Solution: orthogonality and sign changes

We prove the stronger result: all n roots are distinct and lie strictly inside (-1,1).

Step 1: Establish orthogonality to every polynomial of lower degree

Rodrigues’ formula is

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

Let q(x) be any polynomial with degree less than n. Integrating by parts n times gives

\begin{aligned}\int_{-1}^{1}P_n(x)q(x)\,dx &=\frac{1}{2^n n!}\int_{-1}^{1}q(x)\frac{d^n}{dx^n}(x^2-1)^n\,dx\\ &=\frac{(-1)^n}{2^n n!}\int_{-1}^{1}q^{(n)}(x)(x^2-1)^n\,dx=0. \end{aligned}

All boundary terms vanish: (x^2-1)^n=(x-1)^n(x+1)^n and its derivatives up to order n-1 vanish at both endpoints. Also q^{(n)}=0.

Step 2: Count the sign-changing zeros

Let x_1,\ldots,x_k be all the distinct zeros in (-1,1) at which P_n changes sign. Suppose k<n, and define

q(x)=\prod_{j=1}^{k}(x-x_j),

with q=1 if there are no such zeros. At each sign-changing zero both factors change sign, so P_nq does not change sign there. At any other real zero, P_n does not change sign. Thus P_nq has one fixed sign throughout the interval, except where it vanishes.

This product is not the zero polynomial. It is continuous and nonzero on some subinterval, so

\int_{-1}^{1}P_n(x)q(x)\,dx\ne0.

But \deg q=k<n, contradicting Step 1. Hence k\ge n. Since P_n has degree n, it cannot have more than n distinct zeros.

\boxed{P_n\text{ has exactly }n\text{ distinct real roots, all in }(-1,1).}

The degree is exhausted by these roots, so each is simple and no further roots lie at the endpoints, outside the interval, or off the real axis.

Solution: show that the complement is open

Take any x_0\notin A. Then f(x_0)\ne0. Choose

\varepsilon=\frac{|f(x_0)|}{2}>0.

Continuity at x_0 gives a \delta>0 such that

|x-x_0|<\delta\quad\Longrightarrow\quad |f(x)-f(x_0)|<\frac{|f(x_0)|}{2}.

By the reverse triangle inequality,

|f(x)|\ge |f(x_0)|-|f(x)-f(x_0)|>\frac{|f(x_0)|}{2}>0.

Therefore the entire interval (x_0-\delta,x_0+\delta) lies outside A. Every point of the complement has such a neighbourhood; hence the complement is open.

\boxed{A\text{ is closed in }\mathbb R.}

Equivalently, A=f^{-1}(\{0\}), and a continuous map has closed inverse images of closed sets.

Solution: calculate the curl and a potential

The field has continuously differentiable components throughout the simply connected domain \mathbb R^3. Consequently, zero curl is both necessary and sufficient for it to be conservative.

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

Setting each component equal to zero gives

b-a=0,\qquad 1-b=0\quad\Longrightarrow\quad\boxed{a=b=1.}

To check sufficiency directly, find \phi with \nabla\phi=\vec F. Integrating \phi_x=x+z,

\phi=\frac{x^2}{2}+xz+g(y,z).

Now \phi_y=y+z implies g=y^2/2+yz+h(z). Finally,

\phi_z=x+y+h'(z)=x+y\quad\Longrightarrow\quad h'(z)=0. \boxed{\phi(x,y,z)=\frac{x^2+y^2}{2}+xz+yz+C.}

Solution: a repeated eigenvalue with independent eigenvectors

Consider

A=\begin{pmatrix}2&0\\0&2\end{pmatrix}=2I_2.

Its characteristic polynomial is

\det(A-\lambda I)=\begin{vmatrix}2-\lambda&0\\0&2-\lambda\end{vmatrix}=(2-\lambda)^2.

The only eigenvalue is 2, with algebraic multiplicity two, so the eigenvalues are not distinct. Nevertheless,

A\begin{pmatrix}1\\0\end{pmatrix}=2\begin{pmatrix}1\\0\end{pmatrix},\qquad A\begin{pmatrix}0\\1\end{pmatrix}=2\begin{pmatrix}0\\1\end{pmatrix}.

These eigenvectors form a basis. Taking P=I_2 gives P^{-1}AP=A, which is diagonal.

