Solution: prove both implications

1. An isomorphism preserves dimension

Suppose T:V\to W is a bijective linear map. Choose a basis v_1,\ldots,v_n of V. We show that T(v_1),\ldots,T(v_n) is a basis of W.

For linear independence, suppose \sum_i a_iT(v_i)=0. By linearity and injectivity,

T\left(\sum_i a_iv_i\right)=0=T(0) \Longrightarrow\sum_i a_iv_i=0\Longrightarrow a_1=\cdots=a_n=0.

For spanning, take any w\in W. Surjectivity supplies v\in V with T(v)=w. Expressing v=\sum_i b_iv_i gives w=\sum_i b_iT(v_i). Thus the image vectors form a basis containing n vectors, and \dim W=n=\dim V.

2. Equal dimensions give an isomorphism

Conversely, suppose \dim V=\dim W=n. Choose bases v_1,\ldots,v_n and w_1,\ldots,w_n, and define

T\left(\sum_i a_iv_i\right)=\sum_i a_iw_i.

Unique basis coordinates make this well-defined. For vectors v=\sum_i a_iv_i, u=\sum_i b_iv_i and scalars \alpha,\beta\in F,

T(\alpha v+\beta u)=\sum_i(\alpha a_i+\beta b_i)w_i =\alpha T(v)+\beta T(u).

If T(v)=0, independence of the w_i makes all coordinates of v zero, proving injectivity. Every \sum_i c_iw_i is the image of \sum_i c_iv_i, proving surjectivity. If n=0, both spaces consist only of zero and their unique map is also an isomorphism.

\boxed{V\cong W\iff\dim V=\dim W.}

Solution: gradient dotted with a unit direction

The direction is from P=(1,1,1) towards Q=(-3,5,6). Hence

\overrightarrow{PQ}=(-4,4,5),\qquad\|\overrightarrow{PQ}\|=\sqrt{16+16+25}=\sqrt{57},\qquad \widehat{\mathbf u}=\frac{(-4,4,5)}{\sqrt{57}}.

Differentiate the given function:

\nabla f=\left(8e^{2x-y+z},-4e^{2x-y+z},4e^{2x-y+z}\right).

At P, the exponent is 2(1)-1+1=2, giving \nabla f(P)=e^2(8,-4,4). Therefore

\begin{aligned}D_{\widehat{\mathbf u}}f(P) &=\nabla f(P)\cdot\widehat{\mathbf u}\\ &=\frac{e^2}{\sqrt{57}}[8(-4)+(-4)(4)+4(5)]\\ &=\boxed{-\frac{28e^2}{\sqrt{57}}}. \end{aligned}

The negative sign indicates initial decrease in the specified direction. The direction vector must be normalized because the derivative is measured per unit distance.

Solution: compare the tangent-plane normals

Translate to the ellipsoid’s centre and choose its principal axes, obtaining

\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,\qquad a,b,c>0.

A diameter passes through the centre. Its endpoints are therefore P=(x_1,y_1,z_1) and P’=(-x_1,-y_1,-z_1). For F=x^2/a^2+y^2/b^2+z^2/c^2-1, the tangent plane is given by \nabla F(P)\cdot((x,y,z)-P)=0. Thus

\frac{x_1(x-x_1)}{a^2}+\frac{y_1(y-y_1)}{b^2}+\frac{z_1(z-z_1)}{c^2}=0.

Using the ellipsoid equation satisfied by P, the plane becomes

\frac{xx_1}{a^2}+\frac{yy_1}{b^2}+\frac{zz_1}{c^2}=1.

At the opposite endpoint, its equation is

-\frac{xx_1}{a^2}-\frac{yy_1}{b^2}-\frac{zz_1}{c^2}=1, \quad\text{equivalently}\quad \frac{xx_1}{a^2}+\frac{yy_1}{b^2}+\frac{zz_1}{c^2}=-1.

The two distinct planes have the same nonzero normal vector (x_1/a^2,y_1/b^2,z_1/c^2), so they are parallel. Rotations and translations preserve parallelism; hence the result holds for any ellipsoid.

Solution: use the slope as a parameter

For p\ne-1, write the equation as

y=x\phi(p),\qquad\phi(p)=\frac{p^2}{p+1},\qquad p=\frac{dy}{dx}.

Differentiating gives

p=\phi(p)+x\phi'(p)\frac{dp}{dx},\qquad \phi'(p)=\frac{2p(p+1)-p^2}{(p+1)^2}=\frac{p(p+2)}{(p+1)^2}. \frac{p}{p+1}=x\frac{p(p+2)}{(p+1)^2}\frac{dp}{dx}.

On a branch with p\ne0,-1, regarding x as a function of p gives

\frac1x\frac{dx}{dp}=\frac{p+2}{p+1}=1+\frac1{p+1}.

