,\ 6-6t),\qquad0\le t\le1.

Then

dx=2dt,\qquad dy=0,\qquad dz=-6dt.

Also

x+y=2t, \qquad y+z=6-6t.

Therefore

\begin{aligned} I_{CA} &=\int_0^1\left[(2t)(2)+(6-6t)(-6)\right]dt\\ &=\int_0^1(4t-36+36t)\,dt\\ &=\int_0^1(40t-36)\,dt\\ &=\left[20t^2-36t\right]_0^1\\ &=-16. \end{aligned}

Thus

\oint_C\mathbf F\cdot d\mathbf r =I_{AB}+I_{BC}+I_{CA} =1+36-16 =21.

Part II: Evaluate the surface integral of the curl.

First compute

\nabla\times\mathbf F = \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ \partial/\partial x&\partial/\partial y&\partial/\partial z\\ x+y&2x-z&y+z \end{vmatrix}.

Therefore

\begin{aligned} \nabla\times\mathbf F &= \left( \frac{\partial(y+z)}{\partial y} - \frac{\partial(2x-z)}{\partial z} \right)\mathbf i\\ &\quad+ \left( \frac{\partial(x+y)}{\partial z} - \frac{\partial(y+z)}{\partial x} \right)\mathbf j\\ &\quad+ \left( \frac{\partial(2x-z)}{\partial x} - \frac{\partial(x+y)}{\partial y} \right)\mathbf k\\ &=(1-(-1))\mathbf i+(0-0)\mathbf j+(2-1)\mathbf k\\ &=2\mathbf i+\mathbf k. \end{aligned}

The plane through A,B,C is in intercept form

\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1,

or

3x+2y+z=6.

Thus

z=6-3x-2y.

For the upward-oriented normal consistent with (3,2,1),

\mathbf n\,dS = \left(-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right)dx\,dy =(3,2,1)\,dx\,dy.

Hence

(\nabla\times\mathbf F)\cdot\mathbf n\,dS =(2,0,1)\cdot(3,2,1)\,dx\,dy =7\,dx\,dy.

The projection of the triangle on the xy-plane is

D=\{(x,y):x\ge0,\ y\ge0,\ 3x+2y\le6\},

a right triangle with intercepts 2 and 3. Its area is

\operatorname{Area}(D)=\frac12(2)(3)=3.

Thus

\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS =7\iint_Ddx\,dy =7(3) =21.

Both sides are equal:

\oint_C\mathbf F\cdot d\mathbf r = 21 = \iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS.

Final Answer:

\boxed{\oint_C[(x+y)dx+(2x-z)dy+(y+z)dz]=21}

and the surface integral of the curl is also \boxed{21}. Hence Stokes’ theorem is verified.

Solution:

Definition of a wrench:

A system of forces is equivalent to a wrench when it is reduced to a resultant force \mathbf F acting along a certain line, together with a couple whose axis is parallel to \mathbf F. That line is called the central axis. If the parallel couple is p\mathbf F, then p is called the pitch of the wrench.

For a resultant \mathbf F\ne0 and moment \mathbf M about the origin,

p=\frac{\mathbf M\cdot\mathbf F}{|\mathbf F|^2},

and a point \mathbf r_0 on the central axis, chosen perpendicular to \mathbf F, is

\mathbf r_0=\frac{\mathbf F\times\mathbf M}{|\mathbf F|^2}.

Take the positive directions along the three given lines to be the cyclic directions described below. If a force acts in the opposite sense, its corresponding signed magnitude P,Q, or R is simply negative.

Step 1: Direction of the first line.

The line x=0,\ y-z=a is the intersection of the planes with normals

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

A direction vector is their cross product:

(1,0,0)\times(0,1,-1)=(0,1,1).

Hence a unit direction is (0,1,1)/\sqrt2. The point (0,a,0) lies on the line, so

\mathbf F_1=\frac{P}{\sqrt2}(0,1,1), \qquad \mathbf r_1=(0,a,0).

Step 2: Direction of the second line.

For y=0,\ z-x=a, the normals are (0,1,0) and (-1,0,1), so

(0,1,0)\times(-1,0,1)=(1,0,1).

Using the point (0,0,a),

\mathbf F_2=\frac{Q}{\sqrt2}(1,0,1), \qquad \mathbf r_2=(0,0,a).

Step 3: Direction of the third line.