\boxed{A=2I_2\text{ is diagonalizable despite its repeated eigenvalue.}}

Solution: maximum speed in SHM

Write displacement as x(t)=A\sin(\omega t+\phi). Differentiating,

v(t)=A\omega\cos(\omega t+\phi),\qquad |v|_{\max}=A\omega.

The period is T=10\,\mathrm{s}, so

\omega=\frac{2\pi}{T}=\frac{2\pi}{10}=\frac{\pi}{5}\,\mathrm{s}^{-1}.

With A=5\,\mathrm{cm},

v_{\max}=5\times\frac{\pi}{5}=\boxed{\pi\,\mathrm{cm\,s^{-1}}\approx3.1416\,\mathrm{cm\,s^{-1}}.}

In SI units this is 0.031416\,\mathrm{m\,s^{-1}}. The maximum speed occurs at the mean position.

Solution: apply Cauchy–Schwarz

Let \mathbf u,\mathbf v be two nonzero vectors in a real inner-product space. Their angle \theta is defined through

\cos\theta=\frac{\langle\mathbf u,\mathbf v\rangle}{\|\mathbf u\|\,\|\mathbf v\|}.

Cauchy–Schwarz states

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

Both norms are positive. Taking square roots and dividing by their product gives

|\cos\theta|=\frac{|\langle\mathbf u,\mathbf v\rangle|}{\|\mathbf u\|\|\mathbf v\|}\le1. \boxed{-1\le\cos\theta\le1.}

Equality holds precisely when the two nonzero vectors are linearly dependent: their directions are the same or opposite. The angle with a zero vector is not defined.

Solution: construct a basis of the annihilator

Let \dim V=n and \dim W=r. The annihilator is the subspace of the dual space defined by

A(W)=\{\varphi\in V^*: \varphi(w)=0\text{ for every }w\in W\}.

Choose a basis w_1,\ldots,w_r of W, and extend it to a basis of V:

v_1=w_1,\ldots,v_r=w_r,v_{r+1},\ldots,v_n.

Let \varphi_1,\ldots,\varphi_n be the corresponding dual basis, so that \varphi_i(v_j)=\delta_{ij}. Every functional is uniquely expressible as

\varphi=c_1\varphi_1+\cdots+c_n\varphi_n.

For j\le r, evaluation at w_j=v_j gives \varphi(w_j)=c_j. Thus \varphi vanishes on all of W exactly when

c_1=\cdots=c_r=0.

It follows that

A(W)=\operatorname{span}\{\varphi_{r+1},\ldots,\varphi_n\}.

These functionals are linearly independent because they belong to a dual basis. They therefore form a basis of A(W), containing n-r elements.

\boxed{\dim A(W)=n-r=\dim V-\dim W.}

Solution: substitute the plane into the cone

From the plane,

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

Substituting this in the cone gives

\begin{aligned}4x^2-y^2+3z^2 &=4x^2-(z-2x)^2+3z^2\\ &=4x^2-z^2+4xz-4x^2+3z^2\\ &=2z^2+4xz=2z(z+2x)=0. \end{aligned}

Hence either z=0 or z=-2x.

First generator

With z=0, the plane gives y=-2x. Thus

\boxed{L_1:\ (x,y,z)=(t,-2t,0),\quad t\in\mathbb R.}

Second generator

With z=-2x, the plane gives y=-4x. Thus

\boxed{L_2:\ (x,y,z)=(t,-4t,-2t),\quad t\in\mathbb R.}

Equivalently, the lines are the intersections z=0,\ 2x+y=0 and z+2x=0,\ y+4x=0. Substitution verifies the cone: respectively 4t^2-4t^2=0 and 4t^2-16t^2+12t^2=0.

Solution: implicit differentiation at horizontal tangents

Take a>0, the usual convention for this curve parameter. Put

F(x,y)=(x^2+y^2)^2-a^2(x^2-y^2)=0,\qquad S=x^2+y^2.

At a regular point of the curve,

F_x=2x(2S-a^2),\quad F_y=2y(2S+a^2),\quad y’=-\frac{x(2S-a^2)}{y(2S+a^2)}.

Step 1: Locate the horizontal tangents

For a horizontal tangent we need F_x=0 and F_y\ne0. If x=0, the curve equation becomes y^2(y^2+a^2)=0, giving only the origin. The origin is singular, with tangent lines y=\pm x, so it is not a horizontal-tangent point.

