Chapter 5 · Continuity and Differentiability

Exercise 5.6 · Q4: $\frac{dy}{dx}$ when $x=4t$ ⁠, $y=\frac4t$

Exercise 5.6Parametric differentiation · Section 5.6

The question

If $x$ and $y$ are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\dfrac{dy}{dx}$ ⁠.

4. $x = 4t$ ⁠, $y = \dfrac{4}{t}$

Exam answer

\begin{aligned} \frac{dx}{dt} &= 4, \\ \frac{dy}{dt} &= -\frac{4}{t^2} \end{aligned}

\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} = \frac{-4/t^2}{4} \end{aligned}

$\therefore\ \dfrac{dy}{dx} = -\dfrac{1}{t^2}$ $(t \ne 0)$

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

The one idea this problem is built on

The slope is the ratio of the two $t$ -rates:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \quad \text{where } \frac{dx}{dt} \neq 0.

Here $\frac{dx}{dt} = 4$ is never zero, so the only restriction is the one that comes with $y = \frac{4}{t}$ itself: $t \neq 0$ ⁠.

Given, to find, and the plan

Given: $x = 4t$ ⁠, $y = \frac{4}{t}$ ⁠, with parameter $t$ ⁠. Since $\frac{4}{t}$ needs $t \neq 0$ ⁠, the parameter takes every real value except $0$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠, without eliminating $t$ ⁠.
Plan: Differentiate $x$ ⁠; write $y = 4t^{-1}$ and differentiate with the power rule; check $\frac{dx}{dt} \neq 0$ ⁠; divide; picture the curve.

Basics you need

1Power rule, also for negative powers

$\frac{d}{dt}t^n = nt^{n-1}$ holds for negative integers $n$ too (for $t \neq 0$ ⁠). So $\frac{d}{dt}t^{-1} = -t^{-2} = -\frac{1}{t^2}$ ⁠. Basics: derivative rules

2The parametric formula (Section 5.6)

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

3Dividing by a number

Dividing a fraction by $4$ multiplies its denominator by $4$ ⁠: $\frac{-4/t^2}{4} = \frac{-4}{4t^2}$ ⁠.

Solution

Differentiate $x$ with respect to $t$
$\frac{dx}{dt} = \frac{d}{dt}(4t) = 4.$
The parametric formula needs both $t$ -rates. $x = 4t$ grows at the steady rate of 4 units per unit of $t$ ⁠.

$\frac{d}{dt}(ct) = c$ for a constant $c$ ⁠.
Differentiate $y$ with respect to $t$

Write $y = \frac{4}{t} = 4t^{-1}$ ⁠. For $t \neq 0$ ⁠,
$\frac{dy}{dt} = 4\cdot(-1)t^{-2} = -\frac{4}{t^2}.$
Writing $\frac{1}{t}$ as a power $t^{-1}$ lets us use the power rule instead of the quotient rule. (The quotient rule gives the same: $\frac{t\cdot 0 - 4\cdot 1}{t^2} = -\frac{4}{t^2}$ ⁠.)

Power rule with $n = -1$ ⁠. Basics: derivative rules
Check where the formula may be used

$\frac{dx}{dt} = 4 \neq 0$ for every $t$ ⁠. So the formula applies at every $t$ for which the curve is defined, that is every $t \neq 0$ ⁠.

The formula divides by $\frac{dx}{dt}$ ⁠; here that is never a problem. The only excluded value, $t = 0$ ⁠, comes from the equation $y = \frac{4}{t}$ ⁠, not from the formula.

Section 5.6: the formula needs $f'(t) \neq 0$ ⁠. Lesson: parametric differentiation
Divide

For $t \neq 0$ ⁠,
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4/t^2}{4} = -\frac{1}{t^2}.$
The $y$ -rate goes on top. Dividing $-\frac{4}{t^2}$ by $4$ cancels the 4s and leaves $-\frac{1}{t^2}$ ⁠.

