Solution:

We are given

A= \begin{bmatrix} 8&-6&2\\ -6&7&-4\\ 2&-4&3 \end{bmatrix}.

Standard result: The characteristic roots (eigenvalues) of a square matrix A are obtained from

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

Therefore

\begin{vmatrix} 8-\lambda&-6&2\\ -6&7-\lambda&-4\\ 2&-4&3-\lambda \end{vmatrix}=0.

Expand the determinant along the first row:

\begin{aligned} \det(A-\lambda I) =&(8-\lambda) \begin{vmatrix} 7-\lambda&-4\\ -4&3-\lambda \end{vmatrix}\\ &+6 \begin{vmatrix} -6&-4\\ 2&3-\lambda \end{vmatrix} +2 \begin{vmatrix} -6&7-\lambda\\ 2&-4 \end{vmatrix}. \end{aligned}

Calculate the three minors separately:

\begin{aligned} \begin{vmatrix} 7-\lambda&-4\\ -4&3-\lambda \end{vmatrix} &=(7-\lambda)(3-\lambda)-16\\ &=21-10\lambda+\lambda^2-16\\ &=\lambda^2-10\lambda+5, \end{aligned} \begin{aligned} \begin{vmatrix} -6&-4\\ 2&3-\lambda \end{vmatrix} &=-6(3-\lambda)+8\\ &=6\lambda-10, \end{aligned} \begin{aligned} \begin{vmatrix} -6&7-\lambda\\ 2&-4 \end{vmatrix} &=24-2(7-\lambda)\\ &=10+2\lambda. \end{aligned}

Hence

\begin{aligned} \det(A-\lambda I) &=(8-\lambda)(\lambda^2-10\lambda+5) +6(6\lambda-10)+2(10+2\lambda)\\ &=-\lambda^3+18\lambda^2-45\lambda\\ &=-\lambda(\lambda^2-18\lambda+45)\\ &=-\lambda(\lambda-3)(\lambda-15). \end{aligned}

Thus the characteristic equation may be written as

\lambda(\lambda-3)(\lambda-15)=0.

Therefore the characteristic roots are

\lambda_1=0,\qquad \lambda_2=3,\qquad \lambda_3=15.

Final Answer:

\boxed{\lambda(\lambda-3)(\lambda-15)=0}

and the roots are \boxed{0,\ 3,\ 15}.

Solution:

Let

v_1=(1,2,1),\qquad v_2=(2,1,0),\qquad v_3=(1,-1,2).

Definition: A set of vectors is a basis of V_3(\mathbb R) if the vectors are linearly independent and span the space. Since \dim V_3(\mathbb R)=3, any three linearly independent vectors automatically form a basis.

Place the three vectors as columns of a matrix:

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

Compute its determinant:

\begin{aligned} \det M &= 1\begin{vmatrix}1&-1\\0&2\end{vmatrix} -2\begin{vmatrix}2&-1\\1&2\end{vmatrix} +1\begin{vmatrix}2&1\\1&0\end{vmatrix}\\ &= 1(2)-2(4+1)+(0-1)\\ &=2-10-1\\ &=-9. \end{aligned}

Since

\det M=-9\ne0,

the columns v_1,v_2,v_3 are linearly independent.

There are three linearly independent vectors in the three-dimensional vector space V_3(\mathbb R). Hence they span the whole space.

Final Answer:

\boxed{S=\{(1,2,1),(2,1,0),(1,-1,2)\}\text{ is a basis of }V_3(\mathbb R).}

Solution:

Cauchy’s Mean Value Theorem: If f and g are continuous on [a,b], differentiable on (a,b), and g'(x)\ne0 in (a,b), then there exists c\in(a,b) such that

\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}.

Here

f(x)=\sin x,\qquad g(x)=\cos x, \qquad a=0,\quad b=\frac{\pi}{2}.

The hypotheses are satisfied on the given interval. Compute the left-hand side:

\frac{f(\pi/2)-f(0)}{g(\pi/2)-g(0)} = \frac{1-0}{0-1} =-1.

Also

f'(x)=\cos x,\qquad g'(x)=-\sin x.

Therefore, by Cauchy’s theorem,

-1 = \frac{\cos c}{-\sin c} = -\cot c.

Thus

\cot c=1.

Since 0<c<\pi/2, the unique value is

c=\frac{\pi}{4}.

Final Answer:

\boxed{c=\frac{\pi}{4}}

Solution:

The differential equation is

(D^3-7D-6)y=e^{2x}(1+x).

Step 1: Complementary function.

The auxiliary equation is

m^3-7m-6=0.

Since m=3 is a root, factorization gives

m^3-7m-6=(m-3)(m+1)(m+2).

Hence the roots are

m=3,\quad -1,\quad -2,

and therefore

y_c=C_1e^{3x}+C_2e^{-x}+C_3e^{-2x}.

Step 2: Particular integral.

Use the shift rule

\frac{1}{F(D)}e^{ax}V(x) = e^{ax}\frac{1}{F(D+a)}V(x).