Integrating and absorbing signs into the arbitrary constant,

\log|x|=p+\log|p+1|+C_0,\qquad x=C(p+1)e^p.

Substitute into the expression for y:

\boxed{x=C(p+1)e^p,\qquad y=Cp^2e^p.}

Verification and exceptional cases

\frac{dx}{dp}=Ce^p(p+2),\qquad\frac{dy}{dp}=Ce^pp(p+2),\qquad\frac{dy}{dx}=p

on regular graph branches. Also

xp^2-yp-y=Ce^p[(p+1)p^2-p^3-p^2]=0.

The cancelled case p=0 gives the additional solution y\equiv0. To check all constant slopes, let y=cx+d. Substitution gives -cx-(c+1)d=0 for every x, forcing c=d=0. Thus there are no other constant-slope solutions.

The value p=-1 can occur at x=0, but cannot hold on an interval. The parametric family extends regularly through that value. Use regular graph branches away from p=-2, where the parametric derivatives both vanish.

\boxed{\text{The parametric family above, together with }y=0.}

Solution: integrate the area of a semicircle

For the usual interpretation assume a>0. First integrate with respect to x:

I=\int_0^{2a}\left[x\right]_0^{\sqrt{2ay-y^2}}dy =\int_0^{2a}\sqrt{a^2-(y-a)^2}\,dy.

Set y-a=a\sin t. The limits become -\pi/2 and \pi/2, and dy=a\cos t\,dt. Since \cos t\ge0 on this interval,

\begin{aligned}I&=a^2\int_{-\pi/2}^{\pi/2}\cos^2t\,dt\\ &=a^2\left[\frac t2+\frac{\sin2t}{4}\right]_{-\pi/2}^{\pi/2} =\boxed{\frac{\pi a^2}{2}}. \end{aligned}

The region is the half-disk x^2+(y-a)^2\le a^2, x\ge0, confirming the result. At a=0 the integral is zero. If negative a is permitted, the reversed outer limits give -\pi a^2/2.

Solution: Gram–Schmidt construction

Choose a basis u_1,\ldots,u_n. In the complex case, use the convention that the inner product is linear in its first entry; the opposite convention requires the corresponding conjugated projection coefficients.

1. Construct nonzero orthogonal vectors

v_1=u_1,\qquad v_k=u_k-\sum_{j=1}^{k-1}\frac{\langle u_k,v_j\rangle}{\langle v_j,v_j\rangle}v_j \quad(2\le k\le n).

Assume the preceding vectors are nonzero and mutually orthogonal. For i<k,

\begin{aligned}\langle v_k,v_i\rangle &=\langle u_k,v_i\rangle- \sum_{j=1}^{k-1}\frac{\langle u_k,v_j\rangle}{\langle v_j,v_j\rangle}\langle v_j,v_i\rangle\\ &=\langle u_k,v_i\rangle- \frac{\langle u_k,v_i\rangle}{\langle v_i,v_i\rangle}\langle v_i,v_i\rangle=0. \end{aligned}

Inductively the construction gives

\operatorname{span}(v_1,\ldots,v_{k-1})=\operatorname{span}(u_1,\ldots,u_{k-1}).

If v_k=0, its defining equation expresses u_k in this preceding span, contradicting the independence of the original basis. Thus every new vector is nonzero. The defining equation also proves equality of the two spans after the next vector is included.

2. Normalize

e_k=\frac{v_k}{\|v_k\|},\qquad \langle e_i,e_j\rangle=\begin{cases}1,&i=j,\\0,&i\ne j.\end{cases}

The vectors still span V. They are independent: from \sum_i c_ie_i=0, taking the inner product with e_j gives c_j=0 for every j. Hence

\boxed{\{e_1,\ldots,e_n\}\text{ is an orthonormal basis of }V.}

Solution: exclude zero and nonzero eigenvalues

Integration is linear, so T is a linear operator. An eigenvalue \lambda would require a continuous function f\not\equiv0 satisfying

\int_0^x f(t)\,dt=\lambda f(x)\quad\text{for every }x.

Case 1: zero eigenvalue

If \lambda=0, the integral is identically zero. The fundamental theorem of calculus then gives

f(x)=\frac{d}{dx}\int_0^x f(t)\,dt=0,

contradicting the required nonzero eigenfunction.

Case 2: nonzero eigenvalue

If \lambda\ne0, the equation expresses f as a constant multiple of an integral of a continuous function. It follows that f is differentiable. Differentiation gives

\lambda f'(x)=f(x),\qquad f'(x)-\frac1\lambda f(x)=0.

