Solution:

Given:

A= \begin{pmatrix} d & -1 & 0 & 0\\ 0 & d & -1 & 0\\ 0 & 0 & d & -1\\ -6 & 11 & -6 & 1 \end{pmatrix}

We have to determine all values of d for which \operatorname{rank}(A)=3.

Relevant result: For a 4\times4 matrix, rank is equal to 3 precisely when its determinant is zero but at least one minor of order 3 is non-zero.

Step 1: Compute the determinant.

Expand the determinant along the first row:

\det A = d\begin{vmatrix} d & -1 & 0\\ 0 & d & -1\\ 11 & -6 & 1 \end{vmatrix} + \begin{vmatrix} 0 & -1 & 0\\ 0 & d & -1\\ -6 & -6 & 1 \end{vmatrix}.

For the first minor,

\begin{vmatrix} d & -1 & 0\\ 0 & d & -1\\ 11 & -6 & 1 \end{vmatrix} = d\begin{vmatrix}d&-1\\-6&1\end{vmatrix} + \begin{vmatrix}0&-1\\11&1\end{vmatrix} = d(d-6)+11 = d^2-6d+11.

For the second minor, expanding along the first column gives

\begin{vmatrix} 0 & -1 & 0\\ 0 & d & -1\\ -6 & -6 & 1 \end{vmatrix} = (-6) \begin{vmatrix} -1&0\\ d&-1 \end{vmatrix} = -6.

Hence

\det A = d(d^2-6d+11)-6 = d^3-6d^2+11d-6.

Factorizing,

d^3-6d^2+11d-6 =(d-1)(d-2)(d-3).

Therefore, in order that the rank be less than 4, we must have

d=1,\quad d=2,\quad \text{or}\quad d=3.

Step 2: Verify that the rank is exactly 3 for these values.

Consider the leading 3\times3 minor formed by the first three rows and first three columns:

\Delta_3= \begin{vmatrix} d&-1&0\\ 0&d&-1\\ 0&0&d \end{vmatrix} =d^3.

For each of d=1,2,3, d^3\ne0. Therefore a non-zero minor of order 3 exists, and hence

\operatorname{rank}(A)\ge3.

At the same time \det A=0, so \operatorname{rank}(A)<4. Consequently,

\operatorname{rank}(A)=3.

Final Answer:

\boxed{d=1,\ 2,\ 3}

Solution:

Definition: Vectors v_1,v_2,\ldots,v_n are linearly dependent if there exist scalars, not all zero, such that

c_1v_1+c_2v_2+\cdots+c_nv_n=0.

Here

\begin{aligned} v_1&=(1,1,2,4),\\ v_2&=(2,-1,-5,2),\\ v_3&=(1,-1,-4,0),\\ v_4&=(2,1,1,6). \end{aligned}

It is sufficient to exhibit one non-trivial linear relation among these vectors. Consider

v_1-2v_2+3v_3.

Calculate each coordinate explicitly:

\begin{aligned} v_1-2v_2+3v_3 &=(1,1,2,4)-2(2,-1,-5,2)+3(1,-1,-4,0)\\ &=(1,1,2,4)+(-4,2,10,-4)+(3,-3,-12,0)\\ &=(1-4+3,\;1+2-3,\;2+10-12,\;4-4+0)\\ &=(0,0,0,0). \end{aligned}

Thus we have obtained

1\cdot v_1-2\cdot v_2+3\cdot v_3+0\cdot v_4=0.

The coefficients 1,-2,3,0 are not all zero. Hence the defining condition for linear dependence is satisfied.

Final Answer:

The given vectors are linearly dependent, since

\boxed{v_1-2v_2+3v_3=0}

Solution:

We have to evaluate

L=\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{1/x^2}.

As x\to0, \tan x/x\to1 while 1/x^2\to\infty. Hence the limit is of the indeterminate form 1^\infty. We use the logarithmic method.

Let

y=\left(\frac{\tan x}{x}\right)^{1/x^2}.

Taking logarithms,

\log y=\frac{1}{x^2}\log\left(\frac{\tan x}{x}\right).

Using the Maclaurin expansion

\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+O(x^7),

we obtain

\frac{\tan x}{x} =1+\frac{x^2}{3}+\frac{2x^4}{15}+O(x^6).

Now \log(1+u)=u-\frac{u^2}{2}+O(u^3). Therefore, with u=\frac{x^2}{3}+O(x^4),

\log\left(\frac{\tan x}{x}\right) =\frac{x^2}{3}+O(x^4).