Here

F(D)=D^3-7D-6.

Therefore

F(D+2) =(D+2)^3-7(D+2)-6.

Expand:

\begin{aligned} F(D+2) &=D^3+6D^2+12D+8-7D-14-6\\ &=D^3+6D^2+5D-12. \end{aligned}

Hence

y_p = e^{2x} \frac{1}{D^3+6D^2+5D-12}(1+x).

Since the right side is a polynomial of degree one, seek

V=Ax+B

such that

(D^3+6D^2+5D-12)V=1+x.

For V=Ax+B,

DV=A,\qquad D^2V=0,\qquad D^3V=0.

Thus

5A-12(Ax+B)=1+x.

Equating coefficients of x,

-12A=1 \quad\Longrightarrow\quad A=-\frac1{12}.

Equating constant terms,

5A-12B=1.

Substitute A=-1/12:

-\frac5{12}-12B=1 \quad\Longrightarrow\quad -12B=\frac{17}{12} \quad\Longrightarrow\quad B=-\frac{17}{144}.

Therefore

y_p = e^{2x}\left(-\frac{x}{12}-\frac{17}{144}\right).

Final Answer:

\boxed{ y=C_1e^{3x}+C_2e^{-x}+C_3e^{-2x} -e^{2x}\left(\frac{x}{12}+\frac{17}{144}\right)}

Solution:

Given

u=x+y+z,\qquad v=x^2+y^2+z^2,\qquad w=xy+yz+zx.

Compute the three gradients:

\nabla u=(1,1,1), \nabla v=(2x,2y,2z),

and

\nabla w=(y+z,z+x,x+y).

The required scalar triple product is

(\nabla u)\cdot\{(\nabla v)\times(\nabla w)\} = \begin{vmatrix} 1&1&1\\ 2x&2y&2z\\ y+z&z+x&x+y \end{vmatrix}.

Notice that

\nabla w = (x+y+z)(1,1,1)-(x,y,z).

Also

\nabla v=2(x,y,z).

Hence \nabla w is a linear combination of \nabla u and \nabla v:

\nabla w = u\,\nabla u-\frac12\nabla v.

Therefore the three vectors are linearly dependent. The scalar triple product of linearly dependent vectors is zero.

Final Answer:

\boxed{ (\nabla u)\cdot\{(\nabla v)\times(\nabla w)\}=0}

Solution:

Let the side AB be taken as the positive x-direction. The forces act along BC,CA,AB respectively, in the directions indicated by the order of the letters.

In an equilateral triangle, these three directed sides make angles

120^\circ,\qquad 240^\circ,\qquad 0^\circ

with the positive x-axis.

Thus the three force vectors are

\vec F_1 = 13(\cos120^\circ\,\mathbf i+\sin120^\circ\,\mathbf j) = -\frac{13}{2}\mathbf i +\frac{13\sqrt3}{2}\mathbf j, \vec F_2 = 10(\cos240^\circ\,\mathbf i+\sin240^\circ\,\mathbf j) = -5\mathbf i-5\sqrt3\,\mathbf j,

and

\vec F_3=5\mathbf i.

Therefore the resultant is

\begin{aligned} \vec R &=\vec F_1+\vec F_2+\vec F_3\\ &= \left(-\frac{13}{2}-5+5\right)\mathbf i + \left(\frac{13\sqrt3}{2}-5\sqrt3\right)\mathbf j\\ &= -\frac{13}{2}\mathbf i +\frac{3\sqrt3}{2}\mathbf j. \end{aligned}

Its magnitude is

\begin{aligned} R &= \sqrt{ \left(\frac{13}{2}\right)^2+ \left(\frac{3\sqrt3}{2}\right)^2}\\ &= \frac12\sqrt{169+27}\\ &= \frac12\sqrt{196}\\ &=7. \end{aligned}

If \theta is the angle measured counter-clockwise from the direction AB, then the resultant lies in the second quadrant and

\tan(\pi-\theta) = \frac{3\sqrt3}{13}.

Hence

\theta = \pi-\tan^{-1}\left(\frac{3\sqrt3}{13}\right).

Final Answer:

The resultant has magnitude

\boxed{7\text{ kg-wt}}

and direction

\boxed{ \theta=\pi-\tan^{-1}\left(\frac{3\sqrt3}{13}\right)}

from AB in the positive sense.

Solution:

Write the system in matrix form:

AX=B,

where

A= \begin{pmatrix} 1&1&1\\ 1&-1&1\\ 2&2&-1 \end{pmatrix}, \qquad X= \begin{pmatrix}x\\y\\z\end{pmatrix}, \qquad B= \begin{pmatrix}6\\2\\1\end{pmatrix}.

