Chapter 5 · Continuity and Differentiability

Miscellaneous · Q17: $\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}$ for $x=a(\cos t+t\sin t)$ ⁠, $y=a(\sin t-t\cos t)$

Miscellaneous ExerciseSecond order derivative · Parametric form · Miscellaneous Exercise

The question

If $x = a(\cos t + t\sin t)$ and $y = a(\sin t - t\cos t)$ ⁠, find $\dfrac{d^2y}{dx^2}$ ⁠.

Exam answer

Here $x = a(\cos t + t\sin t)$ ⁠, $y = a(\sin t - t\cos t)$ ⁠.

\begin{aligned} \frac{dx}{dt} &= a(-\sin t + \sin t \\ &\qquad + t\cos t) = at\cos t \\ \frac{dy}{dt} &= a(\cos t - \cos t \\ &\qquad + t\sin t) = at\sin t \end{aligned}

For $at\cos t \ne 0$ ⁠,

\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} \\ &= \tan t \end{aligned}

Differentiating w.r.t. $x$ ⁠,

\begin{aligned} \frac{d^2y}{dx^2} &= \sec^2 t\cdot\frac{dt}{dx} \\ &= \sec^2 t\cdot\frac{1}{at\cos t} \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} = \dfrac{\sec^3 t}{at}$ $(at\cos t \ne 0)$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

For a parametric curve, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ is again a function of $t$ ⁠. To get the second derivative we differentiate that function of $t$ with respect to $x$ , which by the chain rule means differentiating it with respect to $t$ and then dividing by $\frac{dx}{dt}$ once more. It is not $\frac{d^2y/dt^2}{d^2x/dt^2}$ ⁠.

Given, to find, and the plan

Given: $x = a(\cos t + t\sin t)$ ⁠, $y = a(\sin t - t\cos t)$ ⁠, with $a$ a non-zero constant.
To find: $\dfrac{d^2y}{dx^2}$ ⁠.
Plan: (1) $\frac{dx}{dt}$ and $\frac{dy}{dt}$ ⁠. (2) $\frac{dy}{dx} = \tan t$ ⁠, noting where $\frac{dx}{dt} \ne 0$ ⁠. (3) $\frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t)\div\frac{dx}{dt}$ ⁠. (4) State where the answer is valid.

Basics you need

1Parametric differentiation

If $x = f(t)$ ⁠, $y = g(t)$ ⁠, then $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ ⁠, provided $f'(t) \ne 0$ (Section 5.6). Lesson: parametric differentiation

2Second order derivative

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ (Section 5.7). Lesson: second order derivative

3Product rule and standard derivatives

$(uv)' = u'v + uv'$ ⁠; $\frac{d}{dt}\sin t = \cos t$ ⁠, $\frac{d}{dt}\cos t = -\sin t$ ⁠, $\frac{d}{dt}\tan t = \sec^2 t$ (Section 5.3, Table 5.3). Basics: derivative rules

4Chain rule

If $u$ depends on $t$ and $t$ on $x$ ⁠: $\frac{du}{dx} = \frac{du}{dt}\cdot\frac{dt}{dx}$ ⁠, and $\frac{dt}{dx} = \frac{1}{dx/dt}$ when $\frac{dx}{dt} \ne 0$ (Theorem 4). Lesson: chain rule

Solution

Differentiate $x$ and $y$ with respect to $t$

Use the product rule on $t\sin t$ and $t\cos t$ ⁠:
$\begin{aligned} \frac{dx}{dt} &= a\big(-\sin t + \sin t + t\cos t\big) \\ &= at\cos t, \\ \frac{dy}{dt} &= a\big(\cos t - \cos t + t\sin t\big) \\ &= at\sin t. \end{aligned}$
$\frac{d}{dt}(t\sin t) = 1\cdot\sin t + t\cos t$ and $\frac{d}{dt}(t\cos t) = 1\cdot\cos t - t\sin t$ ⁠. In each line two terms cancel, which is what makes this curve pleasant to work with.

Product rule (rule (2), Section 5.3) and Table 5.3. Basics: derivative rules
First derivative: $\frac{dy}{dx} = \tan t$

$\frac{dx}{dt} = at\cos t$ is non-zero exactly when $a \ne 0$ ⁠, $t \ne 0$ and $\cos t \ne 0$ ⁠. For such $t$ ⁠,
$\frac{dy}{dx} = \frac{at\sin t}{at\cos t} = \tan t.$
The parametric formula needs $\frac{dx}{dt} \ne 0$ ⁠. Cancelling $at$ is allowed because it is non-zero. At $t = 0$ both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ vanish; where $\cos t = 0$ (for example $t = \frac{\pi}{2}$ ⁠) the tangent is vertical. Those values are excluded.

Section 5.6 (Basic 1). Lesson: parametric differentiation
Second derivative: differentiate $\tan t$ with respect to $x$

$\frac{dy}{dx} = \tan t$ is a function of $t$ ⁠, and $t$ is a function of $x$ ⁠. By the chain rule,
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(\tan t) = \frac{d}{dt}(\tan t)\cdot\frac{dt}{dx} \\ &= \sec^2 t\cdot\frac{1}{at\cos t}. \end{aligned}$
Since $\frac{1}{\cos t} = \sec t$ ⁠,
$\frac{d^2y}{dx^2} = \frac{\sec^3 t}{at}.$
The derivative of $\tan t$ with respect to $t$ is $\sec^2 t$ ⁠, but we need it with respect to $x$ ⁠. The factor $\frac{dt}{dx} = \frac{1}{dx/dt}$ converts the rate per unit $t$ into a rate per unit $x$ ⁠. It is valid because $\frac{dx}{dt} \ne 0$ at the points we kept in Step 2.

