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For a short length $\delta x$, the kinetic energy is $\frac{1}{2}\mu y_t^2\delta x$. For small slope, its increase in length is $[\sqrt{1+y_x^2}-1]\delta x\simeq \frac{1}{2}{y}_x^2\delta x$, so its elastic potential energy is $\frac{1}{2}T{y}_x^2\delta x$. For a travelling wave $y=f(x-vt)$, $y_t=-v{f}^{\prime }$ and $y_x=f^{\prime }$. Using $v^2=T/\mu$ givesWriting a pulse as $f(x)=Ag(x/w)$ yields $E=(T{A}^2/w)\int [g^{\prime }(s){]}^{2}ds$. Thus energy is proportional to amplitude squared for a fixed shape and width. Width must also be retained when comparing pulses on different strings.

For an exact differential $Mdx+Ndy$, equality of the mixed partial derivatives requires $M_y=N_x$. Here these are $\sin x+b\cos y$ and $\sin x+3\cos y$, so $\boxed{b=3}$. Integrating $M$ with respect to $x$ gives $f=-y\cos x+3x\sin y+g(y)$. Comparison with $N$ gives $g^{\prime }=y^2$, and the specified value fixes the integration constant. ThereforeDifferentiating recovers both coefficients, and $f(0,2)=-2+8/3-2/3=0$. The explicit potential function also proves exactness globally, without relying only on a local compatibility test.

The field is the gradient of $\Phi =x^2-3y^2+4z^2$ on all of ${\mathbb{R}}^3$. It is therefore a conservative vector field, and the fundamental theorem for line integrals gives ${\int }_{C}F\cdot dr=\Phi (r_{\mathrm{end}})-\Phi (r_{\mathrm{start}})$. The endpoints are $(3,0,0)$ and $(0,2,-1)$, soThe complicated intermediate path does not affect the work. Directly, $\nabla \times F=0$, consistent with the globally defined potential function.

The gradient equations are $f_x=2x(1-y)=0$ and $f_y=4y-x^2=0$. The critical points are $(0,0)$ and $(\pm 2,1)$. Their Hessian matrix isAt the origin $H=\mathrm{diag}(2,4)$ is positive definite. At either other point its determinant is $-16$, so its eigenvalues have opposite signs. ThusThe minimum value is zero and both saddle values are two. The origin is not a global minimum: at any fixed $y>1$, the term $(1-y)x^2$ makes $f$ arbitrarily negative as $|x|$ increases.

For the contour plot, the local contours around the origin are approximately $x^2+2y^2=c$. The exact saddle contour factors asso it is the union of $y=1$ and $y=x^2/2-1$, crossing at the two saddles. Other levels obey $x^2=(c-2y^2)/(1-y)$ where $y\ne 1$; the right panel of the original figure in part 3(a) shows how these branches join around the minimum and pass through the saddle contour.

A vector line integral along an oriented parametrized curve $r(t)$ is $\int a(r(t))\cdot r^{\prime }(t)dt$; for a force field it is the work along that path. On the upper branch, set $y=t$, $x=t^2$, $0\le t\le 1$. ThensoThe stated $x$ endpoints do not distinguish the two branches of $y^2=x$. If the lower branch is intended, the same parametrization runs from $t=0$ to $t=-1$, giving instead $\boxed{4/3}$. The two values correspond to different terminal points, $(1,1)$ and $(1,-1)$; choosing a branch is necessary to make the question unambiguous.