For z=0,\ x-y=a, the normals are (0,0,1) and (1,-1,0), so

(0,0,1)\times(1,-1,0)=(1,1,0).

Using the point (a,0,0),

\mathbf F_3=\frac{R}{\sqrt2}(1,1,0), \qquad \mathbf r_3=(a,0,0).

Step 4: Resultant force.

\begin{aligned} \mathbf F &=\mathbf F_1+\mathbf F_2+\mathbf F_3\\ &=\frac1{\sqrt2} \bigl(Q+R,\ P+R,\ P+Q\bigr). \end{aligned}

Put

S=(Q+R)^2+(P+R)^2+(P+Q)^2.

Then

|\mathbf F|^2=\frac{S}{2}.

Step 5: Resultant moment about the origin.

For the first force,

\begin{aligned} \mathbf M_1 &=\mathbf r_1\times\mathbf F_1\\ &=(0,a,0)\times\frac{P}{\sqrt2}(0,1,1)\\ &=\frac{aP}{\sqrt2}(1,0,0). \end{aligned}

For the second force,

\begin{aligned} \mathbf M_2 &=(0,0,a)\times\frac{Q}{\sqrt2}(1,0,1)\\ &=\frac{aQ}{\sqrt2}(0,1,0). \end{aligned}

For the third force,

\begin{aligned} \mathbf M_3 &=(a,0,0)\times\frac{R}{\sqrt2}(1,1,0)\\ &=\frac{aR}{\sqrt2}(0,0,1). \end{aligned}

Hence

\mathbf M =\frac{a}{\sqrt2}(P,Q,R).

Step 6: Pitch of the wrench.

First calculate the scalar product:

\begin{aligned} \mathbf M\cdot\mathbf F &=\frac{a}{2} \left[P(Q+R)+Q(P+R)+R(P+Q)\right]\\ &=\frac{a}{2} \left[2PQ+2PR+2QR\right]\\ &=a(PQ+PR+QR). \end{aligned}

Therefore

\begin{aligned} p &=\frac{\mathbf M\cdot\mathbf F}{|\mathbf F|^2}\\ &=\frac{a(PQ+PR+QR)}{S/2}\\ &=\frac{2a(PQ+PR+QR)}{S}. \end{aligned}

Step 7: Find a point on the central axis.

Compute \mathbf F\times\mathbf M:

\begin{aligned} \mathbf F\times\mathbf M &= \frac{a}{2} \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ Q+R&P+R&P+Q\\ P&Q&R \end{vmatrix}\\ &=\frac{a}{2}\Bigl( (P+R)R-(P+Q)Q,\\ &\qquad\qquad (P+Q)P-(Q+R)R,\\ &\qquad\qquad (Q+R)Q-(P+R)P \Bigr). \end{aligned}

Simplify each component:

\begin{aligned} (P+R)R-(P+Q)Q &=(R-Q)(P+Q+R),\\ (P+Q)P-(Q+R)R &=(P-R)(P+Q+R),\\ (Q+R)Q-(P+R)P &=(Q-P)(P+Q+R). \end{aligned}

Thus

\mathbf F\times\mathbf M = \frac{a(P+Q+R)}{2} (R-Q,\ P-R,\ Q-P).

Since |\mathbf F|^2=S/2,

\mathbf r_0 = \frac{a(P+Q+R)}{S} (R-Q,\ P-R,\ Q-P).

Therefore the central axis is the line through \mathbf r_0 parallel to \mathbf F. Since the factor 1/\sqrt2 does not affect direction, its vector equation is

\boxed{ \mathbf r= \frac{a(P+Q+R)}{S} (R-Q,\ P-R,\ Q-P) +t(Q+R,\ P+R,\ P+Q)}.

Equivalently, when the displayed denominators are nonzero, its symmetric equation is

\boxed{ \frac{x-\dfrac{a(P+Q+R)(R-Q)}{S}}{Q+R} = \frac{y-\dfrac{a(P+Q+R)(P-R)}{S}}{P+R} = \frac{z-\dfrac{a(P+Q+R)(Q-P)}{S}}{P+Q}}.

Final Answer:

With

S=(Q+R)^2+(P+R)^2+(P+Q)^2,

the central axis is

\boxed{ \mathbf r= \frac{a(P+Q+R)}{S}(R-Q,\ P-R,\ Q-P) +t(Q+R,\ P+R,\ P+Q)}

and the pitch of the equivalent wrench is

\boxed{p=\frac{2a(PQ+PR+QR)}{S}}.

