Chapter 5 · Continuity and Differentiability

Exercise 5.6 · Q11: Show $\frac{dy}{dx}=-\frac yx$ for $x=\sqrt{a^{\sin^{-1}t}}$ ⁠, $y=\sqrt{a^{\cos^{-1}t}}$

Exercise 5.6Parametric differentiation · Section 5.6

The question

11. If $x = \sqrt{a^{\sin^{-1}t}}$ ⁠, $y = \sqrt{a^{\cos^{-1}t}}$ ⁠, show that $\dfrac{dy}{dx} = -\dfrac{y}{x}$ ⁠.

Exam answer

Here $x = a^{\frac12\sin^{-1}t}$ ⁠, $y = a^{\frac12\cos^{-1}t}$ ( $a > 0$ ⁠, $a \ne 1$ ⁠, $-1 \lt t \lt 1$ ⁠). By the chain rule,

\begin{aligned} \frac{dx}{dt} &= a^{\frac12\sin^{-1}t}\log a\cdot\frac{1}{2\sqrt{1 - t^2}} \\ &= \frac{x\log a}{2\sqrt{1 - t^2}}, \\ \frac{dy}{dt} &= a^{\frac12\cos^{-1}t}\log a\cdot\frac{-1}{2\sqrt{1 - t^2}} \\ &= -\frac{y\log a}{2\sqrt{1 - t^2}} \end{aligned}

\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \\ &= -\frac{y\log a}{2\sqrt{1 - t^2}}\cdot\frac{2\sqrt{1 - t^2}}{x\log a} \end{aligned}

$\therefore\ \dfrac{dy}{dx} = -\dfrac{y}{x}$ ⁠. Hence proved.

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

The one idea this problem is built on

Treat $t$ as a parameter and use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠. Each coordinate is an exponential, $x = a^{\frac12\sin^{-1}t}$ ⁠, so its derivative is itself times $\log a$ times the derivative of the exponent. Because $\sin^{-1}t$ and $\cos^{-1}t$ have derivatives that differ only in sign, the two rates are $x\cdot k$ and $-y\cdot k$ with the same $k$ ⁠, and the ratio is $-\frac{y}{x}$ ⁠.

Given, to find, and the plan

Given: $x = \sqrt{a^{\sin^{-1}t}}$ ⁠, $y = \sqrt{a^{\cos^{-1}t}}$ ⁠. For a power with a real exponent the base must be positive, so $a \gt 0$ ⁠; and $\sin^{-1}t$ ⁠, $\cos^{-1}t$ need $-1 \leq t \leq 1$ ⁠. We also need $a \neq 1$ ⁠: if $a = 1$ ⁠, then $x = y = 1$ for every $t$ ⁠, a single point with no slope.
To show: $\dfrac{dy}{dx} = -\dfrac{y}{x}$ ⁠.
Plan: Rewrite $x$ and $y$ as powers of $a$ ⁠; differentiate each with respect to $t$ by the chain rule; check that $\frac{dx}{dt} \neq 0$ for $-1 \lt t \lt 1$ ⁠; divide.

Basics you need

1Square root of a power

For $a \gt 0$ ⁠, $\sqrt{a^m} = (a^m)^{\frac12} = a^{\frac{m}{2}}$ ⁠. Basics: exponentials and logarithms

2Derivative of $a^u$

$\frac{d}{du}a^u = a^u\log a$ for a positive constant $a$ (Example 28). With the chain rule, $\frac{d}{dt}a^{u} = a^{u}\log a\cdot\frac{du}{dt}$ ⁠. Lesson: exponential and logarithmic functions

3Derivatives of $\sin^{-1}$ and $\cos^{-1}$

For $-1 \lt t \lt 1$ ⁠:

\begin{aligned} \tfrac{d}{dt}\sin^{-1}t &= \tfrac{1}{\sqrt{1 - t^2}}, \\ \tfrac{d}{dt}\cos^{-1}t &= -\tfrac{1}{\sqrt{1 - t^2}}. \end{aligned}

(Section 5.3.3: Example 24 and the table after it.) At $t = \pm1$ these derivatives do not exist. Lesson: derivatives of inverse trig functions

4The parametric formula (Section 5.6)

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠, provided $\frac{dx}{dt} \neq 0$ ⁠. Lesson: parametric differentiation

Solution

Rewrite $x$ and $y$ as powers of $a$

By Basic 1 (since $a \gt 0$ ⁠),
$x = a^{\frac12\sin^{-1}t}, \qquad y = a^{\frac12\cos^{-1}t}.$
A square root of a power is awkward to differentiate; a single power of $a$ is easy, because we know the derivative of $a^u$ ⁠. Note that $x \gt 0$ and $y \gt 0$ ⁠, since a power of a positive number is positive.

