Chapter 5 · Continuity and Differentiability

Example 32: $\frac{dy}{dx}$ when $x=at^2$ ⁠, $y=2at$

ExamplesParametric differentiation · Section 5.6

The question

Find $\dfrac{dy}{dx}$ ⁠, if $x = at^2$ ⁠, $y = 2at$ ⁠.

Exam answer

Here $x = at^2$ and $y = 2at$ ⁠.

Differentiating w.r.t. $t$ ⁠,

\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at}

$\therefore\ \dfrac{dy}{dx} = \dfrac1t$ $(t \ne 0)$ ⁠.

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

The one idea this problem is built on

Both $x$ and $y$ are given in terms of the parameter $t$ ⁠, so the slope comes from the two $t$ -rates:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt},

provided $\frac{dx}{dt} \neq 0$ ⁠. The points $(at^2, 2at)$ trace the parabola $y^2 = 4ax$ ⁠, and the answer $\frac{1}{t}$ tells us how steep the parabola is at the point with parameter $t$ ⁠.

Given, to find, and the plan

Given: $x = at^2$ ⁠, $y = 2at$ ⁠, with $a$ a nonzero constant and $t$ the parameter. (If $a = 0$ ⁠, every $t$ gives the point $(0,0)$ and there is no curve.)
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ ⁠, find where $\frac{dx}{dt} = 0$ ⁠, divide, and simplify. Then identify the curve and check the answer by implicit differentiation.

Basics you need

1Power rule and constant multiples

$\dfrac{d}{dt}(t^n) = n\,t^{n-1}$ (Table 5.3), so $\frac{d}{dt}(t^2) = 2t$ and $\frac{d}{dt}(t) = 1$ ⁠. A constant factor such as $a$ or $2a$ stays in front. Basics: derivative rules

2The parametric formula (Section 5.6)

If $x = f(t)$ ⁠, $y = g(t)$ ⁠, then by the chain rule $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$ ⁠, so

\frac{dy}{dx} = \frac{g'(t)}{f'(t)}, \ \text{ if } f'(t) \neq 0.

Lesson: parametric differentiation

3Cancelling a common factor

$\frac{k\,u}{k\,v} = \frac{u}{v}$ only when $k \neq 0$ ⁠. Before cancelling, check that what you cancel is not zero. Basics: solving for dy/dx

4Slope and vertical tangents

$\frac{dy}{dx}$ is the slope of the tangent. Where the tangent is vertical there is no slope, so $\frac{dy}{dx}$ does not exist. Basics: slope at a point

Solution

Differentiate $x$ with respect to $t$
$\frac{dx}{dt} = a\,\frac{d}{dt}(t^2) = a \cdot 2t = 2at.$
The parametric formula (Basic 2) is built from the two $t$ -rates, so we find each one first. $a$ is a constant, so it stays in front while $t^2$ is differentiated.

Power rule with $n = 2$ ⁠: $\frac{d}{dt}t^2 = 2t$ ⁠. Basics: derivative rules
Differentiate $y$ with respect to $t$
$\frac{dy}{dt} = 2a\,\frac{d}{dt}(t) = 2a.$
$y$ grows at the steady rate $2a$ per unit of $t$ ⁠: $y$ is a linear function of $t$ ⁠.

Power rule with $n = 1$ ⁠: $\frac{d}{dt}t = 1$ ⁠.
Find where the formula may be used

We need $\frac{dx}{dt} \neq 0$ ⁠. Since $a \neq 0$ ⁠,
$2at \neq 0 \iff t \neq 0.$
The excluded value $t = 0$ gives the point $(0, 0)$ ⁠, the vertex of the parabola.

At $t = 0$ the $x$ -coordinate $x = at^2$ has its extreme value (a minimum when $a \gt 0$ ⁠, a maximum when $a \lt 0$ ⁠), so $x$ momentarily stops changing while $y$ still changes at rate $2a \neq 0$ ⁠. The curve is heading straight up or straight down there: the tangent at the vertex is the vertical $y$ -axis, which has no slope (Basic 4).

A product is zero only when one of its factors is zero: $2at = 0$ forces $t = 0$ because $2a \neq 0$ ⁠.
Divide and simplify

For $t \neq 0$ ⁠,
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}.$
So $\dfrac{dy}{dx} = \dfrac{1}{t}$ for every $t \neq 0$ ⁠.

We cancel the common factor $2a$ ⁠, which is allowed because $a \neq 0$ (Basic 3). The result does not depend on $a$ at all: the parabolas $y^2 = 4ax$ for different $a$ are scaled copies of one another, and scaling $x$ and $y$ by the same factor does not change slopes.

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

Eliminate $t$ to see which curve the points lie on:
$y^2 = 4a^2t^2 = 4a\,(at^2) = 4ax.$
Conversely, any point $(x, y)$ with $y^2 = 4ax$ comes from $t = \frac{y}{2a}$ ⁠, since then $2at = y$ and $at^2 = \frac{y^2}{4a} = x$ ⁠. So the curve is the whole parabola $y^2 = 4ax$ ⁠, and each point has exactly one parameter value $t = \frac{y}{2a}$ ⁠.