First compute the determinant:

\begin{aligned} |A| &= 1\begin{vmatrix}-1&1\\2&-1\end{vmatrix} -1\begin{vmatrix}1&1\\2&-1\end{vmatrix} +1\begin{vmatrix}1&-1\\2&2\end{vmatrix}\\ &=(1-2)-(-1-2)+(2+2)\\ &=-1+3+4\\ &=6. \end{aligned}

Since |A|=6\ne0, A^{-1} exists. Direct calculation of the adjugate gives

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

Hence

X=A^{-1}B.

Multiply:

\begin{aligned} x &= -\frac16(6)+\frac12(2)+\frac13(1) =-1+1+\frac13 =\frac13,\\ y &= \frac12(6)-\frac12(2) =3-1 =2,\\ z &= \frac23(6)-\frac13(1) =4-\frac13 =\frac{11}{3}. \end{aligned}

Verification:

\frac13+2+\frac{11}{3}=6, \frac13-2+\frac{11}{3}=2, \frac23+4-\frac{11}{3}=1.

Final Answer:

\boxed{x=\frac13,\qquad y=2,\qquad z=\frac{11}{3}}.

Solution:

The mapping is

f(u_1,u_2,u_3)=(u_1-u_2,\;u_1-u_3).

Definition: A mapping f:V\to W is linear if, for all vectors u,v and scalars \alpha,\beta,

f(\alpha u+\beta v)=\alpha f(u)+\beta f(v).

Let

u=(u_1,u_2,u_3),\qquad v=(v_1,v_2,v_3).

Then

\alpha u+\beta v = (\alpha u_1+\beta v_1,\; \alpha u_2+\beta v_2,\; \alpha u_3+\beta v_3).

Apply f:

\begin{aligned} f(\alpha u+\beta v) =&\big( \alpha u_1+\beta v_1-\alpha u_2-\beta v_2,\\ &\alpha u_1+\beta v_1-\alpha u_3-\beta v_3 \big). \end{aligned}

Collect the \alpha– and \beta-terms:

\begin{aligned} f(\alpha u+\beta v) &= \alpha(u_1-u_2,\;u_1-u_3) + \beta(v_1-v_2,\;v_1-v_3)\\ &= \alpha f(u)+\beta f(v). \end{aligned}

Therefore the defining condition of linearity is satisfied.

Final Answer:

\boxed{f:V_3(\mathbb R)\to V_2(\mathbb R)\text{ is a linear transformation}.}

Solution:

Cauchy-Schwarz inequality: In a real inner-product space, for all vectors u,v,

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

Equivalently,

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

Proof.

If v=0, both sides are zero and the result is immediate. Assume v\ne0.

For any real number t, positive definiteness gives

0\le\|u-tv\|^2.

Expand:

\begin{aligned} \|u-tv\|^2 &=\langle u-tv,u-tv\rangle\\ &=\langle u,u\rangle -2t\langle u,v\rangle +t^2\langle v,v\rangle. \end{aligned}

Choose

t=\frac{\langle u,v\rangle}{\langle v,v\rangle}.

Then

\begin{aligned} 0 &\le \langle u,u\rangle - 2\frac{\langle u,v\rangle^2}{\langle v,v\rangle} + \frac{\langle u,v\rangle^2}{\langle v,v\rangle}\\ &= \langle u,u\rangle - \frac{\langle u,v\rangle^2}{\langle v,v\rangle}. \end{aligned}

Multiplying by the positive quantity \langle v,v\rangle,

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

Taking square roots yields

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

Equality holds precisely when u-tv=0 for the above value of t, i.e. when u and v are linearly dependent.

Final Answer:

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

with equality if and only if the two vectors are linearly dependent.

Solution:

We need

L=\lim_{x\to0}\left(\frac1{x^2}-\cot^2x\right).

Use the standard Maclaurin expansion

\cot x = \frac1x-\frac{x}{3}-\frac{x^3}{45}+O(x^5).

Squaring only to the order needed for the limit,

\begin{aligned} \cot^2x &= \left(\frac1x-\frac{x}{3}+O(x^3)\right)^2\\ &= \frac1{x^2} -\frac23 +O(x^2). \end{aligned}

Therefore

\frac1{x^2}-\cot^2x = \frac23+O(x^2).

Letting x\to0,

Final Answer:

\boxed{\displaystyle \frac23}

Solution:

The astroid is

x^{2/3}+y^{2/3}=a^{2/3}.

Use the standard parametrization

x=a\cos^3\theta,\qquad y=a\sin^3\theta.

Step 1: Radius of curvature.

Differentiate:

\frac{dx}{d\theta} = -3a\cos^2\theta\sin\theta, \frac{dy}{d\theta} = 3a\sin^2\theta\cos\theta.

Differentiate again:

\frac{d^2x}{d\theta^2} = 6a\cos\theta\sin^2\theta -3a\cos^3\theta, \frac{d^2y}{d\theta^2} = 6a\sin\theta\cos^2\theta -3a\sin^3\theta.

The radius of curvature for a parametric curve is

\rho = \frac{\left[ (x')^2+(y')^2 \right]^{3/2}} {|x'y''-y'x''|}.