Using the integrating factor e^{-x/\lambda},

\frac{d}{dx}\left(e^{-x/\lambda}f(x)\right)=0, \qquad f(x)=Ce^{x/\lambda}.

At x=0, the original integral equation gives 0=\lambda f(0). Since \lambda\ne0, C=f(0)=0. Again f\equiv0, a contradiction.

\boxed{T\text{ has no eigenvalues.}}

The proof applies on the real line, or on any nondegenerate interval containing zero on which this operator is defined.

Solution: complete the squares and translate

The condition presupposes a,b,c\ne0. Complete each square:

\begin{aligned} ax^2+2ux&=a\left(x+\frac ua\right)^2-\frac{u^2}{a},\\ by^2+2vy&=b\left(y+\frac vb\right)^2-\frac{v^2}{b},\\ cz^2+2wz&=c\left(z+\frac wc\right)^2-\frac{w^2}{c}. \end{aligned}

The original equation therefore becomes

a\left(x+\frac ua\right)^2+b\left(y+\frac vb\right)^2+c\left(z+\frac wc\right)^2 +d-\frac{u^2}{a}-\frac{v^2}{b}-\frac{w^2}{c}=0.

The assumed relation cancels the constant. With

X=x+\frac ua,\qquad Y=y+\frac vb,\qquad Z=z+\frac wc,

the equation reduces to

\boxed{aX^2+bY^2+cZ^2=0.}

If (X_0,Y_0,Z_0) satisfies it, every scalar multiple does as well, since

a(tX_0)^2+b(tY_0)^2+c(tZ_0)^2 =t^2(aX_0^2+bY_0^2+cZ_0^2)=0.

The entire line joining the translated origin to each point is thus on the surface: this is the cone property. Its vertex in the original coordinates is

\boxed{\left(-\frac ua,-\frac vb,-\frac wc\right).}

Qualification: a nondegenerate real cone requires mixed signs among a,b,c. If all have the same sign, the only real point is the vertex. The stated condition guarantees homogeneity, but does not by itself guarantee a nondegenerate real cone.

Solution: Lagrange multipliers and attainment

Set F=u^2+v^2+w^2, g=u^2/4+v^2/5+w^2/25-1, and h=u+v-w. The constraints form a compact ellipse, so both extrema exist. Their gradients are independent: if \nabla g=t\nabla h, then (u,v,w)=(2t,5t/2,-25t/2). The plane equation forces 17t=0, contradicting the ellipsoid equation.

1. Set up the equations

L=F-\lambda g-2\mu h. u\left(1-\frac\lambda4\right)=\mu,\qquad v\left(1-\frac\lambda5\right)=\mu,\qquad w\left(1-\frac\lambda{25}\right)=-\mu.

Multiply these equations by u,v,w, respectively, and add:

F-\lambda\left(\frac{u^2}{4}+\frac{v^2}{5}+\frac{w^2}{25}\right) =\mu(u+v-w)=0,\qquad F=\lambda.

If \mu=0, at most one coordinate can be nonzero because the three denominators 4,5,25 are distinct. The plane equation forces that coordinate to be zero as well, which is impossible. Hence \mu\ne0, and the stationarity equations also exclude \lambda=4,5,25.

2. Find the possible extreme values

u=\frac{4\mu}{4-\lambda},\quad v=\frac{5\mu}{5-\lambda},\quad w=-\frac{25\mu}{25-\lambda}.

Substitute into the plane constraint and divide by \mu:

\frac4{4-\lambda}+\frac5{5-\lambda}+\frac{25}{25-\lambda}=0. \begin{aligned}0 &=4(5-\lambda)(25-\lambda)+5(4-\lambda)(25-\lambda)+25(4-\lambda)(5-\lambda)\\ &=(500-120\lambda+4\lambda^2)+(500-145\lambda+5\lambda^2)+(500-225\lambda+25\lambda^2)\\ &=34\lambda^2-490\lambda+1500\\ &=2(\lambda-10)(17\lambda-75). \end{aligned} \lambda=10\quad\text{or}\quad\lambda=\frac{75}{17}.

3. Verify both values are attained

At \lambda=10, the ratios are u:v:w=2:3:5. Put (u,v,w)=t(2,3,5):

t^2\left(1+\frac95+1\right)=1 \Longrightarrow t^2=\frac5{19},\qquad F=38t^2=10.

At \lambda=75/17, the ratios are 8:-7:1. Put (u,v,w)=t(8,-7,1):

t^2\left(16+\frac{49}{5}+\frac1{25}\right)=1 \Longrightarrow t^2=\frac{25}{646},\qquad F=114t^2=\frac{75}{17}. \boxed{F_{\max}=10,\qquad F_{\min}=\frac{75}{17}.}

Maxima occur at \pm\sqrt{5/19}(2,3,5); minima at \pm(5/\sqrt{646})(8,-7,1). Compactness and the exhaustive stationarity calculation establish that these are the global extrema.

