Exercise 6.1 · Q11: Where on $6y = x^3 + 2$ does $y$ change 8 times as fast as $x$⁠?

Basics you need

Solution

1. Turn the words into an equation

   Let the particle be at $(x, y)$ at time $t$⁠. The condition "the $y$-coordinate is changing 8 times as fast as the $x$-coordinate" is

   $$\frac{dy}{dt} = 8\,\frac{dx}{dt}.$$

   We assume the particle is actually moving along the curve at that moment, so $\frac{dx}{dt} \ne 0$⁠.

   "Changing 8 times as fast" compares two rates of change, and rates are derivatives with respect to time. The assumption $\frac{dx}{dt} \ne 0$ matters: if $x$ were not changing, then (as Step 2 shows) $y$ would not change either, and "$0 = 8\times 0$" would hold at every point. The question asks for particular points, so it means a moving particle, with $x$ actually changing.
   Section 6.2: $\frac{dy}{dt}$ is the rate of change of $y$ with respect to $t$⁠. Basics: rate of change as a derivative

2. Differentiate the curve's equation with respect to $t$

   The equation $6y = x^3 + 2$ holds at every moment of the motion, so we may differentiate both sides with respect to $t$⁠:

   $$\begin{aligned} \frac{d}{dt}(6y) &= \frac{d}{dt}\left(x^3 + 2\right), \\ 6\,\frac{dy}{dt} &= 3x^2\,\frac{dx}{dt} + 0. \end{aligned}$$

   Dividing by 6,

   $$\frac{dy}{dt} = \frac{x^2}{2}\,\frac{dx}{dt}.$$
   The left side is a constant 6 times $y$⁠, so its rate is 6 times $y$'s rate. On the right, $x^3$ depends on $t$ through $x$⁠, so the chain rule gives $3x^2$ times $\frac{dx}{dt}$⁠, and the constant 2 contributes 0. This one equation holds at every point of the path.
   Chain rule (Theorem 4, Section 5.3.1) and $\frac{d}{dx}\left(x^3\right) = 3x^2$ (Table 5.3). Basics: related rates (chain rule)

3. Impose the condition and solve for $x$

   Put $\frac{dy}{dt} = \frac{x^2}{2}\,\frac{dx}{dt}$ into the condition $\frac{dy}{dt} = 8\,\frac{dx}{dt}$⁠:

   $$\begin{aligned} \frac{x^2}{2}\,\frac{dx}{dt} &= 8\,\frac{dx}{dt}, \\ \left(\frac{x^2}{2} - 8\right)\frac{dx}{dt} &= 0. \end{aligned}$$

   Since $\frac{dx}{dt} \ne 0$ (Step 1), the bracket must be zero:

   $$\frac{x^2}{2} = 8 \;\Longrightarrow\; x^2 = 16 \;\Longrightarrow\; x = 4 \ \text{or}\ x = -4.$$
   A product is zero only when a factor is zero. Writing the equation as a product, instead of silently "cancelling $\frac{dx}{dt}$", shows exactly where the assumption $\frac{dx}{dt} \ne 0$ is used. Then $x^2 = 16$ has two real solutions, and both must be kept.
   $x^2 = 16 \iff (x - 4)(x + 4) = 0 \iff x = 4$ or $x = -4$⁠. Basics: factorising

4. Find the $y$-coordinates from the curve

   Each point must lie on $6y = x^3 + 2$⁠.

   • $x = 4$⁠: $6y = 4^3 + 2 = 64 + 2 = 66$⁠, so $y = \frac{66}{6} = 11$⁠.
   • $x = -4$⁠: $6y = (-4)^3 + 2 = -64 + 2 = -62$⁠, so $y = -\frac{62}{6} = -\frac{31}{3}$⁠.

   The points are $(4, 11)$ and $\left(-4, -\frac{31}{3}\right)$⁠.

   The question asks for points, so we need both coordinates. The cube keeps the sign: $(-4)^3 = -64$⁠, and then $-64 + 2 = -62$⁠, not $-66$⁠.

5. See it as a slope

   Section 6.2 gives $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ when $\frac{dx}{dt} \ne 0$⁠. From Step 2 this ratio is $\frac{x^2}{2}$⁠, which is also what we get by differentiating $y = \frac{x^3 + 2}{6}$ directly with respect to $x$⁠. So the question asks where the curve's slope is 8, and the slope $\frac{x^2}{2}$ equals 8 exactly at $x = \pm 4$⁠. The figure shows the two points with short segments of slope 8.

   Also note that $\frac{x^2}{2} \ge 0$⁠: $\frac{dy}{dt}$ and $\frac{dx}{dt}$ never have opposite signs. So reading "8 times as fast" as a statement about speeds, $\left|\frac{dy}{dt}\right| = 8\left|\frac{dx}{dt}\right|$⁠, gives the same equation $\frac{x^2}{2} = 8$ and the same two points.

   The slope picture explains why the answer does not depend on how fast the particle moves: the ratio of the two rates is fixed by the shape of the curve at that point. The sign remark shows the answer is the same under either reasonable reading of "as fast as".
   Section 6.2: $\frac{dy}{dx} = \frac{dy}{dt}\Big/\frac{dx}{dt}$⁠, provided $\frac{dx}{dt} \ne 0$⁠. Lesson: rate of change of quantities

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