Year 2 Pure · Chapter 13

Implicit and Parametric Differentiation

Specification: Edexcel Pure 7.1, 7.3–7.5 — Differentiation · AQA section G1, G3, G4, G5

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In this chapter

1. Parametric differentiation
2. Implicit differentiation
3. Connected rates of change
4. Concave and convex curves
Practice

Key facts — this chapter

Parametric: if $x$ and $y$ are functions of $t$ , then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
Implicit: differentiate every term with respect to $x$ . A term in $y$ gets an extra factor $\dfrac{dy}{dx}$ by the chain rule, so $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$ .
A product such as $xy$ needs the product rule: $\dfrac{d}{dx}(xy) = x\dfrac{dy}{dx} + y$ .
Connected rates: link the rate you want to the one you know with the chain rule, such as $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ .
A curve is convex where $f''(x) \ge 0$ and concave where $f''(x) \le 0$ . A point of inflection is where it changes from one to the other, so $f''(x)$ changes sign.

Not every curve comes as $y = f(x)$ . Some are parametric, some are given by an equation like $x^2 + y^2 = 25$ that mixes $x$ and $y$ , and some quantities change with time together. The chain rule handles all three. The last section uses the second derivative to describe which way a curve bends.

1. Parametric differentiation

Key result

If $x$ and $y$ are both functions of a parameter $t$ , then

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

wherever $\frac{dx}{dt} \ne 0$ . The answer is in terms of $t$ .

Worked example 13.1 · A tangent to a parametric curve

A curve has parametric equations $x = t^2$ , $y = t^3 - 3t$ . Find the equation of the tangent at the point where $t = 2$ .

Differentiate each with respect to $t$ and divide:
$\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$
At $t = 2$ the gradient is $\tfrac{9}{4}$ , and the point is $(4, 2)$ .
The tangent is
$\begin{aligned} y - 2 &= \tfrac{9}{4}(x - 4) \\ 4y &= 9x - 28 \end{aligned}$

The curve x = t squared, y = t cubed minus 3t, which loops through the origin, with the tangent at the point (4, 2), where t = 2, drawn in orange — x = t², y = t³ − 3t, with its tangent at t = 2.

Worked example 13.2 · A normal to an ellipse

A curve has parametric equations $x = 2\cos\theta$ , $y = 3\sin\theta$ . Find the equation of the normal at the point where $\theta = \tfrac{\pi}{4}$ .

$\frac{dx}{d\theta} = -2\sin\theta$ and $\frac{dy}{d\theta} = 3\cos\theta$ , so
$\frac{dy}{dx} = \frac{3\cos\theta}{-2\sin\theta} = -\tfrac{3}{2}\cot\theta$
At $\theta = \tfrac{\pi}{4}$ the tangent’s gradient is $-\tfrac{3}{2}$ , so the normal’s is $\tfrac{2}{3}$ . The point is $\left(\sqrt2, \tfrac{3\sqrt2}{2}\right)$ .
The normal is $y - \tfrac{3\sqrt2}{2} = \tfrac{2}{3}(x - \sqrt2)$ . Multiplying by $6$ and rearranging gives $6y = 4x + 5\sqrt2$ .

2. Implicit differentiation

An equation like $x^2 + y^2 = 25$ defines $y$ in terms of $x$ without solving for it: the relationship is implicit. You can still differentiate every term with respect to $x$ .

Key result

To differentiate a function of $y$ with respect to $x$ , differentiate it with respect to $y$ , then multiply by $\frac{dy}{dx}$ (the chain rule):

\frac{d}{dx}\big(f(y)\big) = f'(y)\frac{dy}{dx}

Terms that mix $x$ and $y$ , such as $xy$ , need the product rule as well.

Worked example 13.3 · A tangent to a circle

Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$ .

Differentiate every term with respect to $x$ :
$\begin{aligned} 2x + 2y\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -\frac{x}{y} \end{aligned}$
At $(3, 4)$ the gradient is $-\tfrac{3}{4}$ . The tangent is $y - 4 = -\tfrac{3}{4}(x - 3)$ , which is $3x + 4y = 25$ .