Therefore 2S-a^2=0, and

x^2+y^2=\frac{a^2}{2},\qquad x^2-y^2=\frac{(a^2/2)^2}{a^2}=\frac{a^2}{4}.

Adding and subtracting yield

x^2=\frac{3a^2}{8},\qquad y^2=\frac{a^2}{8}.

Step 2: Find the second derivative

Differentiating F_x+F_yy’=0 along the curve gives

F_{xx}+2F_{xy}y’+F_{yy}(y’)^2+F_y y”=0.

At a horizontal tangent y’=0, hence y”=-F_{xx}/F_y. Now

F_{xx}=4S+8x^2-2a^2=2a^2+3a^2-2a^2=3a^2, F_y=2y(2S+a^2)=4a^2y,\qquad y”=-\frac{3}{4y}.

Step 3: Calculate the radius of curvature

\rho=\frac{[1+(y’)^2]^{3/2}}{|y”|}=\frac{4|y|}{3} =\frac{4}{3}\frac{a}{2\sqrt2}=\boxed{\frac{\sqrt2a}{3}.}

All four regular horizontal-tangent points give this value. If the parameter is allowed to be negative, the radius is \sqrt2|a|/3.

Solution: an exponential dominates a power

Since the real logarithm requires x>0, the stated limit is understood as x\to0^+. Set

t=-\log x,\qquad x=e^{-t},\qquad t\to+\infty.

Then

x^m(\log x)^n=e^{-mt}(-t)^n=(-1)^n\frac{t^n}{e^{mt}}.

Apply l’Hôpital’s rule n times to the positive quotient. At each preceding stage both numerator and denominator tend to infinity:

\begin{aligned}\lim_{t\to\infty}\frac{t^n}{e^{mt}} &=\lim_{t\to\infty}\frac{n t^{n-1}}{m e^{mt}} =\cdots=\lim_{t\to\infty}\frac{n!}{m^n e^{mt}}=0. \end{aligned}

The factor (-1)^n changes only the sign, not the limit.

\boxed{\lim_{x\to0^+}x^m(\log_e x)^n=0.}

Solution: rank–nullity by extending a kernel basis

For the finite-dimensional form of the theorem, let \dim V=n and \dim\ker T=k. Choose a basis v_1,\ldots,v_k of the kernel and extend it to a basis v_1,\ldots,v_n of V.

Step 1: Span the image

Every v\in V can be written as \sum_{i=1}^n a_i v_i. By linearity, and because T(v_i)=0 for i\le k,

T(v)=\sum_{i=k+1}^{n}a_iT(v_i).

Thus T(v_{k+1}),\ldots,T(v_n) span the image.

Step 2: Prove independence

Suppose \sum_{i=k+1}^n c_i T(v_i)=0. Then T(\sum_{i=k+1}^n c_i v_i)=0, so that sum belongs to the kernel. Consequently, for some coefficients b_j,

\sum_{i=k+1}^{n}c_i v_i=\sum_{j=1}^{k}b_jv_j.

Rearranging gives a linear relation among the basis vectors of V. Independence forces every c_i and b_j to vanish.

The displayed spanning set is therefore a basis of the image and has n-k vectors.

\boxed{\operatorname{Rank}T+\operatorname{Nullity}T=(n-k)+k=n=\dim V.}

Solution: Euler’s operator applied twice

Work on an open region where the given real expression is defined and twice differentiable, with \cos u\ne0. Let

v=\sin u=\left(\frac{x^{1/3}+y^{1/3}}{x^{1/2}+y^{1/2}}\right)^{1/2}.

Step 1: Determine the degree of homogeneity

For positive scale t,

v(tx,ty)=\left(t^{1/3-1/2}\right)^{1/2}v(x,y) =t^{-1/12}v(x,y).

Define E=x\partial_x+y\partial_y. Euler’s theorem gives Ev=-v/12. The chain rule then yields

\cos u\,Eu=-\frac{1}{12}\sin u\quad\Longrightarrow\quad Eu=xu_x+yu_y=-\frac{1}{12}\tan u.

Step 2: Apply the operator again

E^2u=E\left(-\frac{1}{12}\tan u\right) =-\frac{1}{12}\sec^2u\,Eu =\frac{1}{144}\tan u\sec^2u.

