Chapter 5 · Continuity and Differentiability

Exercise 5.6 · Q1: $\frac{dy}{dx}$ when $x=2at^2$ ⁠, $y=at^4$

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 $\dfrac{dy}{dx}$ ⁠.

1. $x = 2at^2$ ⁠, $y = at^4$

Exam answer

\begin{aligned} \frac{dx}{dt} &= 4at, \\ \frac{dy}{dt} &= 4at^3 \end{aligned}

\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} = \frac{4at^3}{4at} \end{aligned}

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

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

The one idea this problem is built on

When $x$ and $y$ are both given in terms of a third variable $t$ ⁠, the slope $\frac{dy}{dx}$ is the ratio of how fast $y$ changes with $t$ to how fast $x$ changes with $t$ ⁠:

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

So we never need $y$ as a formula in $x$ ⁠. Here the ratio is $\frac{4at^3}{4at}$ ⁠, which is $t^2$ ⁠.

Given, to find, and the plan

Given: $x = 2at^2$ ⁠, $y = at^4$ ⁠, where $t$ is the parameter and $a$ is a constant. We take $a \neq 0$ ⁠: if $a = 0$ ⁠, then $x = y = 0$ for every $t$ ⁠, a single point with no slope to find.
To find: $\dfrac{dy}{dx}$ ⁠, without eliminating $t$ ⁠.
Plan: (1) Differentiate $x$ with respect to $t$ ⁠. (2) Differentiate $y$ with respect to $t$ ⁠. (3) Find the values of $t$ where $\frac{dx}{dt} = 0$ ⁠, which the formula excludes. (4) Divide and simplify. (5) Look at the excluded value on a picture of the curve.

Basics you need

1Power rule

$\frac{d}{dt}t^n = nt^{n-1}$ ⁠, and a constant factor stays in front: $\frac{d}{dt}(ct^n) = cnt^{n-1}$ (Table 5.3). Basics: derivative rules

2The parametric formula (Section 5.6)

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

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

Lesson: parametric differentiation

3Cancelling a common factor

$\frac{pq}{pr} = \frac{q}{r}$ is allowed only when $p \neq 0$ ⁠. Basics: solving for dy/dx

Solution

Differentiate $x$ with respect to $t$
$\frac{dx}{dt} = \frac{d}{dt}\left(2at^2\right) = 2a\cdot 2t = 4at.$
The parametric formula needs the rate of change of $x$ with respect to the parameter. $2a$ is a constant, so it stays in front, and the power rule turns $t^2$ into $2t$ ⁠.

Power rule with $n = 2$ ⁠: $\frac{d}{dt}t^2 = 2t$ (Table 5.3). Basics: derivative rules
Differentiate $y$ with respect to $t$
$\frac{dy}{dt} = \frac{d}{dt}\left(at^4\right) = a\cdot 4t^3 = 4at^3.$
This is the rate of change of $y$ with respect to the same parameter. Again the constant $a$ stays in front and the power rule lowers the power by one.

Power rule with $n = 4$ ⁠: $\frac{d}{dt}t^4 = 4t^3$ ⁠.
Find where the formula may be used

The formula needs $\frac{dx}{dt} \neq 0$ ⁠. Since $a \neq 0$ ⁠,
$4at \neq 0 \iff t \neq 0.$
So we work with $t \neq 0$ ⁠.

The formula divides by $\frac{dx}{dt}$ ⁠, and division by zero is meaningless. Finding the bad values first means we never divide by zero by accident. A product $4at$ is zero only when one of its factors is zero, and $4 \neq 0$ ⁠, $a \neq 0$ ⁠, so only $t = 0$ is excluded.

Section 5.6: $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ "provided $f'(t) \neq 0$ ". Lesson: parametric differentiation
Divide and simplify

For $t \neq 0$ ⁠,
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at^3}{4at} = t^2.$
The common factor $4at$ is not zero (Step 3), so cancelling it is allowed: $\frac{4at\cdot t^2}{4at} = t^2$ ⁠.

The $y$ -rate goes on top and the $x$ -rate underneath, because we want the change in $y$ per unit change in $x$ ⁠. Notice that $a$ has disappeared: the slope at the point with parameter $t$ is $t^2$ whatever the (nonzero) value of $a$ ⁠.

