If $V_{0x}\ne 0$, the first formula in (a) gives $e^{-\gamma t}=1-\gamma x/V_{0x}$ andSubstitute this and $1-e^{-\gamma t}=\gamma x/V_{0x}$ into $z(t)$:For finite physical times $t\ge 0$, the domain is $0\le \gamma x/V_{0x}<1$, valid for either sign of $V_{0x}$. If $V_{0x}=0$, the trajectory is vertical, $x(t)=0$, and in general no single-valued graph $z(x)$ describes it. As a consistency check, the logarithmic expansion at fixed $x$ cancels the apparent $1/\gamma$ terms and gives $z\to (V_{0z}/V_{0x})x-g{x}^2/(2V_{0x}^2)$ as $\gamma \to 0$, the drag-free parabola.

First solve for the velocity components. The integrating factor $e^{\gamma t}$ givesIntegrating once more and imposing the initial position givesThese formulas describe projectile motion with linear drag. Differentiation checks both second-order differential equations and both initial velocities. At large positive time, the horizontal velocity tends to zero and the vertical velocity tends to the terminal value $-g/\gamma$.

Since $f>0$, a level $f=c$ can occur only for $c>0$. Taking its natural logarithm gives $xy=\ln c$. Thus $c>1$ produces rectangular hyperbolas in the first and third quadrants, $0<c<1$ produces them in the second and fourth quadrants, and $c=1$ consists of both coordinate axes.

The gradient $e^{xy}(y,x)$ vanishes only at $(0,0)$, where $f=1$. The Hessian matrix there is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with negative determinant. Equivalently, $f(x,x)>1$ while $f(x,-x)<1$ for nonzero small $x$. Hence the only stationary point is a saddle point at $(0,0,1)$ on the graph; its location in the contour plane is the origin.

The specified unit vector is $\mathbf{u}=(\cos (\pi /3),\sin (\pi /3))=(1/2,\sqrt{3}/2)$; the positive $y$ component selects that direction. The gradient is $\nabla f=e^{xy}(y,x)$, giving $(0,-2)$ at the point. The directional derivative is therefore

For $x>0$, the exponential definitions of the hyperbolic functions give $\cosh x-\sinh x=e^{-x}>0$ and $\sinh x>0$. Taking reciprocals of these positive quantities proves $1/\cosh x<1/\sinh x$. Also $\cosh x>1$ for $x>0$, so division by the positive $\sinh x$ gives $1/\sinh x<\cosh x/\sinh x$. Combining,

At the first endpoint, the Cartesian coordinates are $(r,0,0)$, so its great circle must have $l=0$. At the second endpoint, the plane condition gives $m\sin {\theta }_0\sin {\varphi }_0+n\cos {\theta }_0=0$. Since $\cos {\theta }_0>0$, an equivalent regular plane equation isWhere $\cos \theta \ne 0$, divide by $\cos {\theta }_0\cos \theta$ to obtainThe regular implicit equation includes the equatorial first endpoint; the tangent form is undefined there and must be read by a limit. The shorter connecting arc of this great circle is the shortest path. The endpoints are not antipodal, so this shorter arc is unique. For example, the central angle is $\delta =\arccos (\sin {\theta }_0\cos {\varphi }_0)$, giving spherical distance $r\delta$.

A plane through the origin has equation $lx+my+nz=0$ for some nonzero normal $(l,m,n)$. Dividing that normal by its Euclidean norm makes $l^2+m^2+n^2=1$ without changing the plane. Substitute the spherical coordinates from (a) and cancel the nonzero radius:Its intersection with the sphere is the desired great circle; changing the sign of all three coefficients describes the same circle.

Hold $r$ fixed and differentiate the position vector $\mathbf{X}(\theta ,\varphi )$. Its coordinate tangent vectors areTheir dot product is zero, and their squared Euclidean norms are $r^2$ and $r^2{\sin }^2\theta$. To first order, the displacement is $\delta \mathbf{X}={\mathbf{X}}_{\theta }\delta \theta +{\mathbf{X}}_{\varphi }\delta \varphi +O((|\delta \theta |+|\delta \varphi |{)}^2)$. Squaring givesThis is the line element on a sphere. The chord distance and the short surface arc length agree to the order retained here.

Using the physics convention for spherical coordinates, the polar angle $\theta$ is measured from positive $z$ and the azimuth $\varphi$ from positive $x$ toward positive $y$. The Cartesian coordinates are