Solution:

Let v be the speed of the particle at any instant and let \theta denote the angle described at the centre from the initial position. The radius of the circle is a.

Standard kinematic relations for circular motion:

The tangential acceleration is

a_t=\frac{dv}{dt},

and the normal acceleration is

a_n=\frac{v^2}{a}.

Also, since the arc length s=a\theta,

v=\frac{ds}{dt}=a\frac{d\theta}{dt}.

The given condition is

a_t=\lambda a_n.

Therefore

\frac{dv}{dt}=\lambda\frac{v^2}{a}.

Use the chain rule:

\frac{dv}{dt} = \frac{dv}{d\theta}\frac{d\theta}{dt}.

Since d\theta/dt=v/a,

\frac{dv}{dt} = \frac{dv}{d\theta}\frac{v}{a}.

Equating this with the given tangential acceleration,

\frac{v}{a}\frac{dv}{d\theta} = \lambda\frac{v^2}{a}.

For v\ne0, cancel v/a:

\frac{dv}{d\theta}=\lambda v.

Separate variables:

\frac{dv}{v}=\lambda\,d\theta.

Integrating,

\ln v=\lambda\theta+C.

At \theta=0, the speed is given to be u. Hence

\ln u=C.

Therefore

\ln\frac vu=\lambda\theta \quad\Longrightarrow\quad v=u e^{\lambda\theta}.

Now use

v=a\frac{d\theta}{dt}.

Thus

dt=\frac{a\,d\theta}{v} =\frac{a}{u}e^{-\lambda\theta}\,d\theta.

The particle first returns to the same point after describing one complete revolution, i.e. when \theta increases from 0 to 2\pi. Hence the required time is

\begin{aligned} T &=\frac{a}{u}\int_0^{2\pi}e^{-\lambda\theta}\,d\theta\\ &=\frac{a}{u} \left[ -\frac1\lambda e^{-\lambda\theta} \right]_0^{2\pi}\\ &=\frac{a}{u} \left( -\frac1\lambda e^{-2\pi\lambda} +\frac1\lambda \right)\\ &=\frac{a}{\lambda u} \left(1-e^{-2\pi\lambda}\right). \end{aligned}

Final Answer:

\boxed{ T=\frac{a}{\lambda u}\left(1-e^{-2\pi\lambda}\right)}.

Hence the required result is proved.

Solution:

Given

x\frac{d^2y}{dx^2}-\frac{dy}{dx}=x^2e^x, \qquad x>0.

Divide by x:

\frac{d^2y}{dx^2}-\frac1x\frac{dy}{dx}=xe^x.

The equation contains y' and y'' but not y. Put

v=\frac{dy}{dx}.

Then dv/dx=d^2y/dx^2, and the equation becomes the first-order linear equation

\frac{dv}{dx}-\frac1xv=xe^x.

Step 1: Solve for v.

The integrating factor is

\begin{aligned} \mathrm{I.F.} &=e^{\int(-1/x)\,dx}\\ &=e^{-\ln x}\\ &=\frac1x, \end{aligned}

where x>0. Multiplying the equation by 1/x,

\frac1x\frac{dv}{dx}-\frac{v}{x^2}=e^x.

The left-hand side is

\frac{d}{dx}\left(\frac vx\right) = \frac{xv'-v}{x^2} = \frac1xv'-\frac{v}{x^2}.

Hence

\frac{d}{dx}\left(\frac vx\right)=e^x.

Integrating,

\frac vx=e^x+C_1.

Therefore

v=xe^x+C_1x.

Since v=y',

\frac{dy}{dx}=xe^x+C_1x.

Step 2: Integrate once more to obtain y.

y=\int xe^x\,dx+C_1\int x\,dx+C_2.

For the first integral, use integration by parts:

\begin{aligned} \int xe^x\,dx &=xe^x-\int e^x\,dx\\ &=xe^x-e^x\\ &=e^x(x-1). \end{aligned}

Also

\int x\,dx=\frac{x^2}{2}.

Thus

y=e^x(x-1)+\frac{C_1}{2}x^2+C_2.

Renaming C_1/2 as an arbitrary constant C_1, we may write

y=e^x(x-1)+C_1x^2+C_2.

Verification:

Using the form y=e^x(x-1)+C_1x^2+C_2,

y'=xe^x+2C_1x,

and

y''=(x+1)e^x+2C_1.