Basic 3: dividing a fraction by 4.
What the answer says about the curve

Since $t^2 \gt 0$ for $t \neq 0$ ⁠, the slope $-\frac{1}{t^2}$ is always negative: the curve falls from left to right everywhere. For $t$ close to $0$ the slope is large in size (the curve is steep); for large $|t|$ the slope is close to $0$ (the curve is nearly flat). Points with $t \gt 0$ lie in the first quadrant ( $x \gt 0$ ⁠, $y \gt 0$ ⁠) and points with $t \lt 0$ in the third, so the curve has two separate branches.

The curve $x = 4t$ ⁠, $y = \frac{4}{t}$ (equal scales). At $t = \pm1$ ⁠, the points $(4, 4)$ and $(-4, -4)$ ⁠, the slope is $-1$ ⁠; at $t = 2$ ⁠, the point $(8, 2)$ ⁠, it is $-\frac{1}{4}$ ⁠.

The figure shows both features the formula predicts: the curve always slopes downwards, steeply close to the $y$ -axis, where $|t|$ is small, and gently far out along the $x$ -axis, where $|t|$ is large.

The slope of a curve is the slope of its tangent. Basics: slope at a point

Result

\frac{dy}{dx} = -\frac{1}{t^2} \qquad (t \neq 0).

(Textbook answer: $-\frac{1}{t^2}$ ⁠.)

Check it with numbers

A short chord. At $t = 2$ the point is $(8, 2)$ and the predicted slope is $-\frac14 = -0.25$ ⁠. The points at $t = 1.999$ and $t = 2.001$ are $(7.996000000,\ 2.001000500)$ and $(8.004000000,\ 1.999000500)$ ⁠, so

\frac{\Delta y}{\Delta x} = \frac{-0.002000001}{0.008000000} \approx -0.25000.

Eliminating $t$ ⁠, only as a check. $xy = 4t\cdot\frac{4}{t} = 16$ ⁠, so $y = \frac{16}{x}$ and $\frac{dy}{dx} = -\frac{16}{x^2}$ ⁠. With $x = 4t$ this is $-\frac{16}{16t^2} = -\frac{1}{t^2}$ ⁠. The same answer.

Common mistakes

Sign of the derivative of $\frac{1}{t}$ ⁠. It is $-\frac{1}{t^2}$ ⁠, not $+\frac{1}{t^2}$ and not $\log t$ ⁠.
Dividing the wrong way round. $\frac{4}{-4/t^2} = -t^2$ is $\frac{dx}{dy}$ ⁠, not $\frac{dy}{dx}$ ⁠.
Forgetting $t \neq 0$ ⁠. The curve itself is not defined at $t = 0$ ⁠.
Eliminating $t$ as the solution. $xy = 16$ is a check, not what the question asks for.

What to take away

Differentiate both coordinates with respect to the parameter, then divide.
Restrictions come from two places: the formula ( $\frac{dx}{dt} \neq 0$ ⁠) and the given equations themselves (here $t \neq 0$ ⁠).
Writing $\frac{1}{t}$ as $t^{-1}$ makes the power rule do the work.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dt}t^n = nt^{n-1}$ (here $n = 1$ and $n = -1$ ⁠)	Steps 1, 2	Table 5.3 (Basics: derivative rules)
Quotient rule (alternative in Step 2)	Step 2	Section 5.3, algebra of derivatives
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠, $\frac{dx}{dt} \neq 0$	Steps 3, 4	Section 5.6 (Lesson: parametric differentiation)
$t^2 \gt 0$ for $t \neq 0$	Step 5	Algebra

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Exercise 5.6 · Q4: dydx\frac{dy}{dx}dxdy​ when x=4tx=4tx=4t⁠, y=4ty=\frac4ty=t4​

Basics you need

Solution

Differentiate xxx with respect to ttt

Differentiate yyy with respect to ttt

Check where the formula may be used

Divide

What the answer says about the curve

Every fact used, and where it comes from

Exercise 5.6 · Q4: $\frac{dy}{dx}$ when $x=4t$ ⁠, $y=\frac4t$

Differentiate $x$ with respect to $t$

Differentiate $y$ with respect to $t$