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First method: substitute the trajectory. Along the parametric curve, the temperature is $T(t)=20e^{-{\cos }^{2}t}-(1+\sin t{)}^3$. The chain rule givesSecond method: use the spatial derivatives. The partial derivatives of the scalar field are $T_x=-40x{e}^{-x^2}$ and $T_z=-3(z+1{)}^2$, while the velocity components are $\dot{x}=-\sin t$ and $\dot{z}=\cos t$. The multivariable chain rule therefore giveswhich agrees with the first calculation. This is the gradient dotted with the velocity, rather than a partial derivative at a fixed position.

Work in the small-slope model of a transverse wave on a taut string. Let $T$ be the constant tension, $\rho$ the linear mass density, and take a right-moving sinusoidal displacement $y=A\cos (\kappa x-\omega t)$, with wave number $\kappa =\omega /v$. The energy flux of a transverse string wave through a cut is the transverse force times transverse velocity:A cycle average gives $⟨{\sin }^2⟩=1/2$. Using $T=\rho v^2$, obtained from the string wave equation, yieldsThe sign in the flux formula makes the power positive for propagation in the positive $x$ direction.

The energy per unit equilibrium length is the sum of the kinetic and elastic terms:For a segment of mass $M=\rho \ell$ containing many wavelengths, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq \frac{1}{2}M{\omega }^2{A}^2$. Convert the given quantities to SI units: $M=0.080\mathrm{kg}$, $A=1.6\times 10^{-4}\mathrm{m}$, and $\omega =2\pi (120){\mathrm{s}}^{-1}$. ThenThe many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of wavelengths.

Take the normal vector to point outward from the box. The divergence theorem changes the closed-surface flux into a volume integral. The divergence isThe first term integrates to zero in $x$, since ${\int }_{-1}^2\pi \cos (\pi x)dx=\sin 2\pi -\sin (-\pi )=0$. ThereforeHenceA direct face check gives zero from the two $x$ faces, $45/2$ from the pair of $y$ faces, and $45$ from the pair of $z$ faces. This also checks the outward-orientation signs.

Set $g(x)=\sqrt{8x}+{\sin }^2(2x)$. The chain rule gives $f^{\prime }=-g^{\prime }/g^2$, whileThusThe real domain excludes zero because the original denominator vanishes there; for $x>0$ it is strictly positive.

For a continuous random variable with probability density function $f_X$, its cumulative distribution function isConsequently $F_X^{\prime }=f_X$ wherever the density is continuous, and almost everywhere for an absolutely continuous distribution.

Here $Y=X^2\ge 0$. Thus $F_Y(y)=0$ for $y<0$. For $y\ge 0$, the event $X^2\le y$ is $-\sqrt{y}\le X\le \sqrt{y}$. Symmetry of the Laplace distribution givesTherefore the square of a standard Laplace variable has cumulative distribution functionIt is continuous at zero, so there is no point mass there. As a check, its probability density function for $y>0$ is $e^{-\sqrt{y}}/(2\sqrt{y})$, whose integral becomes ${\int }_0^{\infty }{e}^{-u}du=1$ after setting $u=\sqrt{y}$.

The real logarithm restricts the domain to $x>0$. Differentiating givesSince $x$ cannot vanish in the domain, a stationary point requires $\ln (3x)=-1/2$. The logarithm is strictly increasing, so this equation has exactly one solution:The second derivative is $f^{\prime \prime }(x)=2\ln (3x)+3$, giving $f^{\prime \prime }(x_{\ast })=2>0$. In fact $f^{\prime }$ is negative below $x_{\ast }$ and positive above it, so the point is the unique global minimum on $x>0$. The boundary value $x=0$ is not an additional stationary point, since the original function is not defined there.