Second order derivative (Section 5.7); chain rule (Theorem 4); $\frac{d}{dt}\tan t = \sec^2 t$ (Table 5.3). Lesson: second order derivative
Where the answer is valid, and what the curve is

The formula holds for every $t$ with $at\cos t \ne 0$ ⁠. The answer key states it for $0 < t < \frac{\pi}{2}$ ⁠, where these conditions certainly hold (with $a \ne 0$ ⁠); the same formula is correct for all other allowed $t$ as well.

The curve for $a = 1$ ⁠, $0 \le t \le \pi$ ⁠, with equal scales. It unwinds from the circle of radius $a$ like a taut string (an involute of the circle). At $P$ ( $t = \frac{\pi}{4}$ ⁠) the tangent has slope $\tan\frac{\pi}{4} = 1$ ⁠. At $t = \frac{\pi}{2}$ ⁠, $\cos t = 0$ ⁠: the tangent is vertical and $\frac{dy}{dx}$ does not exist. The segment from $Q = (a\cos t, a\sin t)$ to $P$ has length $at$ and meets the tangent at a right angle.

The book does not give an interval for $t$ ⁠, so the honest statement of the answer includes its conditions. The picture explains the slope: $P - Q = at(\sin t, -\cos t)$ ⁠, so the tangent at $P$ is perpendicular to $QP$ and has direction $(\cos t, \sin t)$ ⁠, slope $\tan t$ ⁠. For $0 < t < \frac{\pi}{2}$ with $a > 0$ ⁠, the second derivative is positive: the curve bends upward there, as in the figure.

$\frac{dx}{dt} \ne 0$ condition of Section 5.6. Basics: slope at a point

Result

\frac{dy}{dx} = \tan t, \qquad \frac{d^2y}{dx^2} = \frac{\sec^3 t}{at},

valid wherever $at\cos t \ne 0$ ⁠. (Textbook answer: $\frac{\sec^3 t}{at}$ ⁠, $0 < t < \frac{\pi}{2}$ ⁠, the same.)

Check it with numbers

Take $a = 2$ ⁠, $t = 0.7$ ⁠. Then $\frac{dx}{dt} = 2(0.7)\cos 0.7 \approx 1.0707791$ and $\frac{dy}{dt} = 2(0.7)\sin 0.7 \approx 0.9019048$ ⁠, ratio $\approx 0.8422884 = \tan 0.7$ ⁠. ✓

Formula: $\frac{\sec^3 0.7}{1.4} \approx 1.5964542$ ⁠. Numerically, the slope $\frac{dy}{dx}$ changes from $\tan(0.7 - h)$ to $\tan(0.7 + h)$ while $x$ changes by $x(0.7+h) - x(0.7-h)$ ⁠; with $h = 0.0001$ the ratio is $\approx 1.5964542$ ⁠. ✓

Common mistakes

Writing $\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2}$ ⁠. Here that would give $\frac{a(\sin t + t\cos t)}{a(\cos t - t\sin t)}$ ⁠, which is wrong.
Stopping at $\sec^2 t$ ⁠. That is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ ⁠, not $\frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠; the division by $\frac{dx}{dt}$ is missing.
Sign slips in the product rule: $\frac{d}{dt}(t\cos t) = \cos t - t\sin t$ ⁠.

What to take away

Parametric second derivative: $\frac{d^2y}{dx^2} = \dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ ⁠.
Always record where $\frac{dx}{dt} \ne 0$ ⁠; both derivatives need it.

Every fact used, and where it comes from

Fact	Used in	Source
Product rule; derivatives of $\sin$ ⁠, $\cos$ ⁠, $\tan$	Steps 1, 3	Rule (2) and Table 5.3, Section 5.3
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠, $\frac{dx}{dt} \ne 0$	Steps 2, 4	Section 5.6 (Lesson: parametric differentiation)
Chain rule	Step 3	Theorem 4, Section 5.3.1
Second order derivative	Step 3	Section 5.7 (Lesson: second order derivative)
Answer	Result	Answers, Miscellaneous Exercise on Chapter 5, Q17
Numerical values, figure coordinates	Check, figure	Recomputed in the site's verification script

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Miscellaneous · Q17: d2ydx2=sec⁡3tat\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}dx2d2y​=atsec3t​ for x=a(cos⁡t+tsin⁡t)x=a(\cos t+t\sin t)x=a(cost+tsint)⁠, y=a(sin⁡t−tcos⁡t)y=a(\sin t-t\cos t)y=a(sint−tcost)

Basics you need

Solution

Differentiate xxx and yyy with respect to ttt

First derivative: dydx=tan⁡t\frac{dy}{dx} = \tan tdxdy​=tant

Second derivative: differentiate tan⁡t\tan ttant with respect to xxx

Where the answer is valid, and what the curve is

Every fact used, and where it comes from

Miscellaneous · Q17: $\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}$ for $x=a(\cos t+t\sin t)$ ⁠, $y=a(\sin t-t\cos t)$

Differentiate $x$ and $y$ with respect to $t$

First derivative: $\frac{dy}{dx} = \tan t$

Second derivative: differentiate $\tan t$ with respect to $x$