Therefore

\begin{aligned} xy''-y' &=x\{(x+1)e^x+2C_1\}-\{xe^x+2C_1x\}\\ &=x(x+1)e^x-xe^x\\ &=x^2e^x, \end{aligned}

which is the given equation.

Final Answer:

\boxed{y=e^x(x-1)+C_1x^2+C_2,\qquad x>0.}

Solution:

The question, as supplied, does not state the rod length or the position of the peg relative to the wall. Therefore a unique numerical angle cannot be obtained. We solve the standard configuration in full by denoting the rod length by L and the horizontal distance of the fixed peg from the vertical wall by d. The vertical height of the peg does not affect the equilibrium angle.

Let the smooth vertical wall be x=0. Let the peg be the fixed point P=(d,h). Let one end A of the uniform rod touch the wall, and let the rod pass over the smooth peg at P. Suppose the rod makes an angle \theta with the horizontal.

Take the unit vector along the rod from A into the region to be

(\cos\theta,\sin\theta).

If AP=s, then the horizontal coordinate of P gives

s\cos\theta=d.

Hence

s=d\sec\theta.

The vertical coordinate of A is consequently

y_A=h-s\sin\theta =h-d\sec\theta\sin\theta =h-d\tan\theta.

Since the rod is uniform, its centre of gravity G is at distance L/2 from A. Thus the height of G is

y_G =y_A+\frac L2\sin\theta =h-d\tan\theta+\frac L2\sin\theta.

Method of potential energy:

For a conservative system with one degree of freedom, an equilibrium position satisfies dU/d\theta=0. It is stable if d^2U/d\theta^2>0 and unstable if d^2U/d\theta^2<0 at the equilibrium position.

If the weight of the rod is W, its potential energy is

U(\theta)=Wy_G =W\left(h-d\tan\theta+\frac L2\sin\theta\right).

Differentiate:

\frac{dU}{d\theta} = W\left(-d\sec^2\theta+\frac L2\cos\theta\right).

At equilibrium,

-d\sec^2\theta+\frac L2\cos\theta=0.

Therefore

\frac L2\cos\theta=d\sec^2\theta.

Multiplying by \cos^2\theta,

\frac L2\cos^3\theta=d.

Hence the equilibrium angle is determined by

\boxed{\cos^3\theta=\frac{2d}{L}}.

A real physical configuration with 0<\theta<\pi/2 therefore requires 0<2d/L<1.

Check by force equilibrium:

Let the reaction at the smooth wall be H, horizontal. Let the reaction of the smooth peg be N, perpendicular to the rod. Taking the peg reaction in the direction (-\sin\theta,\cos\theta), vertical equilibrium gives

N\cos\theta=W\quad\Longrightarrow\quad N=W\sec\theta.

Horizontal equilibrium gives

H=N\sin\theta=W\tan\theta.

Taking moments about the peg, the moment of H is Hd\tan\theta, while the moment arm of the weight is \frac L2\cos\theta-d. Thus

Hd\tan\theta = W\left(\frac L2\cos\theta-d\right).

Substitute H=W\tan\theta:

Wd\tan^2\theta = W\left(\frac L2\cos\theta-d\right).

Cancel W and add d to both sides:

d(\tan^2\theta+1)=\frac L2\cos\theta.

Since 1+\tan^2\theta=\sec^2\theta, this again gives

d\sec^2\theta=\frac L2\cos\theta \quad\Longrightarrow\quad \cos^3\theta=\frac{2d}{L}.

Step 2: Show that the equilibrium is unstable.

Differentiate U'(\theta) once more:

\frac{d^2U}{d\theta^2} = W\left( -2d\sec^2\theta\tan\theta -\frac L2\sin\theta \right).

For the physical position 0<\theta<\pi/2,

d>0,\qquad \sec^2\theta>0,\qquad \tan\theta>0,\qquad \sin\theta>0.

Hence

\frac{d^2U}{d\theta^2}<0.

Thus the potential energy is a local maximum at equilibrium. A small displacement lowers the potential energy, so the rod moves farther away from the equilibrium position rather than returning to it.

Final Answer:

If L is the rod length and d is the horizontal distance of the peg from the wall, the equilibrium position is given by

\boxed{\cos^3\theta=\frac{2d}{L}}.

At this position U''(\theta)<0, so

\boxed{\text{the equilibrium is unstable}.}

Solution:

We have

(1-x^2)y''-2xy'+12y=0.