Now

\begin{aligned} (x')^2+(y')^2 &= 9a^2 \left( \cos^4\theta\sin^2\theta + \sin^4\theta\cos^2\theta \right)\\ &= 9a^2\sin^2\theta\cos^2\theta. \end{aligned}

Also, direct substitution gives

|x'y''-y'x''| = 9a^2|\sin\theta\cos\theta|.

Hence

\rho = \frac{ 27a^3|\sin^3\theta\cos^3\theta|} {9a^2|\sin\theta\cos\theta|} = 3a|\sin\theta\cos\theta|.

Step 2: Perpendicular from the origin to the tangent.

The tangent to the astroid at the parametric point is

\frac{X}{a\cos\theta} + \frac{Y}{a\sin\theta} =1.

Write it as

\frac{X}{\cos\theta} + \frac{Y}{\sin\theta} -a=0.

The perpendicular distance p of the origin from this line is

\begin{aligned} p &= \frac{a}{ \sqrt{\sec^2\theta+\csc^2\theta}}\\ &= \frac{a}{ \sqrt{ \frac{\sin^2\theta+\cos^2\theta} {\sin^2\theta\cos^2\theta}}}\\ &= a|\sin\theta\cos\theta|. \end{aligned}

Comparing with the radius of curvature,

\rho=3p.

Final Answer:

\boxed{\rho=3p}

Hence the radius of curvature is three times the perpendicular from the origin to the tangent.

Solution:

Given

u=x\phi(y/x)+\psi(y/x).

Write

u=v+w,

where

v=x\phi(y/x), \qquad w=\psi(y/x).

Euler’s theorem, second-order form: If F(x,y) is homogeneous of degree n, then

x^2F_{xx}+2xyF_{xy}+y^2F_{yy} = n(n-1)F.

Now

v(tx,ty) = tx\,\phi\left(\frac{ty}{tx}\right) = t\,v(x,y),

so v is homogeneous of degree 1. Therefore

x^2v_{xx}+2xyv_{xy}+y^2v_{yy} = 1(1-1)v =0.

Similarly,

w(tx,ty) = \psi\left(\frac{ty}{tx}\right) = w(x,y),

so w is homogeneous of degree 0. Hence

x^2w_{xx}+2xyw_{xy}+y^2w_{yy} = 0(0-1)w =0.

The differential operator

L=x^2\frac{\partial^2}{\partial x^2} +2xy\frac{\partial^2}{\partial x\partial y} +y^2\frac{\partial^2}{\partial y^2}

is linear. Therefore

L(u)=L(v+w)=L(v)+L(w)=0+0=0.

Final Answer:

\boxed{ x^{2}u_{xx}+2xyu_{xy}+y^{2}u_{yy}=0}.

Hence the required result is proved.

Solution:

We are given the two implicit relations

F_1=u^3+v^3-x-y=0, F_2=u^2+v^2-x^3-y^3=0.

Implicit Jacobian formula: If F_1=F_2=0 determine u,v locally as functions of x,y, then

\frac{\partial(u,v)}{\partial(x,y)} = \frac{ \dfrac{\partial(F_1,F_2)}{\partial(x,y)} }{ \dfrac{\partial(F_1,F_2)}{\partial(u,v)} }.

First compute the numerator:

\begin{aligned} \frac{\partial(F_1,F_2)}{\partial(x,y)} &= \begin{vmatrix} -1&-1\\ -3x^2&-3y^2 \end{vmatrix}\\ &= 3y^2-3x^2\\ &= 3(y^2-x^2). \end{aligned}

Now compute the denominator:

\begin{aligned} \frac{\partial(F_1,F_2)}{\partial(u,v)} &= \begin{vmatrix} 3u^2&3v^2\\ 2u&2v \end{vmatrix}\\ &= 6u^2v-6uv^2\\ &= 6uv(u-v). \end{aligned}

Therefore

\frac{\partial(u,v)}{\partial(x,y)} = \frac{3(y^2-x^2)}{6uv(u-v)}.

Final Answer:

\boxed{ \frac{\partial(u,v)}{\partial(x,y)} = \frac{y^2-x^2}{2uv(u-v)}}

at points where the denominator is nonzero.

Solution:

The solid is the cone

x^2+y^2=z^2

cut by the plane z=1. For the enclosed finite cone,

0\le z\le1.

Use cylindrical coordinates

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

The cone becomes

r^2=z^2.

Since z\ge0, the interior is

0\le r\le z,\qquad 0\le\theta\le2\pi,\qquad 0\le z\le1.

Hence

\iiint_V2z\,dV = \int_0^1\int_0^{2\pi}\int_0^z 2z\,r\,dr\,d\theta\,dz.

Evaluate the r-integral:

\int_0^z2zr\,dr = 2z\left[\frac{r^2}{2}\right]_0^z = z^3.