Hence

\lim_{x\to0}\log y = \lim_{x\to0} \frac{\frac{x^2}{3}+O(x^4)}{x^2} = \frac13.

Exponentiating both sides gives

L=e^{1/3}.

Final Answer:

\boxed{\displaystyle \lim_{x\to0}\left(\frac{\tan x}{x}\right)^{1/x^2}=e^{1/3}}

Solution:

Relevant result: The angle between two smooth surfaces at a common point is the angle between their normal vectors, i.e. between the gradients of their defining functions at that point.

Write the first surface as

F_1(x,y,z)=x\log z-y^2+1=0.

Its gradient is

\nabla F_1= \left(\log z,\,-2y,\,\frac{x}{z}\right).

At (1,1,1), since \log1=0,

\nabla F_1(1,1,1)=(0,-2,1).

Write the second surface as

F_2(x,y,z)=x^2y+z-2=0.

Then

\nabla F_2=(2xy,x^2,1),

and therefore

\nabla F_2(1,1,1)=(2,1,1).

If \theta denotes the acute angle between the surfaces,

\cos\theta = \frac{\left|\nabla F_1\cdot\nabla F_2\right|} {|\nabla F_1|\,|\nabla F_2|}.

Now

\nabla F_1\cdot\nabla F_2 =(0)(2)+(-2)(1)+(1)(1) =-1, |\nabla F_1|=\sqrt{0^2+(-2)^2+1^2}=\sqrt5, \qquad |\nabla F_2|=\sqrt{2^2+1^2+1^2}=\sqrt6.

Thus

\cos\theta=\frac{1}{\sqrt{30}}.

Final Answer:

\boxed{\theta=\cos^{-1}\left(\frac1{\sqrt{30}}\right)}

Solution:

Rodrigues’ formula for Legendre polynomials:

P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n.

We have to prove

P_n(-x)=(-1)^nP_n(x).

Put u=-x. By Rodrigues’ formula,

P_n(-x) = \frac{1}{2^n n!} \left[ \frac{d^n}{du^n}(u^2-1)^n \right]_{u=-x}.

Since u=-x,

\frac{d}{du}=-\frac{d}{dx} \quad\Longrightarrow\quad \frac{d^n}{du^n}=(-1)^n\frac{d^n}{dx^n}.

Also

u^2-1=(-x)^2-1=x^2-1.

Therefore

\begin{aligned} P_n(-x) &= \frac{1}{2^n n!} (-1)^n \frac{d^n}{dx^n}(x^2-1)^n\\ &=(-1)^nP_n(x). \end{aligned}

This also shows the familiar parity property: P_n is even when n is even and odd when n is odd.

Final Answer:

\boxed{P_n(-x)=(-1)^nP_n(x)}

Hence the required result is established.

Solution:

The linear operator is

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

Step 1: Matrix of T.

With respect to the standard basis,

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

Its determinant is

\det[T] = 3\begin{vmatrix}-1&0\\1&1\end{vmatrix} =3(-1) =-3\ne0.

Therefore T is invertible.

Step 2: Determine the inverse transformation.

Let

T(x,y,z)=(u,v,w).

Then

u=3x,\qquad v=x-y,\qquad w=2x+y+z.

From the first equation,

x=\frac{u}{3}.

From v=x-y,

y=x-v=\frac{u}{3}-v.

Finally, from w=2x+y+z,

\begin{aligned} z &=w-2x-y\\ &=w-\frac{2u}{3}-\left(\frac{u}{3}-v\right)\\ &=w-u+v. \end{aligned}

Hence

T^{-1}(u,v,w) = \left(\frac{u}{3},\frac{u}{3}-v,w-u+v\right).

Verification:

Substitution into T gives

T\left(\frac{u}{3},\frac{u}{3}-v,w-u+v\right) =(u,v,w),

so the expression is indeed the inverse.

Final Answer:

\boxed{ T^{-1}(x,y,z) = \left(\frac{x}{3},\frac{x}{3}-y,z-x+y\right)}

Solution:

The system is

\begin{cases} 2x+3y+5z=9,\\ 7x+3y-2z=8,\\ 2x+3y+az=b. \end{cases}

The coefficient matrix is

A= \begin{pmatrix} 2&3&5\\ 7&3&-2\\ 2&3&a \end{pmatrix}.

Step 1: Determine when the coefficient matrix is nonsingular.

