Let $f(x)=\cosh \sqrt{x}$ near $x=2$. The chain rule givesSubstitute $x=2$ into the first three terms of the Taylor expansion:The quadratic coefficient is $f^{\prime \prime }(2)/2$, not $f^{\prime \prime }(2)$. The apparent square root creates no local difficulty at the positive expansion point.

For $n\ge 3$, Taylor's theorem after $n$ terms, counting the constant term, isUnder the usual hypothesis that $f$ is $n-1$ times continuously differentiable on the interval joining $a$ and $x$, and $n$ times differentiable in its interior, the Taylor theorem with Lagrange remainder givesThis is a finite expansion with a controlled remainder; it does not assume an infinite Taylor series converges. For $x=a$ the remainder is zero.

The arc length follows from the speed of the parametrization:Here $\sin (t/2)>0$ inside the given interval. ThereforeThe result scales as a length, while the preceding area scales as $a^2$, as required by dimensional analysis.

Because $dx/dt>0$ along the arch, its area above the baseline is the definite integral $\int ydx$. Changing to the parameter givesAdding or removing the two endpoint points does not change the area. This is the area formula in area and length of a cycloid arch.

The original PDF specifies slope $\sqrt{3}$; the transcription's $\sqrt{-1}$ is erroneous. By part (a), $\cot (t/2)=\sqrt{3}$. Since $t/2\in (0,\pi )$ and cotangent is strictly decreasing there, the unique parameter value is $t=\pi /3$. ThereforeThe tangent line drawn in the preceding figure has the required positive slope.

The cycloid has limiting endpoints $(0,0)$ and $(2\pi a,0)$, with its highest point $(\pi a,2a)$ at $t=\pi$. The transformation $t\mapsto 2\pi -t$ sends $x$ to $2\pi a-x$ and preserves $y$, so the arch is symmetric about $x=\pi a$. Its slope decreases from $+\infty$ through zero to $-\infty$, giving vertical limiting tangent lines at the cusps and a horizontal tangent line at the top. More explicitly,so the arch is a concave function throughout its interior. The endpoint cusps are limiting points, excluded by the open parameter interval.

Differentiate the cycloid parametrization:The half-angle simplification is valid throughout $0<t<2\pi$, where $1-\cos t>0$. In particular the horizontal coordinate increases strictly along the arch.

The heat equation is linear and homogeneous in $u$: the operator $L={\partial }_t-\lambda {\partial }_x^2$ satisfies $L(A{u}_1+B{u}_2)=AL{u}_1+BL{u}_2$. Thus the superposition principle applies to its solutions. Spatial translations are also allowed because the coefficients are constant.

Choose $a=1/(4\lambda )$ in part (b) and multiply by $\sqrt{a}$ to make the initial amplitude one. Taking the two shifts $b=1$ and $b=-1$ then givesAt $t=0$ this reproduces both initial peaks. Each term is an instance of Gaussian initial data for the heat equation, so their sum satisfies the equation. Its extension to negative times is a real smooth solution precisely on the connected intervalAt the lower endpoint the peaks narrow without bound and their heights diverge; there is no finite classical continuation there. For positive times each peak spreads and decreases in height while preserving its integral.

For a Gaussian function, the required derivatives areSubstitute these into the reduced ordinary differential equation and compare the constant and quadratic coefficients:The constant terms give $c=1/(4\lambda )$, and this value also satisfies the quadratic terms. HenceA constant multiple is also a solution by linearity.

Set $T=t+a>0$. The similarity solution uses $y=(x+b)T^{-1/2}$, whose partial derivatives at fixed $x$ or $t$ areApply the multivariable chain rule to $u=T^{-1/2}v(y)$:Substituting in the heat equation and cancelling the positive common factor $T^{-3/2}$ gives

The partial derivatives are $g_x=\sin x$ and $g_y=y$, so all stationary points are $(n\pi ,0)$ for integers $n$. Their Hessian matrix is $\mathrm{diag}(\cos x,1)$. At even multiples of $\pi$ it is a positive-definite matrix, whereas at odd multiples it has one positive and one negative eigenvalue. ConsequentlyThere are no local maxima. Since both $1-\cos x$ and $y^2/2$ are nonnegative, these are also global minima. In the requested open plotting window, the stationary points are $(0,0)$ and $(\pm \pi ,0)$; the local minima at $(\pm 2\pi ,0)$ lie on excluded edges.

For a level set $g=c$, solving for the vertical coordinate givesFor $0<c<2$ these level sets are closed ovals around each local minimum. At $c=0$ the level sets shrink to the local minima. The level set through the saddle points iswhere the two signed branches together describe the same set as $y=\pm 2|\cos (x/2)|$. They cross at $(\pm \pi ,0)$ and reach $(0,\pm 2)$. The crossing slopes at a saddle point are $\pm 1$, as also follows from the local quadratic form $(-\delta x^2+\delta y^2)/2$. For $c>2$, the upper and lower level sets remain separate and extend across the plotting window without meeting the horizontal axis.