Solution: Beta and Gamma functions

Use the trigonometric Beta integral

B(p,q)=2\int_0^{\pi/2}\sin^{2p-1}x\cos^{2q-1}x\,dx,\qquad p,q>0.

It follows from substituting t=\sin^2x in B(p,q)=\int_0^1t^{p-1}(1-t)^{q-1}dt. Here \sqrt{\tan x}=\sin^{1/2}x\cos^{-1/2}x, so match exponents:

2p-1=\frac12\Longrightarrow p=\frac34,\qquad 2q-1=-\frac12\Longrightarrow q=\frac14. I=\frac12B\left(\frac34,\frac14\right) =\frac{\Gamma(3/4)\Gamma(1/4)}{2\Gamma(1)}.

Euler’s reflection identity gives

\Gamma(z)\Gamma(1-z)=\frac\pi{\sin\pi z},\qquad \Gamma(1/4)\Gamma(3/4)=\frac\pi{\sin(\pi/4)}=\pi\sqrt2.

Since \Gamma(1)=1,

\boxed{I=\frac{\pi\sqrt2}{2}=\frac\pi{\sqrt2}.}

The improper integral converges: the integrand behaves like x^{1/2} near zero and (\pi/2-x)^{-1/2} near the other endpoint. Both powers are integrable.

Solution: project the gradient along the unit vector

f_x=2xy^3+y,\qquad f_y=3x^2y^2+x.

At (2,1),

f_x(2,1)=4+1=5,\qquad f_y(2,1)=12+2=14.

Using the standard counterclockwise direction angle from the positive x-axis,

\widehat{\mathbf u}=(\cos(\pi/3),\sin(\pi/3))=\left(\frac12,\frac{\sqrt3}{2}\right). D_{\widehat{\mathbf u}}f(2,1)=\nabla f(2,1)\cdot\widehat{\mathbf u} =5\left(\frac12\right)+14\left(\frac{\sqrt3}{2}\right) =\boxed{\frac52+7\sqrt3}.

Solution: prove the equivalence using neighbourhoods

1. Continuity implies open preimages

Assume f:\mathbb R\to\mathbb R is continuous and G is open. If f^{-1}(G) is empty, it is open. Otherwise take x_0\in f^{-1}(G). Since f(x_0)\in G, some \varepsilon>0 satisfies

(f(x_0)-\varepsilon,f(x_0)+\varepsilon)\subseteq G.

Continuity provides \delta>0 such that

|x-x_0|<\delta\Longrightarrow |f(x)-f(x_0)|<\varepsilon.

Thus (x_0-\delta,x_0+\delta)\subseteq f^{-1}(G). Every point of the preimage is an interior point, so the preimage is open.

2. Open preimages imply continuity

Conversely, assume every open set has an open preimage. Fix x_0\in\mathbb R and \varepsilon>0. The set

G=(f(x_0)-\varepsilon,f(x_0)+\varepsilon)

is open, so U=f^{-1}(G) is open and contains x_0. Some \delta>0 therefore satisfies (x_0-\delta,x_0+\delta)\subseteq U. Hence

|x-x_0|<\delta\Longrightarrow x\in U \Longrightarrow |f(x)-f(x_0)|<\varepsilon.

This proves continuity at x_0, and since the point was arbitrary, on the whole real line.

\boxed{f\text{ continuous}\iff f^{-1}(G)\text{ open for every open }G.}

The notation f^{-1}(G) means preimage, and does not require an inverse function or bijectivity.

Solution: inspect zero and the interior singularity

1. Behaviour at zero

Using \sin x\sim x as x\to0^+,

\frac{\sin^2x}{\sqrt{x}(x-1)}\sim-x^{3/2}.

Since \int_0^\delta x^{3/2}dx=2\delta^{5/2}/5 is finite, there is absolute convergence near zero. The integrand is also continuous at four.

2. Behaviour at one

Let g(x)=\sin^2x/\sqrt{x}. It is continuously differentiable near one and g(1)=\sin^2(1)>0. Write

\frac{g(x)}{x-1}=\frac{g(1)}{x-1}+\frac{g(x)-g(1)}{x-1}.

The second term has finite limit g'(1) and is locally integrable. For the first term, with fixed small \delta>0,

\int_{1-\delta}^{1-\varepsilon}\frac{g(1)}{x-1}\,dx =g(1)(\log\varepsilon-\log\delta)\longrightarrow-\infty, \int_{1+\varepsilon}^{1+\delta}\frac{g(1)}{x-1}\,dx =g(1)(\log\delta-\log\varepsilon)\longrightarrow+\infty.