Subtract the first row from the third row:

R_3\to R_3-R_1.

Then

\det A = \begin{vmatrix} 2&3&5\\ 7&3&-2\\ 0&0&a-5 \end{vmatrix}.

Expanding along the third row,

\det A =(a-5) \begin{vmatrix} 2&3\\ 7&3 \end{vmatrix} =(a-5)(6-21) =-15(a-5).

Therefore, if a\ne5, then \det A\ne0, so the system has exactly one solution for every value of b.

Step 2: Examine the singular case a=5.

When a=5, the first and third equations have identical left-hand sides:

2x+3y+5z=9,\qquad 2x+3y+5z=b.

If b\ne9, these two equations contradict each other. Hence the system is inconsistent and has no solution.

If b=9, the third equation is exactly the same as the first. Thus only two independent equations remain in three unknowns. The coefficient rows

(2,3,5),\qquad (7,3,-2)

are not proportional, so the rank is 2. Since there are three unknowns, one variable is free and infinitely many solutions occur.

Final Answer:

\boxed{ \begin{array}{ll} \text{(i) No solution:} & a=5,\ b\ne9,\\[2mm] \text{(ii) Unique solution:} & a\ne5\ \text{(any }b\text{)},\\[2mm] \text{(iii) Infinitely many solutions:} & a=5,\ b=9. \end{array}}

Solution:

The function is

f(x)= \begin{cases} x,&x\text{ irrational},\\ -x,&x\text{ rational}. \end{cases}

We shall prove that f is continuous at 0 and discontinuous at every a\ne0.

1. Continuity at x=0.

Since 0 is rational, f(0)=0. Moreover, for every real x, irrespective of whether x is rational or irrational,

|f(x)-f(0)|=|f(x)|=|x|.

Given \varepsilon>0, choose \delta=\varepsilon. If |x|<\delta, then

|f(x)-f(0)|=|x|<\delta=\varepsilon.

Thus f is continuous at 0.

2. Discontinuity at every a\ne0.

Both the rational and irrational numbers are dense in \mathbb R. Hence there exists a sequence of rationals \{r_n\} and a sequence of irrationals \{s_n\} such that

r_n\to a,\qquad s_n\to a.

Along the rational sequence,

f(r_n)=-r_n\to-a.

Along the irrational sequence,

f(s_n)=s_n\to a.

For a\ne0, the two limiting values -a and a are different. Therefore \lim_{x\to a}f(x) does not exist, and f cannot be continuous at a.

Final Answer:

\boxed{\text{The function }f\text{ is continuous only at }x=0.}

Solution:

Let

F(x,y)= 2x^3-5x^2y+4xy^2-y^3+6x^2-7xy+y^2-x+5y-3.

For an asymptote not parallel to the y-axis, put

y=mx+c

and require the highest powers of x in F(x,mx+c) to vanish.

Step 1: Determine the possible slopes.

The homogeneous cubic part is

F_3(x,y)=2x^3-5x^2y+4xy^2-y^3.

Substituting y=mx,

F_3(x,mx) =x^3(2-5m+4m^2-m^3).

Hence the slopes satisfy

2-5m+4m^2-m^3=0.

Factorizing,

2-5m+4m^2-m^3 =-(m-2)(m-1)^2.

Thus the possible slopes are

m=2,\qquad m=1\text{ (double root)}.

Step 2: Asymptote corresponding to m=2.

Put y=2x+c in the complete equation. Direct expansion gives

F(x,2x+c) = (-c-4)x^2+(-2c^2-3c+9)x-c^3+c^2+5c-3.

For the curve to approach a straight line as |x|\to\infty, the coefficient of the highest remaining power x^2 must vanish:

-c-4=0 \quad\Longrightarrow\quad c=-4.

Hence one asymptote is

y=2x-4.

Step 3: Asymptotes corresponding to the repeated slope m=1.

Put y=x+c. Because m=1 is a repeated root of the slope equation, the cubic and quadratic-in-x terms cancel automatically. Expansion gives

F(x,x+c) = (c^2-5c+4)x-c^3+c^2+5c-3.

For an asymptote the coefficient of x must vanish:

c^2-5c+4=0.

Therefore

(c-1)(c-4)=0 \quad\Longrightarrow\quad c=1\ \text{or}\ 4.

This gives the two parallel asymptotes

y=x+1,\qquad y=x+4.

There is no vertical asymptote arising from the highest-degree part, since the coefficient of y^3 is non-zero.