The circle x squared plus y squared = 25 with the radius to (3, 4) drawn, of gradient 4 over 3, and the tangent there, of gradient minus 3 over 4, at right angles to it — The tangent at (3, 4) has gradient −x/y = −3/4: at right angles to the radius, as it should be.

Worked example 13.4 · A product term

A curve has equation $x^3 + xy + y^3 = 11$ . Find the gradient of the curve at the point $(1, 2)$ .

Differentiate term by term, using the product rule on $xy$ :
$3x^2 + x\frac{dy}{dx} + y + 3y^2\frac{dy}{dx} = 0$
Collect the $\frac{dy}{dx}$ terms:
$\frac{dy}{dx} = -\frac{3x^2 + y}{x + 3y^2}$
At $(1, 2)$ the gradient is $-\frac{3 + 2}{1 + 12} = -\frac{5}{13}$ .

Worked example 13.5 · Stationary points of an implicit curve

Find the stationary points of the curve $x^2 + xy + y^2 = 12$ .

Differentiate and rearrange:
$\begin{aligned} &2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0 \\ &\frac{dy}{dx} = -\frac{2x + y}{x + 2y} \end{aligned}$
The gradient is $0$ when $2x + y = 0$ , that is $y = -2x$ . Substitute into the curve’s equation:
$x^2 - 2x^2 + 4x^2 = 12 \;\Rightarrow\; x = \pm 2$
The stationary points are $(2, -4)$ and $(-2, 4)$ .

Another route to the exponential derivative

If $y = a^x$ , then $\ln y = x\ln a$ . Differentiating implicitly, $\frac{1}{y}\frac{dy}{dx} = \ln a$ , so $\frac{dy}{dx} = y\ln a = a^x\ln a$ , as in Chapter 12.

3. Connected rates of change

When two quantities are linked, their rates of change are linked too. The chain rule connects them.

Key result

If $V$ is a function of $r$ , and $r$ changes with time $t$ ,

\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}

Write down the rate you know and the rate you want. Then find the link between the variables and differentiate it.

Worked example 13.6 · A balloon

A spherical balloon is inflated at a rate of $50$ cm³ per second. Find the rate at which its radius is increasing when the radius is $10$ cm.

You know $\frac{dV}{dt} = 50$ and want $\frac{dr}{dt}$ . The link is $V = \tfrac{4}{3}\pi r^3$ , so $\frac{dV}{dr} = 4\pi r^2$ , which is $400\pi$ when $r = 10$ .
So
$\frac{dr}{dt} = \frac{dV/dt}{dV/dr} = \frac{50}{400\pi} = \frac{1}{8\pi}$
That is about $0.0398$ cm per second.

4. Concave and convex curves

Key result

A curve is convex where $f''(x) \ge 0$ : its gradient is increasing, and it bends upwards like a bowl.
A curve is concave where $f''(x) \le 0$ : its gradient is decreasing, and it bends downwards like an arch.
A point of inflection is where the curve changes from concave to convex, or the other way. There $f''(x) = 0$ and changes sign.

The cubic y = x cubed minus 6 x squared plus 9 x plus 1, concave for x less than 2 and drawn in orange there, convex for x greater than 2 and drawn in green, with the point of inflection (2, 3) marked and the tangent there crossing the curve — y = x³ − 6x² + 9x + 1 is concave for x < 2 and convex for x > 2. At the point of inflection the tangent crosses the curve.

Green is convex (f″ > 0), orange is concave (f″ < 0). Step x along and watch the tangent: it stays below a convex curve and above a concave one, and at a point of inflection it crosses over.

Worked example 13.7 · Finding a point of inflection

For $f(x) = x^3 - 6x^2 + 9x + 1$ , find where the curve is concave, where it is convex, and its point of inflection.

$f'(x) = 3x^2 - 12x + 9$ and $f''(x) = 6x - 12$ .
$f''(x) \le 0$ for $x \le 2$ , so the curve is concave there. $f''(x) \ge 0$ for $x \ge 2$ , so it is convex there.
$f''(x)$ changes sign at $x = 2$ , and $f(2) = 8 - 24 + 18 + 1 = 3$ . So $(2, 3)$ is a point of inflection.

Worked example 13.8 · Convex everywhere

Show that $f(x) = x\ln x$ is convex for all $x > 0$ .