Expanding the operator explicitly avoids losing the first-derivative terms:

\begin{aligned}E^2u &=x\partial_x(xu_x+yu_y)+y\partial_y(xu_x+yu_y)\\ &=xu_x+yu_y+x^2u_{xx}+2xyu_{xy}+y^2u_{yy}\\ &=Eu+x^2u_{xx}+2xyu_{xy}+y^2u_{yy}. \end{aligned}

Step 3: Subtract and simplify

\begin{aligned}x^2u_{xx}+2xyu_{xy}+y^2u_{yy} &=E^2u-Eu\\ &=\frac{1}{144}\tan u\sec^2u+\frac{1}{12}\tan u\\ &=\frac{1}{144}\tan u(\sec^2u+12)\\ &=\boxed{\frac{1}{144}\tan u(13+\tan^2u)}. \end{aligned}

Solution: determine the feet of the common perpendicular

Write the two lines in parametric form:

L_1:\ \mathbf r=s(2,-3,1),\qquad L_2:\ \mathbf r=(2,1,-2)+t(3,-5,2).

The first direction is (2,-3,1), because the middle fraction is -y/3. Set \mathbf u=(2,-3,1), \mathbf v=(3,-5,2), and \mathbf a=(2,1,-2).

Step 1: Direction and magnitude

\mathbf u\times\mathbf v= \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&-3&1\\3&-5&2\end{vmatrix} =(-1,-1,-1). d=\frac{|\mathbf a\cdot(\mathbf u\times\mathbf v)|}{|\mathbf u\times\mathbf v|} =\frac{|-2-1+2|}{\sqrt3}=\boxed{\frac1{\sqrt3}}.

Step 2: Find the endpoints of the shortest segment

Let A=s\mathbf u and B=\mathbf a+t\mathbf v. The vector B-A must be perpendicular to both lines. Using

\mathbf u^2=14,\quad\mathbf v^2=38,\quad\mathbf u\cdot\mathbf v=23, \quad\mathbf a\cdot\mathbf u=-1,\quad\mathbf a\cdot\mathbf v=-3,

we obtain

-1+23t-14s=0,\qquad -3+38t-23s=0.

Multiply the first equation by 23 and the second by 14:

-23+529t-322s=0,\qquad -42+532t-322s=0.

Subtracting gives 3t=19, hence

t=\frac{19}{3},\qquad s=\frac{23t-1}{14}=\frac{31}{3}. A=\left(\frac{62}{3},-31,\frac{31}{3}\right),\qquad B=\left(21,-\frac{92}{3},\frac{32}{3}\right).

Indeed B-A=(1/3,1/3,1/3), perpendicular to both directions, and its length is \sqrt{3}/3.

Step 3: Equation of the common perpendicular

\boxed{x-\frac{62}{3}=y+31=z-\frac{31}{3}.}

Equivalently, \mathbf r=A+\lambda(1,1,1). The shortest segment corresponds to 0\le\lambda\le1/3.

Solution: separate variables and check the exceptional branches

For real solutions the equation requires y\le3. The value y=2 is impossible for a finite derivative, since it would give 0=4.

Step 1: Nonconstant solutions

On an interval with y<3 and y\ne2, write the two square-root branches as

3(2-y)\frac{dy}{dx}=\pm2\sqrt{3-y},\qquad dx=\pm\frac{3(2-y)}{2\sqrt{3-y}}\,dy.

Put t=\sqrt{3-y}. Then y=3-t^2, dy=-2t\,dt, and 2-y=t^2-1. Thus

\int\frac{3(2-y)}{2\sqrt{3-y}}\,dy =\int -3(t^2-1)\,dt=3t-t^3 =y\sqrt{3-y}.

Therefore the general one-parameter family is

\boxed{x-C=\pm y\sqrt{3-y},\quad\text{or equivalently}\quad (x-C)^2=y^2(3-y).}

Each differentiable local branch is used on its admissible interval; it cannot pass through y=2 with a finite slope. Differentiating the unsquared form checks the equation:

\frac{dx}{dy}=\pm\frac{3(2-y)}{2\sqrt{3-y}} \quad\Longrightarrow\quad 9(2-y)^2\left(\frac{dy}{dx}\right)^2=4(3-y).

Step 2: Singular solution

A constant solution has p=0; substitution gives 4(3-y)=0. Hence

\boxed{y=3.}

This horizontal line satisfies the original equation but is not an entire member of the general family for any fixed C.