Final Answer:

\boxed{y=2x-4,\qquad y=x+1,\qquad y=x+4}

Solution:

We have to extremize

f(x,y,z)=2x+3y+z

subject to

x^2+y^2=5,\qquad x+z=1.

Method of Lagrange multipliers: With two constraints g=0 and h=0, an extremum satisfies

\nabla f=\lambda\nabla g+\mu\nabla h.

Take

g=x^2+y^2-5,\qquad h=x+z-1.

Then

\nabla f=(2,3,1),\quad \nabla g=(2x,2y,0),\quad \nabla h=(1,0,1).

Thus

(2,3,1)=\lambda(2x,2y,0)+\mu(1,0,1),

which gives

2=2\lambda x+\mu,\qquad 3=2\lambda y,\qquad 1=\mu.

From \mu=1,

1=2\lambda x \quad\Longrightarrow\quad x=\frac{1}{2\lambda}.

Also

y=\frac{3}{2\lambda}.

Substitute in x^2+y^2=5:

\frac{1}{4\lambda^2} + \frac{9}{4\lambda^2} =5.

Hence

\frac{10}{4\lambda^2}=5 \quad\Longrightarrow\quad \lambda^2=\frac12, \qquad \lambda=\pm\frac1{\sqrt2}.

For \lambda=1/\sqrt2,

x=\frac1{\sqrt2},\qquad y=\frac3{\sqrt2},\qquad z=1-\frac1{\sqrt2}.

The corresponding value of f is

f = \frac2{\sqrt2} +\frac9{\sqrt2} +1-\frac1{\sqrt2} = 1+5\sqrt2.

For \lambda=-1/\sqrt2,

x=-\frac1{\sqrt2},\qquad y=-\frac3{\sqrt2},\qquad z=1+\frac1{\sqrt2},

and

f=1-5\sqrt2.

Check that these are global extrema: The constraints reduce the feasible set to the circle x^2+y^2=5 with z=1-x, which is compact. A continuous function therefore attains its maximum and minimum there. The two critical values obtained above are consequently the global extrema.

Final Answer:

\boxed{f_{\max}=1+5\sqrt2,\qquad f_{\min}=1-5\sqrt2}

Solution:

We have to evaluate

I=\int_0^2\int_0^{y^2/2} \frac{y}{\sqrt{x^2+y^2+1}}\,dx\,dy.

Step 1: Integrate with respect to x.

For fixed y, put

a^2=y^2+1.

We use the standard integral

\int\frac{dx}{\sqrt{x^2+a^2}} = \log\left(x+\sqrt{x^2+a^2}\right)+C = \operatorname{arsinh}\left(\frac{x}{a}\right)+C.

Therefore

\begin{aligned} \int_0^{y^2/2} \frac{y\,dx}{\sqrt{x^2+y^2+1}} &= y\left[ \operatorname{arsinh} \left(\frac{x}{\sqrt{y^2+1}}\right) \right]_0^{y^2/2}\\ &= y\, \operatorname{arsinh} \left( \frac{y^2}{2\sqrt{y^2+1}} \right). \end{aligned}

Hence

I= \int_0^2 y\, \operatorname{arsinh} \left( \frac{y^2}{2\sqrt{y^2+1}} \right)dy.

Step 2: Simplify the inverse hyperbolic sine.

Let

t=y^2+1.

Then

\frac{y^2}{2\sqrt{y^2+1}} = \frac{t-1}{2\sqrt t} = \frac12\left(\sqrt t-\frac1{\sqrt t}\right).

Using

\sinh(\log u)=\frac12\left(u-\frac1u\right),

with u=\sqrt t, we obtain

\frac12\left(\sqrt t-\frac1{\sqrt t}\right) = \sinh\left(\frac12\log t\right).

Since t=y^2+1\ge1,

\operatorname{arsinh} \left( \frac{y^2}{2\sqrt{y^2+1}} \right) = \frac12\log(y^2+1).

Thus

I=\frac12\int_0^2y\log(y^2+1)\,dy.

Step 3: Evaluate the remaining integral.

Put u=y^2+1. Then du=2y\,dy. When y=0, u=1; when y=2, u=5. Therefore

I = \frac14\int_1^5\log u\,du.

Now

\int\log u\,du=u\log u-u.