Thus

\begin{aligned} \iiint_V2z\,dV &= \int_0^1\int_0^{2\pi}z^3\,d\theta\,dz\\ &= 2\pi\int_0^1z^3\,dz\\ &= 2\pi\left[\frac{z^4}{4}\right]_0^1\\ &= \frac{\pi}{2}. \end{aligned}

Final Answer:

\boxed{\displaystyle \frac{\pi}{2}}

Solution:

Let the sphere have radius R and centre at the origin. Let the half-lengths of an inscribed rectangular solid be x,y,z>0. Then one vertex is (x,y,z), and because the vertex lies on the sphere,

x^2+y^2+z^2=R^2.

The side lengths of the rectangular solid are 2x,2y,2z, so its volume is

V=8xyz.

Use AM-GM on x^2,y^2,z^2:

\frac{x^2+y^2+z^2}{3} \ge (x^2y^2z^2)^{1/3}.

Using the constraint,

\frac{R^2}{3} \ge (xyz)^{2/3}.

Raise both sides to the power 3/2:

xyz \le \left(\frac{R^2}{3}\right)^{3/2}.

Therefore

V = 8xyz \le 8\left(\frac{R^2}{3}\right)^{3/2}.

Equality in AM-GM holds if and only if

x^2=y^2=z^2.

Since x,y,z are positive lengths,

x=y=z=\frac{R}{\sqrt3}.

Thus the three full side lengths are all equal:

2x=2y=2z=\frac{2R}{\sqrt3}.

A rectangular solid with all three side lengths equal is a cube.

Final Answer:

\boxed{\text{The rectangular solid of maximum volume inscribed in a sphere is a cube.}}

Solution:

The equation is

11y^2+14yz+8zx+14xy-6x-16y+2z-2=0.

Write it in matrix form

X^TAX+2d^TX-2=0,

where

X= \begin{pmatrix}x\\y\\z\end{pmatrix}, \qquad A= \begin{pmatrix} 0&7&4\\ 7&11&7\\ 4&7&0 \end{pmatrix}, \qquad d= \begin{pmatrix}-3\\-8\\1\end{pmatrix}.

Step 1: Find the centre.

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

AH+d=0.

Thus

\begin{cases} 7k+4l=3,\\ 7h+11k+7l=8,\\ 4h+7k=-1. \end{cases}

Solving gives

h=\frac13,\qquad k=-\frac13,\qquad l=\frac43.

Translate the origin to the centre by putting

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

After this translation the linear terms disappear, and direct substitution gives

\begin{pmatrix}X&Y&Z\end{pmatrix} A \begin{pmatrix}X\\Y\\Z\end{pmatrix} +1=0.

Step 2: Find the principal directions.

The characteristic equation of A is

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

Its factorization is

(\lambda-18)(\lambda+3)(\lambda+4)=0.

Therefore the eigenvalues are

18,\qquad -3,\qquad -4.

Corresponding mutually orthogonal eigenvectors may be chosen as

(1,2,1),\qquad (1,-1,1),\qquad (1,0,-1).

Normalize them:

e_1=\frac{(1,2,1)}{\sqrt6}, \qquad e_2=\frac{(1,-1,1)}{\sqrt3}, \qquad e_3=\frac{(1,0,-1)}{\sqrt2}.

Introduce principal coordinates

U=\frac{X+2Y+Z}{\sqrt6}, V=\frac{X-Y+Z}{\sqrt3}, W=\frac{X-Z}{\sqrt2}.

Since the transformation is orthogonal, the quadratic form becomes diagonal:

18U^2-3V^2-4W^2+1=0.

Hence

3V^2+4W^2-18U^2=1.

In standard denominator form,

\frac{V^2}{1/3} + \frac{W^2}{1/4} - \frac{U^2}{1/18} =1.

There are two positive square terms and one negative square term. Therefore the surface is a hyperboloid of one sheet. Its centre is (1/3,-1/3,4/3), and its principal axis corresponding to the negative term is in the direction (1,2,1).

Final Answer:

Canonical form:

\boxed{3V^2+4W^2-18U^2=1}

Hence the surface is a hyperboloid of one sheet.

Solution:

The given circle is the intersection of

S\equiv x^2+y^2+z^2+10y-4z-8=0

and the plane

P\equiv x+y+z-3=0.

Every sphere through this circle belongs to the coaxal family

S+\lambda P=0.

Thus the required sphere has equation

x^2+y^2+z^2 +\lambda x +(10+\lambda)y +(-4+\lambda)z -8-3\lambda =0.

Its centre is

C= \left( -\frac{\lambda}{2}, -\frac{10+\lambda}{2}, \frac{4-\lambda}{2} \right).

For the given circle to be a great circle of this sphere, its plane must pass through the centre. Therefore C must satisfy

x+y+z=3.

Substitute the coordinates of C:

-\frac{\lambda}{2} -\frac{10+\lambda}{2} +\frac{4-\lambda}{2} =3.

Multiply by 2:

-\lambda-10-\lambda+4-\lambda=6.

Hence

-3\lambda-6=6 \quad\Longrightarrow\quad \lambda=-4.