The parabola $y^2 = 4ax$ for $a = 1$ (so $x = t^2$ ⁠, $y = 2t$ ⁠). At $t = 1$ ⁠, the point $(1, 2)$ ⁠, the slope is $\frac11 = 1$ ⁠; at $t = -2$ ⁠, the point $(4, -4)$ ⁠, it is $-\frac12$ ⁠. At the vertex ( $t = 0$ ⁠) the tangent is the $y$ -axis, which is vertical.

The picture explains the answer. On the upper half ( $t \gt 0$ when $a \gt 0$ ⁠) the slope $\frac1t$ is positive and gets smaller as $t$ grows: the parabola flattens out far from the vertex. Near the vertex $t$ is small, $\frac1t$ is huge, and the curve is nearly vertical.

Eliminating the parameter: square $y = 2at$ and use $x = at^2$ to replace $at^2$ ⁠.
Check by implicit differentiation

Differentiate $y^2 = 4ax$ with respect to $x$ ⁠:
$2y\,\frac{dy}{dx} = 4a, \ \text{ so } \ \frac{dy}{dx} = \frac{2a}{y}$
wherever $y \neq 0$ ⁠. With $y = 2at$ ⁠: $\dfrac{2a}{2at} = \dfrac1t$ ⁠. The same answer, and $y \neq 0$ is the same condition as $t \neq 0$ ⁠.

A second, independent method guards against slips. It also gives the slope in terms of $y$ ⁠: $\frac{dy}{dx} = \frac{2a}{y}$ ⁠, which is the form often used for tangents to a parabola.

$\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ by the chain rule. Lesson: implicit differentiation

Result

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

Equivalently $\frac{2a}{y}$ ⁠. At $t = 0$ ⁠, the vertex $(0, 0)$ ⁠, the tangent is vertical and $\frac{dy}{dx}$ does not exist. (Book's answer: $\frac1t$ ⁠.)

Check it with numbers

Take $a = 3$ and $t = 2$ ⁠. The point is $(12, 12)$ and the predicted slope is $\frac12 = 0.5$ ⁠.

At $t = 1.999$ ⁠: $(11.988003,\ 11.994)$ ⁠. At $t = 2.001$ ⁠: $(12.012003,\ 12.006)$ ⁠. The chord has

\frac{\Delta y}{\Delta x} = \frac{0.012}{0.024} = 0.5.

In fact the chord between any two parameters $t_1 \neq t_2$ (with $t_1 + t_2 \neq 0$ ⁠) has slope

\frac{2a(t_2 - t_1)}{a(t_2^2 - t_1^2)} = \frac{2}{t_1 + t_2},

and as $t_2$ approaches $t_1$ this approaches $\frac{2}{2t_1} = \frac{1}{t_1}$ ⁠: the derivative found from first principles. Also $\frac{2a}{y} = \frac{6}{12} = 0.5$ ⁠.

Common mistakes

Inverting the fraction. $\frac{dx/dt}{dy/dt} = t$ is $\frac{dx}{dy}$ ⁠, not $\frac{dy}{dx}$ ⁠.
Forgetting $t \neq 0$ ⁠. The formula divides by $2at$ ⁠; at $t = 0$ the tangent is vertical and there is no slope.
Treating $a$ as a variable. $a$ is a fixed constant, so $\frac{d}{dt}(2at) = 2a$ ⁠, not $2a + 2t$ ⁠.

What to take away

$x = at^2$ ⁠, $y = 2at$ is the standard parametrisation of the parabola $y^2 = 4ax$ ⁠, and its slope at parameter $t$ is $\frac1t$ ⁠.
Parametric slope $= \frac{\text{rate of } y}{\text{rate of } x}$ ⁠, valid only where the rate of $x$ is not zero.
Where $\frac{dx}{dt} = 0$ but $\frac{dy}{dt} \neq 0$ ⁠, the tangent is vertical.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dt}t^n = n\,t^{n-1}$	Steps 1, 2	Table 5.3, Section 5.3
$\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ ⁠, $f'(t) \neq 0$	Steps 3, 4	Section 5.6 (Lesson: parametric differentiation)
Cancelling needs a nonzero factor	Step 4	Algebra
$y^2 = 4ax$ is a parabola with vertex at the origin	Step 5	Class XI, Conic Sections
Chain rule; implicit differentiation	Step 6	Theorem 4 and Section 5.3.2 (Lesson: implicit differentiation)

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Example 32: dydx\frac{dy}{dx}dxdy​ when x=at2x=at^2x=at2⁠, y=2aty=2aty=2at

Basics you need

Solution

Differentiate xxx with respect to ttt

Differentiate yyy with respect to ttt

Find where the formula may be used

Divide and simplify

Identify the curve

Check by implicit differentiation

Every fact used, and where it comes from

Example 32: $\frac{dy}{dx}$ when $x=at^2$ ⁠, $y=2at$

Differentiate $x$ with respect to $t$

Differentiate $y$ with respect to $t$