Hence

\begin{aligned} I &= \frac14 [u\log u-u]_1^5\\ &= \frac14 \{(5\log5-5)-(0-1)\}\\ &= \frac{5\log5-4}{4}. \end{aligned}

Final Answer:

\boxed{\displaystyle I=\frac{5\log5-4}{4}}

Solution:

The equation is

y=xy'+(y')^2.

Put

p=\frac{dy}{dx}.

Then

y=xp+p^2.

This is a Clairaut equation of the standard form

y=xp+f(p),

with f(p)=p^2.

Step 1: Differentiate with respect to x.

\frac{dy}{dx} = p+x\frac{dp}{dx}+2p\frac{dp}{dx}.

Since dy/dx=p,

p=p+(x+2p)\frac{dp}{dx}.

Therefore

(x+2p)\frac{dp}{dx}=0.

Thus there are two cases.

Case 1: dp/dx=0.

Then p=C, where C is an arbitrary constant. Substitution into the original equation gives

y=Cx+C^2.

This is the one-parameter general solution.

Case 2: x+2p=0.

Then

p=-\frac{x}{2}.

Substituting this value of p into y=xp+p^2,

\begin{aligned} y &= x\left(-\frac{x}{2}\right) + \left(-\frac{x}{2}\right)^2\\ &= -\frac{x^2}{2}+\frac{x^2}{4}\\ &= -\frac{x^2}{4}. \end{aligned}

This is the singular solution. It is also the envelope of the family y=Cx+C^2, because differentiating that family with respect to C gives x+2C=0.

Final Answer:

\boxed{\text{General solution: }y=Cx+C^2} \boxed{\text{Singular solution: }y=-\frac{x^2}{4}}

Solution:

We have

(D^2-2D+2)y=e^x\tan x, \qquad D=\frac{d}{dx}.

Step 1: Complementary function.

The auxiliary equation is

m^2-2m+2=0.

Thus

m=\frac{2\pm\sqrt{4-8}}{2} =1\pm i.

Therefore

y_c=e^x(C_1\cos x+C_2\sin x).

Step 2: Remove the factor e^x from the right-hand side.

Put

y=e^xv.

Using the shift rule

F(D)\{e^{ax}v\} = e^{ax}F(D+a)v,

we get

(D^2-2D+2)(e^xv) = e^x\{(D+1)^2-2(D+1)+2\}v.

Simplifying the operator,

(D+1)^2-2(D+1)+2 = D^2+2D+1-2D-2+2 = D^2+1.

Hence the differential equation becomes

e^x(v''+v)=e^x\tan x,

or

v''+v=\tan x.

Step 3: Particular integral by variation of parameters.

For v''+v=f(x), a fundamental pair is

v_1=\cos x,\qquad v_2=\sin x,

with Wronskian

W= \begin{vmatrix} \cos x&\sin x\\ -\sin x&\cos x \end{vmatrix} =1.

Thus

v_p = -\cos x\int\sin x\,\tan x\,dx + \sin x\int\cos x\,\tan x\,dx.

Evaluate the first integral:

\begin{aligned} \int\sin x\tan x\,dx &= \int\frac{\sin^2x}{\cos x}\,dx\\ &= \int\frac{1-\cos^2x}{\cos x}\,dx\\ &= \int(\sec x-\cos x)\,dx\\ &= \log|\sec x+\tan x|-\sin x. \end{aligned}

The second integral is

\int\cos x\tan x\,dx = \int\sin x\,dx = -\cos x.

Therefore

\begin{aligned} v_p &= -\cos x\{\log|\sec x+\tan x|-\sin x\} +\sin x(-\cos x)\\ &= -\cos x\log|\sec x+\tan x| +\sin x\cos x-\sin x\cos x\\ &= -\cos x\log|\sec x+\tan x|. \end{aligned}

Thus

v=C_1\cos x+C_2\sin x-\cos x\log|\sec x+\tan x|.

Since y=e^xv,

Final Answer:

\boxed{ y=e^x\left[ C_1\cos x+C_2\sin x -\cos x\log|\sec x+\tan x| \right]}

Solution:

The scalar function is

\phi(x,y,z)=xy^2+yz^3.

Relevant result: The directional derivative of \phi at a point in the direction of a unit vector \hat n is

D_{\hat n}\phi=\nabla\phi\cdot\hat n.

Step 1: Find \nabla\phi at (2,-1,1).

\frac{\partial\phi}{\partial x}=y^2, \qquad \frac{\partial\phi}{\partial y}=2xy+z^3, \qquad \frac{\partial\phi}{\partial z}=3yz^2.