Substitute \lambda=-4 in the sphere family:

x^2+y^2+z^2-4x+6y-8z+4=0.

Complete squares:

(x-2)^2+(y+3)^2+(z-4)^2=25.

Thus the centre is (2,-3,4), which indeed lies on x+y+z=3, and the radius is 5.

Final Answer:

\boxed{x^2+y^2+z^2-4x+6y-8z+4=0}

or equivalently

\boxed{(x-2)^2+(y+3)^2+(z-4)^2=25}.

Solution:

The direction ratios of the generators are

(1,-2,3).

Let (X,Y,0) be a point on the guiding ellipse

X^2+2Y^2=1.

The generator through (X,Y,0) is parallel to (1,-2,3). Hence a general point (x,y,z) on it is

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

From z=3t,

t=\frac z3.

Therefore

X=x-\frac z3, \qquad Y=y+\frac{2z}{3}.

Since (X,Y,0) lies on the ellipse,

\left(x-\frac z3\right)^2 + 2\left(y+\frac{2z}{3}\right)^2 =1.

Multiplying by 9 gives the polynomial form

(3x-z)^2+2(3y+2z)^2=9.

Final Answer:

\boxed{(3x-z)^2+2(3y+2z)^2=9}.

Solution:

Let

p=\frac{dy}{dx}.

The differential equation is

p^2(2-3y)^2=4(1-y).

For non-constant solutions with y<1, take square roots:

p(2-3y)=\pm2\sqrt{1-y}.

Since p=dy/dx, invert:

\frac{dx}{dy} = \pm\frac{2-3y}{2\sqrt{1-y}}.

Put

u=\sqrt{1-y}.

Then

y=1-u^2, \qquad dy=-2u\,du,

and

2-3y = 2-3(1-u^2) = -1+3u^2.

Thus

dx = \pm \frac{-1+3u^2}{2u} (-2u\,du) = \pm(1-3u^2)\,du.

Integrating,

x = C\pm(u-u^3).

Now

u-u^3 = u(1-u^2) = y\sqrt{1-y}.

Hence the two branches of the general integral are

x-C=\pm y\sqrt{1-y}.

Equivalently, eliminating the sign,

(x-C)^2=y^2(1-y).

Singular/constant solution.

In the original equation, if y=1, then p=0 and both sides are zero:

0^2(2-3)^2=4(1-1)=0.

Thus y=1 is also a solution. It is lost when we divide by \sqrt{1-y}.

Final Answer:

General integral:

\boxed{(x-C)^2=y^2(1-y)}

with the two signs represented by x-C=\pm y\sqrt{1-y}, together with the constant solution

\boxed{y=1}.

Solution:

The equation is

x^2y''-(x^2+2x)y'+(x+2)y=x^3e^x.

For x\ne0, divide by x^2:

y'' -\left(1+\frac2x\right)y' +\left(\frac1x+\frac2{x^2}\right)y = xe^x. \tag{1}

Step 1: Complementary function.

Observe that y_1=x is a solution of the homogeneous equation, because

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

For an equation y''+P(x)y'+Q(x)y=0, if one solution y_1 is known, a second solution is

y_2 = y_1\int \frac{e^{-\int P\,dx}}{y_1^2}\,dx.

Here

P(x)=-1-\frac2x.

Therefore

\int P\,dx=-x-2\log|x|,

and

e^{-\int P\,dx}=e^xx^2.

Since y_1=x,

y_2 = x\int\frac{e^xx^2}{x^2}\,dx = x\int e^x\,dx = xe^x.

Hence

y_c=C_1x+C_2xe^x.

Step 2: Particular integral by variation of parameters.

For equation (1),

R(x)=xe^x.

With y_1=x and y_2=xe^x, the Wronskian is

\begin{aligned} W &= \begin{vmatrix} x&xe^x\\ 1&(x+1)e^x \end{vmatrix}\\ &= x(x+1)e^x-xe^x\\ &=x^2e^x. \end{aligned}

The variation-of-parameters formula is

y_p = -y_1\int\frac{y_2R}{W}\,dx + y_2\int\frac{y_1R}{W}\,dx.

Compute the first integrand:

\frac{y_2R}{W} = \frac{(xe^x)(xe^x)}{x^2e^x} = e^x.

Compute the second:

\frac{y_1R}{W} = \frac{x(xe^x)}{x^2e^x} =1.

Thus

\begin{aligned} y_p &= -x\int e^x\,dx + xe^x\int1\,dx\\ &= -xe^x+x^2e^x\\ &= x(x-1)e^x. \end{aligned}

Final Answer:

\boxed{ y=C_1x+C_2xe^x+x(x-1)e^x}.