An improper integral with an interior singularity exists only if the one-sided integrals converge separately. These do not.

\boxed{\displaystyle\int_0^4\frac{\sin^2x}{\sqrt{x}(x-1)}\,dx\text{ diverges.}}

The two infinities cannot be cancelled in an ordinary improper integral. A symmetric Cauchy principal value is a different notion and can exist here, but is not ordinary convergence.

Solution: bounds from the alternating Bessel series

Interpretation: “smallest root” here means the smallest positive root. The function is even, so negative roots must not be included in that ordering.

J_0(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(k!)^2}\left(\frac{x^2}{4}\right)^k.

Put t=x^2/4 and A_k=t^k/(k!)^2. The ratio of successive term magnitudes is

\frac{A_{k+1}}{A_k}=\frac{t}{(k+1)^2}.

1. Exclude all roots up to two

For 0<x\le2, we have 0<t\le1. The alternating tail after the cubic term starts positive and has decreasing magnitudes. Consequently,

J_0(x)\ge1-t+\frac{t^2}{4}-\frac{t^3}{36} =1-t+\frac{t^2(9-t)}{36}>0.

Here 1-t\ge0 and the last term is strictly positive. Also J_0(0)=1. In particular, J_0(2)\ge2/9>0.

2. Show a negative value at the other endpoint

At x=\sqrt8, t=2. Starting with k=1, the magnitudes strictly decrease. Truncate after the positive term indexed by four: the remaining alternating tail starts negative, giving

\begin{aligned}J_0(\sqrt8) &\le1-2+\frac{2^2}{(2!)^2}-\frac{2^3}{(3!)^2}+\frac{2^4}{(4!)^2}\\ &=1-2+1-\frac29+\frac1{36}=-\frac7{36}<0. \end{aligned}

3. Locate the first positive root

The power series defines a continuous function. Opposite signs at two and \sqrt8 guarantee a zero strictly between them. Since there is no zero in [0,2], the smallest positive zero lies in that interval.

\boxed{2<j_{0,1}<\sqrt8.}

Solution: reduce to constant coefficients

The source contains \log x, so work on x>0. Put t=\log x, Y(t)=y(e^t). Then

xy’=Y’,\qquad x^2y”=Y”-Y’.

The equation becomes

Y”-2mY’+(m^2+n^2)Y=n^2e^{mt}t.

Set Y=e^{mt}Z(t) and compute

Y’=e^{mt}(mZ+Z’),\qquad Y”=e^{mt}(m^2Z+2mZ’+Z”).

Substitution cancels the terms involving m:

Z”+n^2Z=n^2t.

Case 1: n is nonzero

The complementary solution is Z_c=C_1\cos nt+C_2\sin nt. A particular solution is Z_p=t, since its second derivative is zero. Returning to the original variable,

\boxed{y=x^m[C_1\cos(n\log x)+C_2\sin(n\log x)+\log x],\quad n\ne0.}

Case 2: n is zero

Now Z”=0, so Z=C_1+C_2t. Hence

\boxed{y=x^m(C_1+C_2\log x),\quad n=0.}

The repeated-root case must be given separately; substituting zero into the sine-and-cosine family would lose one independent homogeneous solution.

Solution: Frobenius series, including the logarithmic solution

Divide by x(1-x):

y”+\frac1xy’-\frac1{x(1-x)}y=0.

Here xP(x)=1 and x^2Q(x)=-x/(1-x) are analytic at zero. Thus zero is a regular singular point.

1. Indicial equation and recurrence

y=\sum_{n=0}^{\infty}a_nx^{n+r},\qquad a_0\ne0.

Group the differential operator as

L[y]=xy”+y’-(x^2y”+xy’+y).

The first pair contributes \sum_{n\ge0}(n+r)^2a_nx^{n+r-1}; the bracket contributes \sum_{n\ge0}[(n+r)^2+1]a_nx^{n+r}. The lowest-power coefficient gives

r^2a_0=0\Longrightarrow r=0\text{ is a repeated indicial root}.

For n\ge1, comparison of the coefficients of x^{n+r-1} gives

(n+r)^2a_n=[(n+r-1)^2+1]a_{n-1}.

Set r=0 and choose a_0=1. Then

a_n=\frac{(n-1)^2+1}{n^2}a_{n-1},\qquad a_1=1,\quad a_2=\frac12,\quad a_3=\frac5{18},\quad a_4=\frac{25}{144},\quad a_5=\frac{17}{144}. \boxed{A(x)=1+x+\frac{x^2}{2}+\frac{5x^3}{18}+\frac{25x^4}{144}+\frac{17x^5}{144}+\cdots.}

For all n, the coefficients are specified by

a_n=\frac{\displaystyle\prod_{j=0}^{n-1}(j^2+1)}{(n!)^2},

where the empty product is one.