To see the envelope candidates, let \Phi=(x-C)^2-y^2(3-y). Then

\Phi_C=-2(x-C)=0\quad\Longrightarrow\quad x=C, \qquad y^2(3-y)=0.

The candidate y=3 is valid. The other candidate y=0 is extraneous as a constant solution: it gives 0=12 in the original equation. Thus it must be rejected. Nonconstant general-solution branches may still cross y=0.

Solution: reduce to Bessel’s equation of order one-half

For x\ne0, the differential equation is

y”+\frac2x y’+\frac12y=0.

First take x>0 and put y=x^{-1/2}u(x). Direct differentiation gives

y’=-\frac12x^{-3/2}u+x^{-1/2}u’, y”=\frac34x^{-5/2}u-x^{-3/2}u’+x^{-1/2}u”.

Substitution and multiplication by x^{5/2} yield

x^2u”+xu’+\left(\frac{x^2}{2}-\frac14\right)u=0.

Now set z=x/\sqrt2 and u(x)=U(z). Since u’=U_z/\sqrt2 and u”=U_{zz}/2,

z^2U_{zz}+zU_z+\left(z^2-\left(\frac12\right)^2\right)U=0.

This is Bessel’s equation with order \nu=1/2. Therefore

\boxed{y=x^{-1/2}\left[C_1J_{1/2}\left(\frac{x}{\sqrt2}\right)+C_2Y_{1/2}\left(\frac{x}{\sqrt2}\right)\right],\quad x>0.}

Using the half-order formulas

J_{1/2}(z)=\sqrt{\frac2{\pi z}}\sin z,\qquad Y_{1/2}(z)=-\sqrt{\frac2{\pi z}}\cos z,

and absorbing numerical factors into the constants gives the equivalent real form

\boxed{y=\frac{A\sin(x/\sqrt2)+B\cos(x/\sqrt2)}{x},\quad x\ne0.}

This real form also applies on negative intervals. For a direct check, set v=xy; then the original left side becomes v”+v/2, which vanishes for the displayed sine and cosine combination. If a finite smooth extension through x=0 is required, set B=0 and use the limit of the sine term; it has y'(0)=0, as the original equation requires.

Solution: variation of parameters

The equation is y”-2y’+2y=e^x\tan x. Work on an interval avoiding the zeros of \cos x.

Step 1: Complementary solution

r^2-2r+2=0\quad\Longrightarrow\quad r=1\pm i. y_c=e^x(C_1\cos x+C_2\sin x),\quad y_1=e^x\cos x,\quad y_2=e^x\sin x.

Their derivatives are y_1’=e^x(\cos x-\sin x) and y_2’=e^x(\sin x+\cos x), so

W=y_1y_2′-y_2y_1’=e^{2x}(\cos^2x+\sin^2x)=e^{2x}.

Step 2: Vary the coefficients

For y_p=u_1y_1+u_2y_2, impose u_1’y_1+u_2’y_2=0. The other equation is u_1’y_1’+u_2’y_2’=e^x\tan x. Solving this two-by-two system gives

u_1’=-\frac{y_2e^x\tan x}{W}=-\sin x\tan x =-\frac{\sin^2x}{\cos x}=\cos x-\sec x, u_2’=\frac{y_1e^x\tan x}{W}=\cos x\tan x=\sin x.

Integrating, omitting constants already represented by y_c,

u_1=\sin x-\log|\sec x+\tan x|,\qquad u_2=-\cos x.

Step 3: Simplify the particular integral

\begin{aligned}y_p &=e^x\cos x\,[\sin x-\log|\sec x+\tan x|]-e^x\sin x\cos x\\ &=-e^x\cos x\log|\sec x+\tan x|. \end{aligned} \boxed{y=e^x\left[C_1\cos x+C_2\sin x-\cos x\log|\sec x+\tan x|\right].}

Check: writing y=e^xv reduces the equation to v”+v=\tan x. For v_p=-\cos x\,L, where L=\log|\sec x+\tan x| and L’=\sec x, one finds v_p’=\sin x\,L-1 and v_p”=\cos x\,L+\tan x. Thus v_p”+v_p=\tan x.