Solution:

Generating function of Bessel functions:

\exp\left[\frac{x}{2}\left(t-\frac1t\right)\right] = \sum_{n=-\infty}^{\infty}J_n(x)t^n. \tag{1}

Differentiate (1) with respect to t. Multiplying immediately by t, the left-hand side gives

t\frac{\partial}{\partial t} \exp\left[\frac{x}{2}\left(t-\frac1t\right)\right] = \frac{x}{2}\left(t+\frac1t\right) \exp\left[\frac{x}{2}\left(t-\frac1t\right)\right].

For the right-hand side,

t\frac{\partial}{\partial t} \sum_{n=-\infty}^{\infty}J_n(x)t^n = \sum_{n=-\infty}^{\infty}nJ_n(x)t^n.

Substituting the generating function back into the left side,

\sum_{n=-\infty}^{\infty}nJ_n(x)t^n = \frac{x}{2} \left(t+\frac1t\right) \sum_{n=-\infty}^{\infty}J_n(x)t^n.

Expand the right side:

\frac{x}{2} \sum_{n=-\infty}^{\infty} J_n(x) \left(t^{n+1}+t^{n-1}\right).

The coefficient of t^n on the right is

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

Equating coefficients of t^n on both sides,

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

Multiplying by 2 gives the required recurrence relation.

Final Answer:

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

Solution:

Let f(x,y,z) and g(x,y,z) be scalar fields.

We have

\nabla g = g_x\mathbf i+g_y\mathbf j+g_z\mathbf k.

Therefore

f\nabla g = fg_x\mathbf i+fg_y\mathbf j+fg_z\mathbf k.

Taking divergence,

\begin{aligned} \operatorname{div}(f\nabla g) &= \frac{\partial}{\partial x}(fg_x) + \frac{\partial}{\partial y}(fg_y) + \frac{\partial}{\partial z}(fg_z). \end{aligned}

Apply the ordinary product rule to each term:

\begin{aligned} \operatorname{div}(f\nabla g) =& f_xg_x+fg_{xx} +f_yg_y+fg_{yy}\\ &+f_zg_z+fg_{zz}. \end{aligned}

Group the terms:

\operatorname{div}(f\nabla g) = f(g_{xx}+g_{yy}+g_{zz}) +(f_xg_x+f_yg_y+f_zg_z).

Recognize

\nabla^2g=g_{xx}+g_{yy}+g_{zz}

and

\nabla f\cdot\nabla g = f_xg_x+f_yg_y+f_zg_z.

Final Answer:

\boxed{ \operatorname{div}(f\nabla g) = f\nabla^2g+\nabla f\cdot\nabla g}.

Hence proved.

Solution:

Let

\vec a=(a_1,a_2,a_3).

Using index notation with the Levi-Civita symbol, the i-th component of \nabla\times(\nabla\times\vec a) is

[\nabla\times(\nabla\times\vec a)]_i = \varepsilon_{ijk}\partial_j \left( \varepsilon_{klm}\partial_l a_m \right).

Use the identity

\varepsilon_{ijk}\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}.

Therefore

\begin{aligned} [\nabla\times(\nabla\times\vec a)]_i &= (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) \partial_j\partial_l a_m\\ &= \partial_j\partial_i a_j - \partial_j\partial_j a_i. \end{aligned}

Assuming the required second partial derivatives are continuous, mixed derivatives may be interchanged. Hence

\partial_j\partial_i a_j = \partial_i(\partial_j a_j).

Thus

[\nabla\times(\nabla\times\vec a)]_i = [\nabla(\nabla\cdot\vec a)]_i - [\nabla^2\vec a]_i.

Since this holds for each component i=1,2,3,

Final Answer:

\boxed{ \nabla\times(\nabla\times\vec a) = \nabla(\nabla\cdot\vec a)-\nabla^2\vec a}.

Solution:

Write the line integral as

\oint_C(P\,dx+Q\,dy),

where

P=\cos x\sin y-xy, \qquad Q=\sin x\cos y.

Green’s theorem: If C is the positively oriented boundary of a region D, then

\oint_C(P\,dx+Q\,dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)dA.

Here D is the unit disc x^2+y^2\le1. Compute the derivatives:

\frac{\partial Q}{\partial x} = \cos x\cos y, \frac{\partial P}{\partial y} = \cos x\cos y-x.

Therefore

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

Hence

\oint_C(P\,dx+Q\,dy) = \iint_Dx\,dA.

The disc D is symmetric about the y-axis, while x is odd under x\mapsto-x. Therefore the double integral vanishes:

\iint_Dx\,dA=0.

Final Answer:

\boxed{0}

Solution:

Let the weight of each of the six equal heavy rods be W. The regular hexagon is suspended from A. By symmetry its vertical diagonal AD is vertical, while the light rods BF and CE are horizontal.

Equivalent joint loading: The weight W of each uniform side rod acts at its midpoint. For the equilibrium of the hinged framework, this load is statically equivalent to two vertical loads W/2 at the two ends of that rod. Since two heavy rods meet at every vertex, each vertex therefore carries an equivalent downward load W.

The total weight is 6W, so the upward support reaction at A is 6W. We now use joint equilibrium. Let tension be positive; a negative answer therefore denotes thrust (compression).