Therefore

\nabla\phi(2,-1,1) = (1,-3,-3).

Step 2: Find the required normal direction.

The surface is

x\log z-y^2=-4,

or

F(x,y,z)=x\log z-y^2+4=0.

A normal to the surface is \nabla F:

\nabla F = \left(\log z,-2y,\frac{x}{z}\right).

At (-1,2,1),

\nabla F(-1,2,1) =(0,-4,-1).

Its magnitude is

|\nabla F| = \sqrt{0^2+(-4)^2+(-1)^2} = \sqrt{17}.

Hence the unit normal in the direction of \nabla F is

\hat n=\frac{(0,-4,-1)}{\sqrt{17}}.

Step 3: Take the scalar product.

\begin{aligned} D_{\hat n}\phi &= (1,-3,-3)\cdot \frac{(0,-4,-1)}{\sqrt{17}}\\ &= \frac{0+12+3}{\sqrt{17}}\\ &= \frac{15}{\sqrt{17}}. \end{aligned}

If the opposite unit normal is chosen, the directional derivative changes sign.

Final Answer:

\boxed{\displaystyle D_{\hat n}\phi=\frac{15}{\sqrt{17}}}

Solution:

The force field is

\vec F= (y^2\cos x+z^3)\mathbf i +(2y\sin x-4)\mathbf j +(3xz^2+2)\mathbf k.

Let

P=y^2\cos x+z^3,\qquad Q=2y\sin x-4,\qquad R=3xz^2+2.

Step 1: Show that the field is conservative.

Compute the cross partial derivatives:

\frac{\partial P}{\partial y}=2y\cos x, \qquad \frac{\partial Q}{\partial x}=2y\cos x, \frac{\partial P}{\partial z}=3z^2, \qquad \frac{\partial R}{\partial x}=3z^2, \frac{\partial Q}{\partial z}=0, \qquad \frac{\partial R}{\partial y}=0.

Thus

\nabla\times\vec F=\vec0.

Since the field is continuously differentiable throughout \mathbb R^3, a simply connected domain, it is conservative.

Step 2: Find a potential function \Phi.

We require

\nabla\Phi=\vec F.

Integrate \Phi_x=P with respect to x:

\Phi = \int(y^2\cos x+z^3)\,dx = y^2\sin x+xz^3+g(y,z).

Differentiate with respect to y:

\Phi_y=2y\sin x+g_y.

Comparing with Q=2y\sin x-4,

g_y=-4.

Hence

g(y,z)=-4y+h(z).

Now

\Phi_z=3xz^2+h'(z).

Comparing with R=3xz^2+2,

h'(z)=2 \quad\Longrightarrow\quad h(z)=2z+C.

Therefore a potential is

\Phi(x,y,z)=y^2\sin x+xz^3-4y+2z.

Step 3: Work done from A to B.

For a conservative field,

W=\Phi(B)-\Phi(A).

At A=(0,1,1),

\Phi(A) = 1^2\sin0+0\cdot1^3-4(1)+2(1) =-2.

At B=(\pi/2,-1,2),

\begin{aligned} \Phi(B) &= (-1)^2\sin\frac{\pi}{2} +\frac{\pi}{2}(2^3) -4(-1)+2(2)\\ &= 1+4\pi+4+4\\ &= 9+4\pi. \end{aligned}

Hence

W=(9+4\pi)-(-2)=11+4\pi.

Final Answer:

The field is conservative, with potential

\boxed{\Phi=y^2\sin x+xz^3-4y+2z+C}

and the work done from A to B is

\boxed{11+4\pi}

Solution:

The homogeneous second-degree equation is

2x^2-6y^2-12z^2+18yz+2zx+xy=0.

Criterion: A homogeneous quadratic equation in x,y,z represents a pair of planes through the origin when its quadratic form is reducible into two real linear factors. We can verify this directly by factorization.

Group the terms as follows:

\begin{aligned} &(x+2y-2z)(2x-3y+6z)\\ &=2x^2-3xy+6xz +4xy-6y^2+12yz -4xz+6yz-12z^2\\ &=2x^2+xy+2xz-6y^2+18yz-12z^2. \end{aligned}

This is exactly the given quadratic. Hence the equation represents the pair of planes

x+2y-2z=0

and

2x-3y+6z=0.

Step 2: Find the angle between the planes.