$(a^m)^n = a^{mn}$ for $a \gt 0$ ⁠. Basics: exponentials and logarithms
Differentiate $x$ with respect to $t$

Chain rule with outer function $a^u$ and inner function $u = \frac12\sin^{-1}t$ ⁠, for $-1 \lt t \lt 1$ ⁠:
$\begin{aligned} \frac{dx}{dt} &= a^{\frac12\sin^{-1}t}\cdot\log a\cdot\frac{1}{2\sqrt{1 - t^2}} \\ &= \frac{x\log a}{2\sqrt{1 - t^2}}. \end{aligned}$
The outer function $a^u$ differentiates to $a^u\log a$ (Basic 2), evaluated at the inner function; that is $x\log a$ ⁠. The inner function $\frac12\sin^{-1}t$ differentiates to $\frac12\cdot\frac{1}{\sqrt{1 - t^2}}$ (Basic 3). We must keep $t$ strictly between $-1$ and $1$ ⁠, where this exists.

Chain rule (Theorem 4, Section 5.3.1); Example 28; Section 5.3.3. Lesson: chain rule
Differentiate $y$ with respect to $t$

Same method, inner function $u = \frac12\cos^{-1}t$ ⁠:
$\begin{aligned} \frac{dy}{dt} &= a^{\frac12\cos^{-1}t}\cdot\log a\cdot\frac{-1}{2\sqrt{1 - t^2}} \\ &= -\frac{y\log a}{2\sqrt{1 - t^2}}. \end{aligned}$
Everything is the same as in Step 2 except the sign of the inner derivative, because $\frac{d}{dt}\cos^{-1}t = -\frac{1}{\sqrt{1 - t^2}}$ ⁠.

Basic 3, second line.
Check that the formula may be used

For $-1 \lt t \lt 1$ ⁠: $x \gt 0$ (Step 1), $\sqrt{1 - t^2} \gt 0$ ⁠, and $\log a \neq 0$ because $a \neq 1$ ⁠. So
$\frac{dx}{dt} = \frac{x\log a}{2\sqrt{1 - t^2}} \neq 0.$
The formula divides by $\frac{dx}{dt}$ ⁠, so we must be sure it is never zero. Each factor is nonzero, so the product is nonzero. This is where the condition $a \neq 1$ is really used: $\log 1 = 0$ ⁠.

$\log a = 0 \iff a = 1$ ⁠. Basics: exponentials and logarithms
Divide

For $-1 \lt t \lt 1$ ⁠,
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\dfrac{y\log a}{2\sqrt{1 - t^2}}}{\dfrac{x\log a}{2\sqrt{1 - t^2}}} = -\frac{y}{x}.$
The common factor $\frac{\log a}{2\sqrt{1 - t^2}}$ is not zero (Step 4), so it cancels. This is what we had to show.

The two rates are the same expression multiplied by $x$ and by $-y$ ⁠, so their ratio is $-\frac{y}{x}$ ⁠. Unlike Exercises 1 to 10, the required answer is in terms of $x$ and $y$ ⁠, and since $x \gt 0$ it is always defined.

Parametric formula, Section 5.6. Lesson: parametric differentiation
Picture the curve

As $t$ runs from $-1$ to $1$ ⁠, $\sin^{-1}t$ runs from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ and $\cos^{-1}t$ from $\pi$ down to $0$ ⁠, so for $a \gt 1$ the point moves from $\big(a^{-\frac{\pi}{4}}, a^{\frac{\pi}{2}}\big)$ down to $\big(a^{\frac{\pi}{4}}, 1\big)$ ⁠. At the two ends the derivatives of $\sin^{-1}t$ and $\cos^{-1}t$ do not exist, so we have shown $\frac{dy}{dx} = -\frac{y}{x}$ at every point except the two end points.

The curve for $a = 2$ (equal scales): an arc of the dashed curve $xy = 2^{\frac{\pi}{4}}$ ⁠, from $t = -1$ to $t = 1$ ⁠. At $t = 0$ the point is $\big(1, 2^{\frac{\pi}{4}}\big)$ ⁠; at $t = \frac{1}{\sqrt2}$ ⁠, where $\sin^{-1}t = \cos^{-1}t = \frac{\pi}{4}$ ⁠, it is $\big(2^{\frac{\pi}{8}}, 2^{\frac{\pi}{8}}\big)$ and the slope is $-\frac{y}{x} = -1$ ⁠.

The picture shows a piece of a rectangular hyperbola $xy = \text{constant}$ ⁠, whose slope is always $-\frac{y}{x}$ ⁠. The check below explains why the product $xy$ is constant.

Principal ranges: $\sin^{-1}t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ ⁠, $\cos^{-1}t \in [0, \pi]$ (Chapter 2, Section 2.2).

Result

\frac{dy}{dx} = -\frac{y}{x},

at every point with $-1 \lt t \lt 1$ (where $a \gt 0$ ⁠, $a \neq 1$ ⁠), as was to be shown.