2. Obtain a second, independent solution

The repeated indicial root gives only one analytic solution. Seek

y_2=A(x)\log|x|+B(x),\qquad B(x)=\sum_{n=0}^{\infty}b_nx^n.

For x\ne0,

(A\log|x|)’=A’\log|x|+\frac Ax, (A\log|x|)”=A”\log|x|+\frac{2A’}x-\frac A{x^2}.

Substitution into the original operator, with L[A]=0, cancels the logarithmic terms and gives

L[A\log|x|]=2(1-x)A’,\qquad L[B]=-2(1-x)A’.

Equating coefficients of x^{n-1} yields

n^2b_n-[(n-1)^2+1]b_{n-1} =-2[na_n-(n-1)a_{n-1}],\qquad n\ge1.

Choose b_0=0; another value only adds a multiple of A. Successive calculations give

\begin{aligned} b_1&=-2a_1=-2,\\ 4b_2-2b_1&=-2(2a_2-a_1)=0 &&\Longrightarrow b_2=-1,\\ 9b_3-5b_2&=-2(3a_3-2a_2)=\frac13 &&\Longrightarrow b_3=-\frac{14}{27},\\ 16b_4-10b_3&=-2(4a_4-3a_3)=\frac5{18} &&\Longrightarrow b_4=-\frac{265}{864}. \end{aligned} \boxed{y_2=A(x)\log|x|-2x-x^2-\frac{14x^3}{27}-\frac{265x^4}{864}+\cdots.}

3. State the complete local solution

The logarithmic behaviour ensures that y_2 is independent of A. The Frobenius expansions converge on 0<|x|<1, separately on each side of zero; the next singular point of the equation is one. Thus

\boxed{y=C_1A(x)+C_2y_2(x).}

Only the first solution is analytic at zero, so an analytic solution must have C_2=0. The two coefficient recurrences specify every term of the full solution.

Solution: compute the polar radius of curvature

Assume a\ne0 and use the positive radial coordinate r. For a polar curve,

R=\frac{(r^2+(r’)^2)^{3/2}}{|r^2+2(r’)^2-rr”|},\qquad r’=\frac{dr}{d\theta}.

This follows from x=r\cos\theta, y=r\sin\theta: differentiation gives x’^2+y’^2=r^2+(r’)^2 and x’y”-y’x”=r^2+2(r’)^2-rr”.

1. Differentiate the given curve

2rr’=2a^2\sec(2\theta)\tan(2\theta)=2r^2\tan(2\theta),\qquad r’=r\tan(2\theta). r”=r’\tan(2\theta)+2r\sec^2(2\theta) =r[\tan^2(2\theta)+2\sec^2(2\theta)].

2. Substitute

r^2+(r’)^2=r^2\sec^2(2\theta), \begin{aligned}r^2+2(r’)^2-rr” &=r^2[1+2\tan^2(2\theta)-\tan^2(2\theta)-2\sec^2(2\theta)]\\ &=-r^2\sec^2(2\theta). \end{aligned}

On the real curve, \sec(2\theta)=r^2/a^2>0. Therefore

R=\frac{r^3\sec^3(2\theta)}{r^2\sec^2(2\theta)} =r\sec(2\theta)=\boxed{\frac{r^3}{a^2}}.

For fixed a, the factor 1/a^2 is constant, proving R\propto r^3.

Solution: use x as the parameter

Parametrize by x=t, y=t^2/4, z=t, with 0\le t\le2. This traverses the curve from (0,0,0) to (2,1,2). Then

dx=dt,\qquad dy=\frac t2dt,\qquad dz=dt.

The three terms of the differential form become

(x+y)dx=\left(t+\frac{t^2}{4}\right)dt,\quad -x^2dy=-\frac{t^3}{2}dt,\quad (y+z)dz=\left(\frac{t^2}{4}+t\right)dt. \begin{aligned}I &=\int_0^2\left(2t+\frac{t^2}{2}-\frac{t^3}{2}\right)dt\\ &=\left[t^2+\frac{t^3}{6}-\frac{t^4}{8}\right]_0^2\\ &=4+\frac86-\frac{16}{8}=\boxed{\frac{10}{3}}. \end{aligned}

Reversing the direction of traversal would reverse the sign.

Solution: evaluate the surface and boundary integrals

Orient the upper hemisphere outward. The boundary circle C:x^2+y^2=16,z=0 is then traversed counterclockwise when viewed from above. The source’s z>0 is understood as the upper cap, including its limiting boundary when applying Stokes’s theorem.