Since the coefficient 1-x^2 is nonzero at x=0, the point x=0 is an ordinary point. Therefore we seek a power-series solution about x=0.

Step 1: Assume a power series.

y=\sum_{n=0}^{\infty}a_nx^n.

Differentiate term by term:

y'=\sum_{n=1}^{\infty}na_nx^{n-1}.

Shift the index by writing n\mapsto n+1:

y'=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n.

Differentiating again,

y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}.

Shift the index by n\mapsto n+2:

y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n.

Step 2: Substitute all series into the differential equation.

Write

(1-x^2)y''-2xy'+12y = y''-x^2y''-2xy'+12y.

The first term is

y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n.

For the second term,

x^2y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^n.

For the third term,

2xy' = 2\sum_{n=1}^{\infty}na_nx^n.

Finally,

12y=12\sum_{n=0}^{\infty}a_nx^n.

Thus

\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - \sum_{n=2}^{\infty}n(n-1)a_nx^n - 2\sum_{n=1}^{\infty}na_nx^n + 12\sum_{n=0}^{\infty}a_nx^n =0.

The same coefficient formula also works for n=0,1, because the missing factors n(n-1) or n are then zero. Therefore, for every n\ge0, the coefficient of x^n is

(n+2)(n+1)a_{n+2} -n(n-1)a_n -2na_n +12a_n=0.

Combine the terms containing a_n:

-n(n-1)-2n+12 =-n^2+n-2n+12 =12-n(n+1).

Hence

(n+2)(n+1)a_{n+2} +\{12-n(n+1)\}a_n=0.

Therefore the recurrence relation is

\boxed{ a_{n+2} = \frac{n(n+1)-12}{(n+1)(n+2)}a_n = \frac{(n-3)(n+4)}{(n+1)(n+2)}a_n}.

Step 3: Compute the even coefficients.

For n=0,

a_2=\frac{(0-3)(0+4)}{(1)(2)}a_0 =\frac{-12}{2}a_0 =-6a_0.

For n=2,

\begin{aligned} a_4 &=\frac{(2-3)(2+4)}{(3)(4)}a_2\\ &=-\frac{6}{12}a_2\\ &=-\frac12(-6a_0)\\ &=3a_0. \end{aligned}

For n=4,

\begin{aligned} a_6 &=\frac{(4-3)(4+4)}{(5)(6)}a_4\\ &=\frac{8}{30}(3a_0)\\ &=\frac45a_0. \end{aligned}

For n=6,

\begin{aligned} a_8 &=\frac{(6-3)(6+4)}{(7)(8)}a_6\\ &=\frac{30}{56}\cdot\frac45a_0\\ &=\frac{15}{28}\cdot\frac45a_0\\ &=\frac37a_0. \end{aligned}

For n=8,

\begin{aligned} a_{10} &=\frac{(8-3)(8+4)}{(9)(10)}a_8\\ &=\frac{60}{90}\cdot\frac37a_0\\ &=\frac23\cdot\frac37a_0\\ &=\frac27a_0. \end{aligned}

Thus the even series begins

y_{\text{even}} = a_0\left( 1-6x^2+3x^4+\frac45x^6+\frac37x^8+\frac27x^{10}+\cdots \right).

Step 4: Compute the odd coefficients.

For n=1,

\begin{aligned} a_3 &=\frac{(1-3)(1+4)}{(2)(3)}a_1\\ &=-\frac{10}{6}a_1\\ &=-\frac53a_1. \end{aligned}

For n=3,

a_5 = \frac{(3-3)(3+4)}{(4)(5)}a_3 =0.

Then the recurrence gives a_7=a_9=\cdots=0. Hence the odd series terminates:

y_{\text{odd}} = a_1\left(x-\frac53x^3\right).

The two arbitrary starting coefficients a_0 and a_1 give two linearly independent solutions. Therefore the general power-series solution about x=0 is

y= a_0\left( 1-6x^2+3x^4+\frac45x^6+\frac37x^8+\frac27x^{10}+\cdots \right) + a_1\left(x-\frac53x^3\right).

Because the nearest singular points of the differential equation are x=\pm1, the power series about 0 has radius of convergence 1. Thus the displayed series is valid for |x|<1.

Final Answer:

\boxed{ y= C_1\left( 1-6x^2+3x^4+\frac45x^6+\frac37x^8+\frac27x^{10}+\cdots \right) + C_2\left(x-\frac53x^3\right),\quad |x|<1.}