Their normal vectors are

\mathbf n_1=(1,2,-2), \qquad \mathbf n_2=(2,-3,6).

The acute angle \theta between the planes is the acute angle between their normals:

\cos\theta = \frac{|\mathbf n_1\cdot\mathbf n_2|} {|\mathbf n_1|\,|\mathbf n_2|}.

Now

\mathbf n_1\cdot\mathbf n_2 = (1)(2)+(2)(-3)+(-2)(6) = 2-6-12=-16.

Also

|\mathbf n_1| = \sqrt{1+4+4}=3, \qquad |\mathbf n_2| = \sqrt{4+9+36}=7.

Therefore

\cos\theta = \frac{16}{21}.

Final Answer:

The pair of planes is

\boxed{x+2y-2z=0,\qquad 2x-3y+6z=0}

and the acute angle between them is

\boxed{\theta=\cos^{-1}\left(\frac{16}{21}\right)}.

Solution:

The hyperboloid is

2x^2-6y^2+3z^2=5.

The two given planes

3x-3y+6z-5=0

and

x+9y-3z=0

intersect in a straight line. We require the tangent planes to the hyperboloid that contain this line.

Step 1: Write the general plane through the given line.

Every plane through the intersection of the two given planes is

(3x-3y+6z-5) +\lambda(x+9y-3z)=0.

Therefore

(3+\lambda)x+(-3+9\lambda)y+(6-3\lambda)z=5.

Set

l=3+\lambda,\qquad m=-3+9\lambda,\qquad n=6-3\lambda.

Step 2: Obtain the tangency condition.

For the quadric

2x^2-6y^2+3z^2=5,

the tangent plane at (x_0,y_0,z_0) is

2xx_0-6yy_0+3zz_0=5.

Comparing this with a plane

lx+my+nz=5,

we must have

x_0=\frac l2,\qquad y_0=-\frac m6,\qquad z_0=\frac n3.

For this point to lie on the hyperboloid,

2\left(\frac l2\right)^2 -6\left(-\frac m6\right)^2 +3\left(\frac n3\right)^2 =5.

Hence the condition of tangency is

\frac{l^2}{2}-\frac{m^2}{6}+\frac{n^2}{3}=5.

Step 3: Substitute l,m,n.

\frac{(3+\lambda)^2}{2} - \frac{(-3+9\lambda)^2}{6} + \frac{(6-3\lambda)^2}{3} =5.

Multiply by 6:

3(3+\lambda)^2 -(-3+9\lambda)^2 +2(6-3\lambda)^2 =30.

Expanding,

3(9+6\lambda+\lambda^2) -(9-54\lambda+81\lambda^2) +2(36-36\lambda+9\lambda^2) =30.

Thus

27+18\lambda+3\lambda^2 -9+54\lambda-81\lambda^2 +72-72\lambda+18\lambda^2 =30.

Therefore

90-60\lambda^2=30,

so

60(1-\lambda^2)=0 \quad\Longrightarrow\quad \lambda=\pm1.

Case 1: \lambda=1.

4x+6y+3z=5.

Case 2: \lambda=-1.

2x-12y+9z=5.

Verification of contact points:

For the first plane, the point of contact obtained from (l/2,-m/6,n/3) is (2,-1,1), and

2(2)^2-6(-1)^2+3(1)^2=8-6+3=5.

For the second plane, the point of contact is (1,2,3), and

2(1)^2-6(2)^2+3(3)^2=2-24+27=5.

Final Answer:

\boxed{4x+6y+3z=5}

and

\boxed{2x-12y+9z=5}.

Solution:

We have to prove the finite-dimensional Riesz representation theorem.

Statement: Let V be a finite-dimensional inner product space over F, and let g:V\to F be linear. Then there exists a unique vector y\in V such that

g(x)=\langle x,y\rangle \qquad\text{for every }x\in V.

We use the convention that the inner product is linear in its first argument. For a real inner product space the conjugation signs below simply disappear.

Step 1: Choose an orthonormal basis.

Since V is finite-dimensional, by the Gram-Schmidt process it possesses an orthonormal basis

\{e_1,e_2,\ldots,e_n\}.

Every vector x\in V can be written uniquely as

x=\sum_{i=1}^{n}x_i e_i.

Because the basis is orthonormal,

x_i=\langle x,e_i\rangle.

Step 2: Construct the representing vector.

Define

y=\sum_{i=1}^{n}\overline{g(e_i)}\,e_i.

In the real case this is simply y=\sum g(e_i)e_i.