Joint D.

By symmetry the forces in rods CD and DE are equal; call each T. Each side makes an angle 30^\circ with the horizontal. Vertical equilibrium gives

2T\sin30^\circ=W.

Since \sin30^\circ=1/2,

T=W.

Thus

T_{CD}=T_{DE}=W.

Joint C.

At C, the force T_{CD}=W acts along CD, the force in BC is vertical, the force in the light rod CE is horizontal, and the equivalent joint load W acts downward.

Vertical equilibrium:

T_{BC}-W-W\sin30^\circ=0.

Therefore

T_{BC} = W+\frac W2 = \frac{3W}{2}.

Horizontal equilibrium, taking tension in CE as positive toward E, gives

-T_{CE}-W\cos30^\circ=0.

Thus

T_{CE} = -W\frac{\sqrt3}{2}.

The negative sign means the member is in compression. Therefore the thrust in CE is

\frac{\sqrt3}{2}W.

Joint B.

At B, BC pulls vertically downward with magnitude 3W/2, the joint load W acts downward, AB makes 30^\circ with the horizontal, and BF is horizontal.

Vertical equilibrium gives

T_{AB}\sin30^\circ-\frac{3W}{2}-W=0.

Hence

\frac{T_{AB}}2=\frac{5W}{2} \quad\Longrightarrow\quad T_{AB}=5W.

Horizontal equilibrium gives

-T_{BF}-T_{AB}\cos30^\circ=0.

Thus

T_{BF} = -5W\frac{\sqrt3}{2}.

Again the negative sign indicates compression. Hence the thrust in BF is 5\sqrt3W/2.

Final Answer:

If each heavy rod has weight W, then

\boxed{\text{Thrust in }BF=\frac{5\sqrt3}{2}W}

and

\boxed{\text{Thrust in }CE=\frac{\sqrt3}{2}W}.

Solution:

Let the particle have mass m. Take the downward direction as positive. Let its speed at time t be v.

The resistance is proportional to the speed, so write it as kv, where k>0. Since the particle moves downward, the resistance acts upward.

The equation of motion is therefore

m\frac{dv}{dt}=mg-kv.

Rearrange:

\frac{dv}{dt}+\frac{k}{m}v=g.

This is a first-order linear differential equation. Its integrating factor is

e^{kt/m}.

Multiplying through,

\frac{d}{dt} \left( ve^{kt/m} \right) = ge^{kt/m}.

Integrate:

ve^{kt/m} = \frac{mg}{k}e^{kt/m}+C.

Thus

v=\frac{mg}{k}+Ce^{-kt/m}.

The particle starts from rest, so v(0)=0. Hence

0=\frac{mg}{k}+C,

and therefore

C=-\frac{mg}{k}.

Consequently

v(t) = \frac{mg}{k} \left(1-e^{-kt/m}\right).

The acceleration is

a=\frac{dv}{dt} = ge^{-kt/m}.

Thus the acceleration starts at g and decreases exponentially to zero.

As t\to\infty,

v(t)\to\frac{mg}{k}.

This limiting speed is the terminal velocity:

V_T=\frac{mg}{k}.

If s denotes the downward distance fallen from the initial position, then

\frac{ds}{dt} = \frac{mg}{k} \left(1-e^{-kt/m}\right).

Using s(0)=0, integration gives

\begin{aligned} s(t) &= \frac{mg}{k} \left[ t-\frac{m}{k} \left(1-e^{-kt/m}\right) \right]. \end{aligned}

Final Answer:

\boxed{ v(t)=\frac{mg}{k}\left(1-e^{-kt/m}\right)} \boxed{ a(t)=ge^{-kt/m}} \boxed{ s(t)= \frac{mg}{k} \left[ t-\frac{m}{k}(1-e^{-kt/m}) \right]}

and the terminal velocity is \boxed{mg/k}.

Solution:

For a planet moving in an elliptical orbit under an inverse-square attraction toward the Sun at a focus, the greatest speed occurs at perihelion and the least speed occurs at aphelion.

Let

v_p=30\text{ km/s}, \qquad v_a=29.2\text{ km/s}.

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

r_p=a(1-e), \qquad r_a=a(1+e).

At perihelion and aphelion the velocity is perpendicular to the radius vector. Conservation of angular momentum therefore gives

v_pr_p=v_ar_a.

Substitute the expressions for r_p and r_a:

v_pa(1-e)=v_aa(1+e).

Cancel a:

v_p(1-e)=v_a(1+e).

Expand:

v_p-v_pe=v_a+v_ae.

Hence

v_p-v_a=e(v_p+v_a).

Therefore

e=\frac{v_p-v_a}{v_p+v_a}.

Substitute the numerical values:

e = \frac{30-29.2}{30+29.2} = \frac{0.8}{59.2} = \frac1{74}.

Thus

e\approx0.01351.

Final Answer:

\boxed{e=\frac1{74}\approx0.0135}.