By the product rule, $f'(x) = \ln x + 1$ , so $f''(x) = \frac{1}{x}$ . This is positive for every $x > 0$ , so the curve is convex there.

Common mistake

$f''(x) = 0$ is not enough on its own. For $y = x^4$ , $f''(x) = 12x^2$ is $0$ at $x = 0$ , but it is positive on both sides, so the curve is convex throughout and has no point of inflection. Always check that $f''(x)$ changes sign.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [2 marks]

A curve has parametric equations $x = 3t^2$ , $y = 2t^3$ . Find $\dfrac{dy}{dx}$ in terms of $t$ .

Show solution

\frac{dy}{dx} = \frac{6t^2}{6t} = t

2 [2 marks]

Find $\dfrac{dy}{dx}$ for the curve $y^2 = 4x$ , in terms of $y$ .

Show solution

Differentiating implicitly, $2y\dfrac{dy}{dx} = 4$ , so $\dfrac{dy}{dx} = \dfrac{2}{y}$ .

3 [2 marks]

Find $\dfrac{dy}{dx}$ for the curve $x^2y = 8$ , in terms of $x$ and $y$ .

Show solution

By the product rule, $2xy + x^2\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{2y}{x}$ .

4 [2 marks]

Find the values of $x$ for which the curve $y = x^3 - 3x$ is concave.

Show solution

$\dfrac{d^2y}{dx^2} = 6x$ , which is $\le 0$ when $x \le 0$ . So the curve is concave for $x \le 0$ .

Standard

5 [5 marks]

A curve has parametric equations $x = t + \dfrac{1}{t}$ , $y = t - \dfrac{1}{t}$ , $t > 0$ .

Find $\dfrac{dy}{dx}$ in terms of $t$ . [2]
Find the equation of the tangent at the point where $t = 2$ . [3]

Show solution

(a) $\dfrac{dx}{dt} = 1 - \dfrac{1}{t^2}$ and $\dfrac{dy}{dt} = 1 + \dfrac{1}{t^2}$ . Multiplying top and bottom by $t^2$ :

\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}

(b) At $t = 2$ the gradient is $\tfrac{5}{3}$ and the point is $\left(\tfrac{5}{2}, \tfrac{3}{2}\right)$ . The tangent is $y - \tfrac{3}{2} = \tfrac{5}{3}\left(x - \tfrac{5}{2}\right)$ , which is $3y = 5x - 8$ .

6 [5 marks]

A curve has equation $x^2 + 3xy - y^2 = 3$ .

Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ . [3]
Find the equation of the normal at the point $(1, 1)$ . [2]

Show solution

(a) Differentiating implicitly, with the product rule on $3xy$ :

\begin{aligned} &2x + 3x\frac{dy}{dx} + 3y - 2y\frac{dy}{dx} = 0 \\ &\frac{dy}{dx} = -\frac{2x + 3y}{3x - 2y} \end{aligned}

(b) At $(1, 1)$ the gradient is $-5$ , so the normal has gradient $\tfrac{1}{5}$ : $y - 1 = \tfrac{1}{5}(x - 1)$ , which is $5y = x + 4$ .

7 [3 marks]

The edges of a cube are increasing at $0.1$ cm per second. Find the rate at which its volume is increasing when the edges are $5$ cm long.

Show solution

$V = x^3$ , so $\dfrac{dV}{dx} = 3x^2 = 75$ when $x = 5$ . Then

\begin{aligned} \frac{dV}{dt} &= \frac{dV}{dx} \times \frac{dx}{dt} \\ &= 75 \times 0.1 = 7.5 \text{ cm}^3 \text{ per second} \end{aligned}

8 [5 marks]

For $f(x) = x^4 - 6x^2 + 5$ , find the values of $x$ for which the curve $y = f(x)$ is concave, and the coordinates of its points of inflection.

Show solution

$f'(x) = 4x^3 - 12x$ and $f''(x) = 12x^2 - 12 = 12(x - 1)(x + 1)$ .