Solution: Stokes’ theorem with the clockwise orientation

Orientation convention: interpret clockwise as seen from the side of the positive normal (1,4,1) looking towards the plane (equivalently, clockwise in its projection viewed from positive z). The required oriented normal is then the negative one. Clockwise without a viewing side is otherwise ambiguous.

Step 1: Curl and surface element

\nabla\times\vec F= \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\\partial_x&\partial_y&\partial_z\\y&-z^3&x^2\end{vmatrix} =(3z^2,-2x,-1).

The triangular surface has vertices A=(4,0,0), B=(0,1,0), and C=(0,0,4). Parameterize it as

\mathbf r(x,y)=(x,y,4-x-4y),\qquad 0\le x\le4,\quad0\le y\le1-\frac{x}{4}. \mathbf r_x=(1,0,-1),\quad\mathbf r_y=(0,1,-4),\quad \mathbf r_x\times\mathbf r_y=(1,4,1).

For the positive normal the integrand is 3z^2-8x-1. Clockwise reverses that orientation, so Stokes’ theorem gives

\oint_C\vec F\cdot d\vec r =\int_0^4\int_0^{(4-x)/4}\left[-3(4-x-4y)^2+8x+1\right]dy\,dx.

Step 2: Carry out both integrations

Put A_x=4-x temporarily. The inner integral is

\begin{aligned}\int_0^{A_x/4}[-3(A_x-4y)^2+8x+1]dy &=\left[\frac{(A_x-4y)^3}{4}+(8x+1)y\right]_0^{A_x/4}\\ &=-\frac{(4-x)^3}{4}+\frac{(8x+1)(4-x)}4. \end{aligned} \begin{aligned}\oint_C\vec F\cdot d\vec r &=\frac14\int_0^4\left[-(4-x)^3-8x^2+31x+4\right]dx\\ &=\frac14\left[-64-\frac{512}{3}+248+16\right]\\ &=\boxed{\frac{22}{3}}. \end{aligned}

Independent orientation check

The positively oriented circuit is A\to B\to C\to A. Its three line integrals are

\int_{AB}\vec F\cdot d\vec r=\int_0^1-4t\,dt=-2, \int_{BC}\vec F\cdot d\vec r=\int_0^1 64t^3\,dt=16, \int_{CA}\vec F\cdot d\vec r=\int_0^1-64t^2\,dt=-\frac{64}{3}.

Their sum is -22/3; reversing the circuit to clockwise gives 22/3. If clockwise is intended as viewed from the opposite side, the answer has the opposite sign.

Solution: normalize the direction vector

The gradient is

\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)=(2x,6y,4z).

The specified direction vector is \mathbf a=(2,-1,-2). Its magnitude and unit vector are

|\mathbf a|=\sqrt{4+1+4}=3,\qquad \widehat{\mathbf a}=\frac13(2,-1,-2).

The directional derivative at a general point is

D_{\widehat{\mathbf a}}f=\nabla f\cdot\widehat{\mathbf a} =\frac{4x-6y-8z}{3}.

At (1,-3,2), \nabla f=(2,-18,8). Hence

D_{\widehat{\mathbf a}}f(1,-3,2) =\frac{(2)(2)+(-18)(-1)+(8)(-2)}{3} =\frac{4+18-16}{3}=\boxed{2.}

Using the unnormalized vector would give 6, which is not the derivative per unit distance in the specified direction.

Solution: solve the equation and exhibit a basis

Let L=D^3-6D^2+11D-6. Since L is linear, if Ly_1=Ly_2=0, then L(ay_1+by_2)=0 for real constants a,b. The zero function is also a solution, so the solution set is a real vector space.

Step 1: Solve the characteristic equation

r^3-6r^2+11r-6=(r-1)(r-2)(r-3)=0.

The three distinct roots give the general solution

y=c_1e^x+c_2e^{2x}+c_3e^{3x}.

Thus the three functions span the solution space.

Step 2: Verify linear independence

Their Wronskian is

\begin{aligned}W(x)&=\begin{vmatrix}e^x&e^{2x}&e^{3x}\\e^x&2e^{2x}&3e^{3x}\\e^x&4e^{2x}&9e^{3x}\end{vmatrix}\\ &=e^{6x}\begin{vmatrix}1&1&1\\1&2&3\\1&4&9\end{vmatrix} =e^{6x}[(18-12)-(9-3)+(4-2)]\\ &=2e^{6x}\ne0. \end{aligned}

Therefore the three solutions are linearly independent on every interval. They form a basis:

\boxed{\mathcal B=\{e^x,e^{2x},e^{3x}\},\qquad\dim_{\mathbb R}V=3.}

Solution: resolve the forces along two axes

Choose the positive horizontal axis along the force of magnitude Q. The forces of magnitudes P and R make angles 120^\circ with it.