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

1. Curl and surface integral

For \mathbf v=(3x-y,-2yz^2,-2y^2z),

\begin{aligned}\nabla\times\mathbf v &=\left(\frac{\partial(-2y^2z)}{\partial y}-\frac{\partial(-2yz^2)}{\partial z}, \frac{\partial(3x-y)}{\partial z}-\frac{\partial(-2y^2z)}{\partial x}, \frac{\partial(-2yz^2)}{\partial x}-\frac{\partial(3x-y)}{\partial y}\right)\\ &=(-4yz+4yz,0,0-(-1))=(0,0,1). \end{aligned}

Use spherical coordinates on the hemisphere:

\mathbf r(\phi,\theta)=(4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi), \quad0\le\phi\le\frac\pi2,\quad0\le\theta\le2\pi.

The vertical component of the outward unit normal is \cos\phi, and the surface element is dS=16\sin\phi\,d\phi\,d\theta. Hence

\begin{aligned}\iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS &=16\int_0^{2\pi}\int_0^{\pi/2}\cos\phi\sin\phi\,d\phi\,d\theta\\ &=16(2\pi)\left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2}=16\pi. \end{aligned}

2. Boundary line integral

x=4\cos t,\qquad y=4\sin t,\qquad z=0,\qquad0\le t\le2\pi.

On this circle the second and third field components vanish. Thus

\begin{aligned}\oint_C\mathbf v\cdot d\mathbf r &=\int_0^{2\pi}(12\cos t-4\sin t)(-4\sin t)dt\\ &=-48\int_0^{2\pi}\sin t\cos t\,dt+16\int_0^{2\pi}\sin^2t\,dt\\ &=0+16\pi=16\pi. \end{aligned} \boxed{\displaystyle\oint_C\mathbf v\cdot d\mathbf r =\iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS=16\pi.}

Both sides agree, verifying Stokes’s theorem.

Solution: project the sphere centre onto the plane

Completing squares gives

(x+1)^2+(y-1)^2+(z-2)^2=19+1+1+4=25.

The sphere has centre C=(-1,1,2) and radius R=5. The plane normal is \mathbf n=(1,2,2), with \|\mathbf n\|^2=9.

1. Circle centre

The centre H of the intersection circle is the foot of the perpendicular from C to the plane. The plane expression at C is -1+2(1)+2(2)+7=12. Therefore

\begin{aligned}H &=C-\frac{\mathbf n\cdot C+7}{\|\mathbf n\|^2}\mathbf n\\ &=(-1,1,2)-\frac{12}{9}(1,2,2)\\ &=\boxed{\left(-\frac73,-\frac53,-\frac23\right)}. \end{aligned}

As a check, the plane expression there is -7/3-10/3-4/3+7=0.

2. Circle radius

CH=\frac{|12|}{\sqrt{1+4+4}}=4.

For any intersection point P, the triangle CHP is right-angled at H, because CH is normal to the plane. Hence

HP^2=CP^2-CH^2=25-16=9. \boxed{\text{Centre }\left(-\frac73,-\frac53,-\frac23\right),\qquad\text{radius }3.}

Solution: show zero resultant and zero moment

Direction qualification: the forces must all point along inward normals, or all along outward normals, with a common proportionality constant. Reversing only one force generally destroys equilibrium. The statement is understood with this consistent direction choice.

Label the vertices A,B,C counterclockwise, with position vectors \mathbf a,\mathbf b,\mathbf c. Let J denote counterclockwise rotation through 90^\circ in the plane: J(x,y)=(-y,x).

1. Resultant force

The three forces can be written, with a common constant k, as

\mathbf F_{AB}=kJ(\mathbf b-\mathbf a),\quad \mathbf F_{BC}=kJ(\mathbf c-\mathbf b),\quad \mathbf F_{CA}=kJ(\mathbf a-\mathbf c).

Each is perpendicular to its side and has magnitude equal to |k| times that side’s length. Positive k gives inward normals; negative k gives outward normals. By linearity of rotation,

\mathbf F_{AB}+\mathbf F_{BC}+\mathbf F_{CA} =kJ[(\mathbf b-\mathbf a)+(\mathbf c-\mathbf b)+(\mathbf a-\mathbf c)]=\mathbf0.

2. Resultant moment

Each force acts along a side’s perpendicular bisector. These lines meet at the circumcentre, so every force has zero moment about that common point.

For an explicit moment calculation, the planar scalar cross product satisfies \mathbf r\times J\mathbf s=\mathbf r\cdot\mathbf s. The midpoint of AB is (\mathbf a+\mathbf b)/2, giving

M_{AB}=\frac{\mathbf a+\mathbf b}{2}\times kJ(\mathbf b-\mathbf a) =\frac k2(\|\mathbf b\|^2-\|\mathbf a\|^2).