$f''(x) \le 0$ for $-1 \le x \le 1$ , so the curve is concave there, and convex for $|x| \ge 1$ . $f''(x)$ changes sign at $x = \pm 1$ , where $f(\pm 1) = 1 - 6 + 5 = 0$ . So the points of inflection are $(1, 0)$ and $(-1, 0)$ .

9 [4 marks]

Show that the curve $y = xe^x$ has exactly one point of inflection, and find its coordinates.

Show solution

By the product rule, $\dfrac{dy}{dx} = e^x(x + 1)$ , and again, $\dfrac{d^2y}{dx^2} = e^x(x + 2)$ .

$e^x > 0$ , so $\dfrac{d^2y}{dx^2}$ has the sign of $x + 2$ . It changes sign only at $x = -2$ , so there is exactly one point of inflection, at $(-2, -2e^{-2})$ .

Exam style

10 [7 marks]

A curve has parametric equations $x = \cos t$ , $y = \sin 2t$ , $0 \le t \le \pi$ .

Find $\dfrac{dy}{dx}$ in terms of $t$ . [2]
Find the coordinates of the points where the tangent is horizontal. [3]
Find the coordinates of the points where the tangent is vertical. [2]

Show solution

(a)

\frac{dy}{dx} = \frac{2\cos 2t}{-\sin t} = -\frac{2\cos 2t}{\sin t}

(b) Horizontal when $\cos 2t = 0$ . With $0 \le 2t \le 2\pi$ , $2t = \tfrac{\pi}{2}$ or $\tfrac{3\pi}{2}$ , so $t = \tfrac{\pi}{4}$ or $\tfrac{3\pi}{4}$ . The points are $\left(\tfrac{\sqrt2}{2}, 1\right)$ and $\left(-\tfrac{\sqrt2}{2}, -1\right)$ .

(c) Vertical when $\dfrac{dx}{dt} = -\sin t = 0$ (and $\dfrac{dy}{dt} \ne 0$ ), so $t = 0$ or $\pi$ : the points $(1, 0)$ and $(-1, 0)$ .

11 [7 marks]

A curve has equation $2x^2 + y^2 - 2xy = 25$ .

Show that $\dfrac{dy}{dx} = \dfrac{y - 2x}{y - x}$ . [4]
Find the exact coordinates of the points where the tangent is parallel to the $x$ -axis. [3]

Show solution

(a) Differentiating implicitly, with the product rule on $2xy$ :

\begin{aligned} 4x + 2y\frac{dy}{dx} - 2x\frac{dy}{dx} - 2y &= 0 \\ (2y - 2x)\frac{dy}{dx} &= 2y - 4x \end{aligned}

Dividing by $2(y - x)$ gives $\dfrac{dy}{dx} = \dfrac{y - 2x}{y - x}$ .

(b) The gradient is $0$ when $y = 2x$ . Substituting, $2x^2 + 4x^2 - 4x^2 = 25$ , so $x^2 = \tfrac{25}{2}$ and $x = \pm\tfrac{5\sqrt2}{2}$ . The points are $\left(\tfrac{5\sqrt2}{2}, 5\sqrt2\right)$ and $\left(-\tfrac{5\sqrt2}{2}, -5\sqrt2\right)$ .

12 [7 marks]

The area of a circular ripple increases at a constant $12$ cm² per second.

Find the rate at which the radius is increasing when the radius is $3$ cm. [3]
Find the rate at which the radius is increasing when the area is $100\pi$ cm². [2]
Find the rate at which the circumference is increasing when the radius is $3$ cm. [2]

Show solution

(a) $A = \pi r^2$ , so $\dfrac{dA}{dr} = 2\pi r$ . Then

\frac{dr}{dt} = \frac{dA/dt}{dA/dr} = \frac{12}{2\pi r}

which is $\dfrac{12}{6\pi} = \dfrac{2}{\pi} = 0.637$ cm per second when $r = 3$ .

(b) $\pi r^2 = 100\pi$ gives $r = 10$ , so $\dfrac{dr}{dt} = \dfrac{12}{20\pi} = 0.191$ cm per second (3 s.f.).

(c) $C = 2\pi r$ , so $\dfrac{dC}{dt} = 2\pi\dfrac{dr}{dt} = 2\pi \times \dfrac{2}{\pi} = 4$ cm per second.