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This homework set is out of 40 points (with the possibility of additional extra credit).

1. (10 points) Let \(C\) be the circle of radius \(r\) which is centered at the origin. Find the periodic function \(f\) which parameterizes \(C\) with respect to arc length in the clockwise direction and satisfies \(f(0) = (a,\sqrt{r^2-a^2})\).
   
   Solution If \(c\) is a constant, then \(f(s) = (r \sin (\frac{\theta}{r} + c),r \cos(\frac{\theta}{r}+c))\) is a unit speed clockwise parameterization of the circle centered at the origin. If we set \(c = \sin^{-1}(a)/r\) then \(f(0) = (a,\sqrt{r^2-a^2})\).
2. (10 points) Define \(E := \{ (x,y,z) \in \mathbb{R}^3 \mid x^2 + z^2 = 1 \textrm{ and } y^2 + z^2 = 1\}\). Show that \(E\) is a union of two regular closed curves. Note: \(E\) is the intersection of two unit cylinders, one centered on the \(x\)-axis, the other on the \(y\)-axis.
   
   Solution Observe that if \((x,y,y) \in E\), then \(x = \pm y\). Define \(\alpha(t) = (\cos(t),\cos(t),\sin(t))\) and \(\beta(t) = (\cos(t),-\cos(t),\sin(t))\). If \((x,y,z) \in E\), then there is a \(t\) such that \((x,z) = (\cos(t),\sin(t))\). Then either \(\alpha(t) = (x,y,z)\) or \(\beta(t) = (x,y,z)\). For any \(t\), \(\|\dot \alpha\|^2 = \|\dot \beta\|^2 = 1 + \cos^2 (t) > 0\) and therefore both \(\alpha\) and \(\beta\) are regular.
3. (10 points) Give an informal but convincing argument that \(E\) is not a regular curve. Extra credit (5 points) if you are able to make your argument completely mathematically rigorous.
   
   Solution Suppose that \(\gamma(t) = (x(t),y(t),z(t))\) is a regular curve contained in \(E\). Since for every \(t\), \(x(t) = \pm y(t)\), we have that \(\dot x(t) = \pm \dot y(t)\). If we set \[ g(t) = \frac{\dot \gamma(t)}{\|\dot \gamma\|} \cdot \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}} \] then \(g\) is continuous and takes values in \(\{0,1\}\). By the Intermediate Value Theorem, \(g\) must be constant. This implies either \(x(t) = y(t)\) for all \(t\) or \(x(t) = -y(t)\) for all \(t\). Since \((1,-1,0)\) and \((1,1,0)\) are both in \(E\), one of these points is not in the range of \(\gamma\).
4. (10 points) Suppose that \(f(x,y)\) is a function from \(\mathbb{R}^2\) to \(\mathbb{R}\) such that \(\nabla f\) exists and is nonzero at all points in the domain. Show that if \(\gamma(t)\) is a solution to the differential equation \(\nabla f \cdot \gamma'(t) = 0\), then \(\gamma(t)\) parameterizes part of a level curve of \(f\). Hint: This is the same as showing that \(f(\gamma(t))\) is constant. It may be helpful to denote the coordinates of \(\gamma(t)\) by \(x(t)\) and \(y(t)\).
   
   Solution By the chain rule, the derivative of \(f(\gamma(t))\) is \(\nabla f (\gamma(t)) \cdot \gamma'(t)\). Since this derivative is \(0\) if and only if \(f(\gamma(t))\) is constant, it follows that any \(\gamma\) satisfying this differential equation is a level curve of \(f\).

Challenge problem (15 points): If \(C\) and \(f\) are as in the first problem, define \(g(s) = s - |\!|f(s) - f(0)|\!|\). Find the first nonzero term in the Maclaurin series expansion of \(g\) and relate its coefficient to the curvature of \(C\). Remark: \(g\) measures the discrepancy between the distance from \(f(0)\) to \(f(s)\) measured in the plane and measured along \(C\). The point of this problem is to relate the two responses to the question "what does curvature measure?" given in the first lecture.
Note: The challenge problem may be substituted for any of the regular problems from this homework set.