91Ó°ÊÓ

We are given a mapping \(\mathbf{f} = (f_1, f_2, f_3)\) of \(\mathbb{R}^2\) into \(\mathbb{R}^3\), where $$ \begin{array}{l} f_1(s,t) = (b + a\cos{s})\cos{t},\\ f_2(s,t) = (b + a\cos{s})\sin{t},\\ f_3(s,t) = a\sin{s}. \end{array} $$

First, find the partial derivatives of \(f_1\) and \(f_3\) with respect to \(s\) and \(t\). For \(f_1\), we have $$ \frac{\partial f_1}{\partial s} = -a\sin{s}\cos{t},\ \frac{\partial f_1}{\partial t} = -(b+a\cos{s})\sin{t}, $$ so the gradient of \(f_1\) is $$ \nabla f_1 = \left(-a\sin{s}\cos{t}, -(b+a\cos{s})\sin{t}\right). $$ Similarly, for \(f_3\) we have $$ \frac{\partial f_3}{\partial s} = a\cos{s},\ \frac{\partial f_3}{\partial t} = 0, $$ and the gradient of \(f_3\) is $$ \nabla f_3 = \left(a\cos{s}, 0\right). $$

In order to find the 4 points where \((\nabla f_1)\left(\mathbf{f}^{-1}(\mathbf{p})\right) = \mathbf{0}\), we need to solve the following system of equations: $$ \begin{cases} -a\sin{s}\cos{t}=0,\\ -(b+a\cos{s})\sin{t}=0. \end{cases} $$ From the first equation, we can see that either \(\sin s = 0\) or \(\cos t = 0\). If \(\sin s = 0\), then \(s=0\) or \(s=\pi\). If \(\cos t = 0\), then \(t=\pi/2\) or \(t=3\pi/2\). Combining these conditions, we get four points \((s,t)\): \((0,\pi/2)\), \((0,3\pi/2)\), \((\pi,\pi/2)\), and \((\pi,3\pi/2)\).

Now apply the mapping to the points found in Step 3 to get the points in the range K: $$ \begin{aligned} \mathbf{p}_1 &= \mathbf{f}(0,\pi/2) = (a, b, 0),\\ \mathbf{p}_2 &= \mathbf{f}(0,3\pi/2) = (a, -b, 0),\\ \mathbf{p}_3 &= \mathbf{f}(\pi,\pi/2) = (-a, b, 0),\\ \mathbf{p}_4 &= \mathbf{f}(\pi,3\pi/2) = (-a, -b, 0). \end{aligned} $$

In order to find the set of points, we need to solve \(a\cos{s} = 0\). This occurs when \(s=\pi/2\) or \(s=3\pi/2\). Hence, for these two values of \(s\), the points in the range K will be $$ \mathbf{q} = (0, (b+a\cos{s})\sin{t}, 0), $$ and the set of all \(\mathbf{q}\) will be a circle in the xy-plane with radii \(b-a\) and \(b+a\).

Compute the second partial derivatives of \(f_1\): $$ \frac{\partial^2 f_1}{\partial s^2} = -a\cos{s}\cos{t},\ \frac{\partial^2 f_1}{\partial t^2} = -(b+a\cos{s})\cos{t},\ \frac{\partial^2 f_1}{\partial s\partial t} = a\sin{s}\sin{t}. $$ For points \((s,t)\) found in Step 3, we can compute the determinant of the Hessian matrix: $$ \begin{aligned} D(0,\pi/2) &= \left(-a\cos{(0)}\cos{(\pi/2)}\right)\left(-(b+a\cos{(0)})\cos{(\pi/2)}\right) - \left(a\sin{(0)}\sin{(\pi/2)}\right)^2 = 0,\\ D(0,3\pi/2) &= 0,\\ D(\pi,\pi/2) &= -a(b-a) > 0,\\ D(\pi,3\pi/2) &= a(b+a) > 0. \end{aligned} $$ From these results, we can conclude that \(\mathbf{p}_1\) and \(\mathbf{p}_2\) are saddle points, \(\mathbf{p}_3\) corresponds to a local minimum, and \(\mathbf{p}_4\) corresponds to a local maximum. For the points \(\mathbf{q}\) found in Step 5, we have \(f_3=0\). Since \(f_3\) does not depend on \(t\), its critical points will be along the circles found in the xy-plane (without any maxima or minima).

We are given a new function \(g(t)=\mathbf{f}(t,\lambda t)\) where \(\lambda\) is an irrational number. Compute the derivative of \(\mathbf{g}\) with respect to \(t\): $$ \mathbf{g}'(t) = \frac{d\mathbf{f}(t,\lambda t)}{dt} = \left(-a\sin t\cos (\lambda t)+\lambda(b+a\cos t)(-\sin(\lambda t)), -(b+a\cos t)\sin(\lambda t)-\lambda a\sin t\sin(\lambda t), a\cos t\right). $$ Compute the square of the magnitude of \(\mathbf{g}'(t)\): $$ \left|\mathbf{g}'(t)\right|^2=(-a\sin t\cos (\lambda t)+\lambda(b+a\cos t)(-\sin(\lambda t)))^2+(-(b+a\cos t)\sin(\lambda t)-\lambda a\sin t\sin(\lambda t))^2+(a\cos t)^2 $$ Simplify expression: $$ \left|\mathbf{g}'(t)\right|^2 = a^{2} + \lambda^{2}(b+a\cos{t})^{2}. $$ Since \(g(t)\) is a mapping of \(\mathbb{R}\) onto a subset of K, it is a one-to-one mapping onto a dense subset of K.