Basic 3: cancel only factors that are not zero.
Look at the curve and at $t = 0$

Since $x = 2at^2$ and $y = at^4$ use only even powers, $t$ and $-t$ give the same point. Reassuringly, they also give the same slope, $t^2 = (-t)^2$ ⁠: a slope belongs to a point of the curve, so it must not depend on which of the two parameter values we use.

At $t = 0$ both rates are zero, $\frac{dx}{dt} = 4a\cdot 0 = 0$ and $\frac{dy}{dt} = 4a\cdot 0 = 0$ ⁠, and the point is the origin. The formula says nothing there. Near $t = 0$ the slope $t^2$ is close to $0$ ⁠, so the curve flattens out as it comes into the origin. For $a \gt 0$ ⁠, $x = 2at^2 \geq 0$ ⁠, so the origin is the left end of the curve.

The curve for $a = 1$ (equal scales on both axes). At $(2, 1)$ ⁠, where $t = \pm 1$ ⁠, the slope is $t^2 = 1$ ⁠; at $(4, 4)$ ⁠, where $t = \pm\sqrt2$ ⁠, it is $t^2 = 2$ ⁠. At $t = 0$ both $t$ -rates vanish.

A picture shows what the formula means: as $|t|$ grows, the point moves out along the curve and the slope $t^2$ grows, so the curve gets steeper. It also shows why $t = 0$ is special: the moving point slows to a stop at the origin and turns back.

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

Result

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

(Textbook answer: $t^2$ ⁠.)

Check it with numbers

A short chord. Take $a = 1$ and $t = 1.5$ ⁠: the point is $(4.5000000,\ 5.0625000)$ and the predicted slope is $1.5^2 = 2.25$ ⁠. The points at $t = 1.499$ and $t = 1.501$ are $(4.494002000,\ 5.049013494)$ and $(4.506002000,\ 5.076013506)$ ⁠, so

\frac{\Delta y}{\Delta x} = \frac{0.027000012}{0.012000000} \approx 2.25000.

Eliminating $t$ ⁠, only as a check. From $x = 2at^2$ ⁠, $t^2 = \frac{x}{2a}$ ⁠, so $y = a(t^2)^2 = \frac{x^2}{4a}$ ⁠. Then $\frac{dy}{dx} = \frac{2x}{4a} = \frac{x}{2a}$ ⁠, and putting $x = 2at^2$ back gives $\frac{2at^2}{2a} = t^2$ ⁠. The same answer.

Common mistakes

Dividing the wrong way round. $\frac{dx/dt}{dy/dt} = \frac{1}{t^2}$ is $\frac{dx}{dy}$ ⁠, not $\frac{dy}{dx}$ ⁠. The $y$ -rate goes on top.
Eliminating the parameter. The question says "without eliminating the parameter". Turning the equations into $y = \frac{x^2}{4a}$ is fine as a check, but not as the solution.
Forgetting the condition. The formula needs $\frac{dx}{dt} \neq 0$ ⁠, so state $t \neq 0$ ⁠.
Expecting the answer in $x$ ⁠. In parametric problems the answer is normally given in terms of the parameter, as the book's Note in Section 5.6 says.

What to take away

Parametric slope: differentiate both coordinates with respect to the parameter, then divide, $y$ -rate over $x$ -rate.
Always find where $\frac{dx}{dt} = 0$ before dividing, and cancel only factors that are not zero.
If two parameter values give the same point, they must give the same slope; this is a quick sense check on the answer.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dt}t^n = nt^{n-1}$ ⁠; constant factors stay in front	Steps 1, 2	Table 5.3, Section 5.3 (Basics: derivative rules)
$\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ ⁠, $f'(t) \neq 0$	Steps 3, 4	Section 5.6 (Lesson: parametric differentiation)
A product is zero only if a factor is zero	Step 3	Algebra
Slope of a curve = slope of its tangent	Step 5	Class XI, Limits and Derivatives (the derivative as the slope of the tangent) (Basics: slope at a point)
$\frac{dy}{dx}$ is expressed in terms of the parameter only	Common mistake 4	Note after Example 33, Section 5.6

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Exercise 5.6 · Q1: dydx\frac{dy}{dx}dxdy​ when x=2at2x=2at^2x=2at2⁠, y=at4y=at^4y=at4

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

Look at the curve and at t=0t = 0t=0

Every fact used, and where it comes from

Exercise 5.6 · Q1: $\frac{dy}{dx}$ when $x=2at^2$ ⁠, $y=at^4$

Differentiate $x$ with respect to $t$

Differentiate $y$ with respect to $t$

Look at the curve and at $t = 0$