They must lie on opposite sides of this axis: if both were on the same side, their transverse components would have the same nonzero sign and could not balance. Thus take their angles as 120^\circ and 240^\circ.

Horizontal equilibrium

Q+P\cos120^\circ+R\cos240^\circ=0 \quad\Longrightarrow\quad Q-\frac P2-\frac R2=0. 2Q=P+R.

Vertical equilibrium

P\sin120^\circ+R\sin240^\circ=0 \quad\Longrightarrow\quad\frac{\sqrt3}{2}P-\frac{\sqrt3}{2}R=0.

Hence P=R. Substitution into the horizontal equation gives 2Q=2P.

\boxed{P=Q=R.}

As a check, the three directions are separated by 120^\circ; three equal vectors in those directions sum to zero.

Solution: total energy of SHM

For displacement x, SHM gives v^2=\omega^2(A^2-x^2). Taking potential energy to be zero at the equilibrium position,

K=\frac12mv^2=\frac12m\omega^2(A^2-x^2),\qquad U=\frac12m\omega^2x^2.

The sum is independent of displacement:

E=K+U=\frac12m\omega^2A^2.

Convert to SI units and substitute

m=2.5\,\mathrm g=2.5\times10^{-3}\,\mathrm{kg},\quad A=2\,\mathrm{cm}=2\times10^{-2}\,\mathrm m, f=10\,\mathrm{Hz},\qquad\omega=2\pi f=20\pi\,\mathrm{s^{-1}}. \begin{aligned}E &=\frac12(2.5\times10^{-3})(20\pi)^2(2\times10^{-2})^2\\ &=\frac12(2.5\times10^{-3})(400\pi^2)(4\times10^{-4})\\ &=2\times10^{-4}\pi^2\,\mathrm J\\ &=\boxed{1.9739\times10^{-3}\,\mathrm J\ \text{(approximately)}}. \end{aligned}

In CGS units, using 1\,\mathrm J=10^7\,\mathrm{erg},

\boxed{E=2000\pi^2\,\mathrm{erg}\approx1.9739\times10^4\,\mathrm{erg}.}

Solution: derive the central-force law from the orbit

Let the particle have mass m, and let h=r^2\dot\theta denote its constant angular momentum per unit mass. We take a nonradial orbit, so h\ne0, and work away from r=0. Write u=1/r.

Step 1: Obtain the radial acceleration formula

Because \dot\theta=hu^2, differentiation with respect to time gives

\dot r=\frac{d(u^{-1})}{d\theta}\dot\theta =-u^{-2}u’\,hu^2=-hu’, \ddot r=-hu”\dot\theta=-h^2u^2u”,\qquad r\dot\theta^2=h^2u^3.

Here primes mean differentiation with respect to \theta. Hence the outward-positive radial acceleration is

a_r=\ddot r-r\dot\theta^2=-h^2u^2(u”+u).

Step 2: Differentiate the given orbit

r=a\sin\theta\quad\Longrightarrow\quad u=\frac1a\csc\theta, u’=-\frac1a\csc\theta\cot\theta, u”=\frac1a\left(\csc\theta\cot^2\theta+\csc^3\theta\right).

Using 1+\cot^2\theta=\csc^2\theta,

u”+u=\frac1a\left[\csc\theta(\cot^2\theta+1)+\csc^3\theta\right] =\frac2a\csc^3\theta=2a^2u^3.

Step 3: Force and its direction

F_r=ma_r=-mh^2u^2(2a^2u^3) =\boxed{-\frac{2mh^2a^2}{r^5}.} \boxed{\vec F=-\frac{2mh^2a^2}{r^5}\,\widehat{\mathbf r}.}

The negative sign means attraction towards the force centre, and the magnitude varies as the inverse fifth power of distance. Per unit mass the force is -2h^2a^2/r^5. If L=mh is the angular momentum, the same radial force is -2L^2a^2/(mr^5). The orbit alone does not determine the numerical coefficient without specifying the angular momentum.