The cyclic sum telescopes:

\begin{aligned}M_{AB}+M_{BC}+M_{CA} =\frac k2[&\|\mathbf b\|^2-\|\mathbf a\|^2 +\|\mathbf c\|^2-\|\mathbf b\|^2\\ &+\|\mathbf a\|^2-\|\mathbf c\|^2]=0. \end{aligned} \boxed{\text{Resultant force}=0,\quad\text{resultant moment}=0; \quad\text{equilibrium follows.}}

Solution: compare circular speed with energy gained in falling

Let f(r)>0 be the magnitude of the inward acceleration. Assume it is continuous and the work integral from infinity converges. Falling from infinity means starting from rest there.

1. Circular-orbit speed

The centripetal acceleration is supplied by the attraction:

\frac{v_c^2}{r}=f(r),\qquad v_c^2=rf(r).

2. Speed acquired in falling

The radial equation is \ddot r=-f(r). Multiply by \dot r and integrate from infinity to r:

\frac12v_f^2=-\int_\infty^r f(s)\,ds=\int_r^\infty f(s)\,ds, \qquad v_f^2=2\int_r^\infty f(s)\,ds.

3. Enforce equality at every distance

rf(r)=2\int_r^\infty f(s)\,ds.

Differentiating with respect to r gives

f(r)+rf'(r)=-2f(r),\qquad rf'(r)=-3f(r).

Dividing by rf(r) and integrating,

\frac{f'(r)}{f(r)}=-\frac3r,\qquad \log f(r)=-3\log r+\log K. \boxed{f(r)=\frac K{r^3},\qquad K>0.}

4. Verify sufficiency

v_c^2=r\frac K{r^3}=\frac K{r^2}, v_f^2=2K\int_r^\infty s^{-3}\,ds =2K\left[-\frac1{2s^2}\right]_r^\infty=\frac K{r^2}.

The two speeds are indeed equal to \sqrt K/r. Thus the inverse-cube law is both necessary and sufficient. For a mass m, the corresponding force magnitude is mK/r^3.

Solution: combine the two resultant conditions

Let \mathbf R=\mathbf P+\mathbf Q, where the nonzero forces have magnitudes P,Q and angle \theta\in[0,\pi]. Suppose \mathbf R\perp\mathbf P; the phrase “half the other” means R=Q/2.

Perpendicularity gives

0=\mathbf R\cdot\mathbf P=P^2+PQ\cos\theta, \qquad\cos\theta=-\frac P Q.

Using the squared magnitude of the resultant,

R^2=P^2+Q^2+2PQ\cos\theta=P^2+Q^2-2P^2=Q^2-P^2.

Since R=Q/2,

\frac{Q^2}{4}=Q^2-P^2\Longrightarrow P^2=\frac{3Q^2}{4}\Longrightarrow\frac P Q=\frac{\sqrt3}{2}. \cos\theta=-\frac{\sqrt3}{2}\Longrightarrow\boxed{\theta=150^\circ=\frac{5\pi}{6}}.

The obtuse angle ensures that the component of \mathbf Q along \mathbf P opposes and cancels it.

Solution: obtain the linear restoring force

In equilibrium the fixed ends lie at distances a and b on either side of the particle along the straight, taut wire. Give the particle a small transverse displacement y. Neglect the wire’s mass, and use the equilibrium tension T in both segments to first order.

1. Resolve the tensions

If the displaced segments make angles \alpha,\beta with the original straight line,

\sin\alpha=\frac{y}{\sqrt{a^2+y^2}},\qquad \sin\beta=\frac{y}{\sqrt{b^2+y^2}}.

Both transverse components act towards the equilibrium line, giving

F_y=-T\left(\frac{y}{\sqrt{a^2+y^2}}+\frac{y}{\sqrt{b^2+y^2}}\right).

For |y|\ll a,b,

\frac{y}{\sqrt{a^2+y^2}}=\frac ya\left(1+\frac{y^2}{a^2}\right)^{-1/2} =\frac ya+O(y^3),

and similarly for the other segment. Retaining the linear terms gives

F_y=-T\left(\frac1a+\frac1b\right)y=-\frac{T(a+b)}{ab}y.

The axial tension components cancel to first order, because both cosines differ from one only by second-order terms.

2. Identify the oscillation frequency

Newton’s law gives

m\ddot y=-\frac{T(a+b)}{ab}y,\qquad \ddot y+\omega^2y=0,\qquad\omega^2=\frac{T(a+b)}{mab}.

This is simple harmonic motion, with y=A\cos\omega t+B\sin\omega t. Therefore its period is

\boxed{\tau=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{mab}{T(a+b)}}.}

This is the required period under the small-displacement, taut-wire approximation.