Now, for arbitrary x=\sum x_i e_i,

\begin{aligned} \langle x,y\rangle &= \left\langle \sum_{i=1}^{n}x_i e_i, \sum_{j=1}^{n}\overline{g(e_j)}e_j \right\rangle\\ &= \sum_{i=1}^{n}\sum_{j=1}^{n} x_i\,g(e_j)\, \langle e_i,e_j\rangle. \end{aligned}

Since the basis is orthonormal, \langle e_i,e_j\rangle=\delta_{ij}. Hence all terms with i\ne j vanish, and

\langle x,y\rangle = \sum_{i=1}^{n}x_i g(e_i).

By linearity of g,

g(x) = g\left(\sum_{i=1}^{n}x_i e_i\right) = \sum_{i=1}^{n}x_i g(e_i).

Therefore

g(x)=\langle x,y\rangle

for every x\in V. Thus existence is proved.

Step 3: Prove uniqueness.

Suppose y_1 and y_2 both represent g. Then for every x\in V,

\langle x,y_1\rangle = g(x) = \langle x,y_2\rangle.

Hence

\langle x,y_1-y_2\rangle=0 \qquad\text{for every }x\in V.

Take x=y_1-y_2. Then

\langle y_1-y_2,y_1-y_2\rangle = \|y_1-y_2\|^2 =0.

Positive definiteness of the inner product implies

y_1-y_2=0,

so y_1=y_2. Therefore the representing vector is unique.

Final Answer:

There exists a unique y\in V such that

\boxed{g(x)=\langle x,y\rangle\quad\forall x\in V.}

The required theorem is proved.

Solution:

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

Standard kinematical formulae:

The tangential and normal accelerations are

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

The problem states that these accelerations are always equal. Therefore

v\frac{dv}{ds} = \frac{v^2}{\rho}.

For non-zero speed, divide by v^2:

\frac1v\frac{dv}{ds} = \frac1\rho.

Step 1: Use the prescribed law of velocity.

Since the velocity varies as e^{\tan^{-1}(s/c)}, write

v=K e^{\tan^{-1}(s/c)},

where K is a positive constant.

Taking logarithms,

\log v=\log K+\tan^{-1}\left(\frac{s}{c}\right).

Differentiate with respect to s:

\frac1v\frac{dv}{ds} = \frac{1/c}{1+s^2/c^2} = \frac{c}{c^2+s^2}.

Comparing with (1/v)(dv/ds)=1/\rho,

\frac1\rho=\frac{c}{c^2+s^2},

so

\rho=\frac{c^2+s^2}{c}.

Step 2: Obtain the tangent angle.

The curvature is

\frac1\rho=\frac{d\psi}{ds}.

Hence

\frac{d\psi}{ds} = \frac{c}{c^2+s^2}.

Integrating,

\psi = \tan^{-1}\left(\frac{s}{c}\right)+\psi_0.

Choose the x-axis along the tangent at the fixed point s=0. Then \psi_0=0, so

\psi=\tan^{-1}\left(\frac{s}{c}\right).

Therefore

\cos\psi=\frac{c}{\sqrt{c^2+s^2}}, \qquad \sin\psi=\frac{s}{\sqrt{c^2+s^2}}.

Step 3: Determine the Cartesian coordinates.

Since

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

we have

\frac{dx}{ds} = \frac{c}{\sqrt{c^2+s^2}}.

Integrating from s=0 and choosing x=0 there,

x = c\,\operatorname{arsinh}\left(\frac{s}{c}\right).

Similarly,

\frac{dy}{ds} = \frac{s}{\sqrt{c^2+s^2}}.

Integrating and choosing y=0 at s=0,

y = \sqrt{c^2+s^2}-c.

Step 4: Eliminate s.

From

\frac{x}{c} = \operatorname{arsinh}\left(\frac{s}{c}\right),

we get

\frac{s}{c} = \sinh\left(\frac{x}{c}\right).

Therefore

\sqrt{c^2+s^2} = c\sqrt{1+\sinh^2(x/c)} = c\cosh(x/c).

Substituting in the expression for y,

y = c\cosh\left(\frac{x}{c}\right)-c.

Thus, apart from a translation and rotation of axes, the path is a catenary.

Final Answer:

\boxed{y+c=c\cosh\left(\frac{x}{c}\right)}

or equivalently \boxed{y=c[\cosh(x/c)-1]} for the chosen origin and tangent direction.