Check it with numbers

A short chord. Take $a = 2$ ⁠, $t = 0.5$ ⁠: the point is $(1.1989732,\ 1.4375367)$ and the predicted slope is $-\frac{y}{x} \approx -1.1989732$ ⁠. The points at $t = 0.499$ and $t = 0.501$ are $(1.198493621,\ 1.438111897)$ and $(1.199453252,\ 1.436961325)$ ⁠, so

\frac{\Delta y}{\Delta x} = \frac{-0.001150572}{0.000959632} \approx -1.19897.

A second route (eliminating $t$ ⁠, only as a check). Let $\alpha = \sin^{-1}t$ ⁠, so $-\frac{\pi}{2} \leq \alpha \leq \frac{\pi}{2}$ and $\sin\alpha = t$ ⁠. Then $\frac{\pi}{2} - \alpha$ lies in $[0, \pi]$ ⁠, the principal range of $\cos^{-1}$ ⁠, and $\cos(\frac{\pi}{2} - \alpha) = \sin\alpha = t$ ⁠, so $\cos^{-1}t = \frac{\pi}{2} - \alpha$ ⁠. Hence $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$ and

xy = a^{\frac12(\sin^{-1}t + \cos^{-1}t)} = a^{\frac{\pi}{4}},

a constant. Differentiating $xy = a^{\frac{\pi}{4}}$ with respect to $x$ by the product rule, $y + x\frac{dy}{dx} = 0$ ⁠, so $\frac{dy}{dx} = -\frac{y}{x}$ ⁠. The same answer. (At $t = 0.5$ with $a = 2$ ⁠: $xy \approx 1.7235679$ ⁠, which is $2^{\frac{\pi}{4}}$ ⁠.)

Common mistakes

Treating $a^u$ like $u^n$ ⁠. The variable is in the exponent: $\frac{d}{du}a^u = a^u\log a$ ⁠, not $ua^{u-1}$ ⁠.
Forgetting the $\frac12$ ⁠. The square root halves the exponent; the $\frac12$ appears in both rates and cancels, but it must be there.
Sign of $\frac{d}{dt}\cos^{-1}t$ ⁠. It is $-\frac{1}{\sqrt{1 - t^2}}$ ⁠; this minus sign is the minus sign in the answer.
Ignoring the conditions. $a \gt 0$ for the powers, $a \neq 1$ so that $\frac{dx}{dt} \neq 0$ ⁠, and $-1 \lt t \lt 1$ for the derivatives of $\sin^{-1}$ and $\cos^{-1}$ ⁠.

What to take away

Rewrite roots of powers as single powers before differentiating.
$\frac{d}{dt}a^{u} = a^{u}\log a\,\frac{du}{dt}$ ⁠: the function reappears in its own derivative, which is why the answer can be written with $x$ and $y$ ⁠.
$\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$ explains the result: $xy$ is constant, so the curve is part of a rectangular hyperbola.

Every fact used, and where it comes from

Fact	Used in	Source
$\sqrt{a^m} = a^{\frac{m}{2}}$ for $a \gt 0$ ⁠; $\log a = 0 \iff a = 1$	Steps 1, 4	Section 5.4 (Basics: exponentials and logarithms)
$\frac{d}{du}a^u = a^u\log a$ ⁠, $a \gt 0$	Steps 2, 3	Example 28, Section 5.5 (Lesson: exponential and logarithmic functions)
Chain rule	Steps 2, 3	Theorem 4, Section 5.3.1 (Lesson: chain rule)
$\frac{d}{dt}\sin^{-1}t = \frac{1}{\sqrt{1 - t^2}}$ ⁠, $\frac{d}{dt}\cos^{-1}t = -\frac{1}{\sqrt{1 - t^2}}$ on $(-1, 1)$	Steps 2, 3	Example 24 and the table in Section 5.3.3 (Lesson: derivatives of inverse trig functions)
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠, $\frac{dx}{dt} \neq 0$	Steps 4, 5	Section 5.6 (Lesson: parametric differentiation)
Principal ranges of $\sin^{-1}$ and $\cos^{-1}$	Step 6, check	Chapter 2, Section 2.2
$\cos(\frac{\pi}{2} - \alpha) = \sin\alpha$ ⁠; product rule	Check	Class XI, Trigonometric Functions; Section 5.3

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Exercise 5.6 · Q11: Show dydx=−yx\frac{dy}{dx}=-\frac yxdxdy​=−xy​ for x=asin⁡−1tx=\sqrt{a^{\sin^{-1}t}}x=asin−1t​⁠, y=acos⁡−1ty=\sqrt{a^{\cos^{-1}t}}y=acos−1t​

Basics you need

Solution

Rewrite xxx and yyy as powers of aaa

Differentiate xxx with respect to ttt

Differentiate yyy with respect to ttt

Check that the formula may be used

Divide

Picture the curve

Every fact used, and where it comes from

Exercise 5.6 · Q11: Show $\frac{dy}{dx}=-\frac yx$ for $x=\sqrt{a^{\sin^{-1}t}}$ ⁠, $y=\sqrt{a^{\cos^{-1}t}}$

Rewrite $x$ and $y$ as powers of $a$

Differentiate $x$ with respect to $t$

Differentiate $y$ with respect to $t$