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Worksheet 9.6 Interactive Problem Set

The problems below are auto-graded. Each one is deliberately chosen so that a correct answer requires a full argument rather than pattern matching: the distractors are the answers you get from the most common shortcuts.

1. True or False.

    If every directional derivative \(D_{\mathbf u}f(P_0)\) exists at a point \(P_0\text{,}\) then \(f\) must be continuous at \(P_0\text{.}\)

  • True.

  • False, and the failure is dramatic. Let
    \begin{equation*} f(x,y) = \frac{x^2 y}{x^4 + y^2}\ \text{ for } (x,y)\ne(0,0), \qquad f(0,0) = 0 . \end{equation*}
    For a unit vector \(\mathbf u = \langle a,b\rangle\) with \(b \ne 0\text{,}\)
    \begin{equation*} D_{\mathbf u}f(0,0) = \lim_{t\to 0}\frac{f(ta,tb)}{t} = \lim_{t\to 0}\frac{a^2 b}{t^2a^4 + b^2} = \frac{a^2}{b}, \end{equation*}
    and \(D_{\mathbf u}f(0,0) = 0\) when \(b = 0\text{.}\) So every directional derivative exists. Yet along the parabola \(y = mx^2\text{,}\)
    \begin{equation*} f(x,mx^2) = \frac{m}{1+m^2}, \end{equation*}
    which depends on \(m\) — different parabolas approach different values, so \(\lim_{(x,y)\to(0,0)}f\) does not exist and \(f\) is not continuous at the origin. Existence of directional derivatives is a statement about behavior along straight lines only; continuity requires control along every approach.

  • False.

  • False, and the failure is dramatic. Let
    \begin{equation*} f(x,y) = \frac{x^2 y}{x^4 + y^2}\ \text{ for } (x,y)\ne(0,0), \qquad f(0,0) = 0 . \end{equation*}
    For a unit vector \(\mathbf u = \langle a,b\rangle\) with \(b \ne 0\text{,}\)
    \begin{equation*} D_{\mathbf u}f(0,0) = \lim_{t\to 0}\frac{f(ta,tb)}{t} = \lim_{t\to 0}\frac{a^2 b}{t^2a^4 + b^2} = \frac{a^2}{b}, \end{equation*}
    and \(D_{\mathbf u}f(0,0) = 0\) when \(b = 0\text{.}\) So every directional derivative exists. Yet along the parabola \(y = mx^2\text{,}\)
    \begin{equation*} f(x,mx^2) = \frac{m}{1+m^2}, \end{equation*}
    which depends on \(m\) — different parabolas approach different values, so \(\lim_{(x,y)\to(0,0)}f\) does not exist and \(f\) is not continuous at the origin. Existence of directional derivatives is a statement about behavior along straight lines only; continuity requires control along every approach.

2. Which Identity Fails?

Exactly one of the following is not valid for all vectors \(\vec a, \vec b, \vec c\) in \(\R^3\text{.}\) Which one?
  • \(\vec a\cdot(\vec b\times\vec c) = (\vec a\times\vec b)\cdot\vec c\)
  • This one is valid. Both sides equal the determinant with rows \(\vec a,\vec b,\vec c\text{,}\) so the dot and the cross may be interchanged.
  • \(\vec a\times(\vec b\times\vec c) = (\vec a\times\vec b)\times\vec c\)
  • Correct — the cross product is not associative. Take \(\vec a = \vec b = \mathbf i\) and \(\vec c = \mathbf j\text{.}\) Then
    \begin{equation*} \mathbf i\times(\mathbf i\times\mathbf j) = \mathbf i\times\mathbf k = -\mathbf j, \qquad (\mathbf i\times\mathbf i)\times\mathbf j = \vec 0\times\mathbf j = \vec 0 . \end{equation*}
    The left side lies in the plane of \(\vec b\) and \(\vec c\text{;}\) the right side lies in the plane of \(\vec a\) and \(\vec b\text{.}\) There is no reason for those to agree.
  • \(\vec a\times(\vec b + \vec c) = \vec a\times\vec b + \vec a\times\vec c\)
  • This one is valid — the cross product is distributive over addition. (It is only associativity that fails.)
  • \(\vec a\times(\vec b\times\vec c) = (\vec a\cdot\vec c)\,\vec b - (\vec a\cdot\vec b)\,\vec c\)
  • This one is valid: it is the “BAC-CAB” triple product expansion. It looks the most exotic of the four, which is exactly why it is worth remembering that it is true.

3. Distance Between Skew Lines.

Consider the two lines
\begin{align*} L_1: \amp \quad \vec r = \langle 1,0,2\rangle + t\,\langle 2,-1,1\rangle,\\ L_2: \amp \quad \vec r = \langle 0,3,-1\rangle + s\,\langle 1,2,-1\rangle . \end{align*}
They are skew. Let \(\vec u,\vec v\) be the direction vectors and let \(\vec w\) join a point of \(L_1\) to a point of \(L_2\text{.}\)
The absolute value of the scalar triple product \(\left|\vec w\cdot(\vec u\times\vec v)\right|\) equals .
The distance between the two lines, to three decimal places, is .

4. Why a Magnetic Field Cannot Change a Particle’s Speed.

Arrange the steps below into a correct proof that the magnetic force does no work, and therefore that a charged particle in a magnetic field moves at constant speed.

5. Which of These Are Hyperboloids?

None of the equations below is in standard position. Complete the square where necessary, then select every equation whose surface is a hyperboloid, of one sheet or of two.
  • \(x^2 + y^2 - z^2 - 2x = 0\)
  • Correct. Completing the square gives \((x-1)^2 + y^2 - z^2 = 1\text{:}\) one negative square, equal to a positive constant, so a hyperboloid of one sheet.
  • \(x^2 + y^2 + 4z^2 - 4y = 0\)
  • No. This is \(x^2 + (y-2)^2 + 4z^2 = 4\) — every square is positive, so it is an ellipsoid.
  • \(x^2 - y^2 - z^2 = 1\)
  • Correct. Two negative squares against a positive constant gives a hyperboloid of two sheets.
  • \(x^2 + y^2 - z^2 = 0\)
  • No — and this is the trap. It has the same signs as the first option, but it equals \(0\) rather than a positive constant. That degenerates the hyperboloid into an elliptic cone, the surface the two families of hyperboloids approach asymptotically.
  • \(z + 1 = x^2 - y^2\)
  • No. The word “hyperbolic” is a red herring: this is linear in \(z\text{,}\) so it is a hyperbolic paraboloid (a saddle) with vertex at \((0,0,-1)\text{,}\) not a hyperboloid.

6. Classify the Critical Points.

7. Find the Error.

8. The Roles of the Two Products.

In two or three sentences, explain why a magnetic field can bend a charged particle’s path into a circle but can never make it move faster. Your answer should say what job the cross product does and what job the dot product does, and should make clear which one is responsible for the speed staying constant.
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