NDA Maths · Application of Derivatives

Tangents, Rates of Change & Approximations

The derivative read geometrically (slope of the tangent), dynamically (a rate of change), and as a tool for estimating small changes via differentials.

All Application of Derivatives notes NDA Maths strategy

Why this matters

These are the most direct uses of f′(x): the slope of a tangent or normal, how fast one quantity changes with another, and a quick linear estimate of a small change. They are reliably easy marks once you read the derivative the right way.

Concept 1 of 2: Tangent and normal to a curve

The slope of the tangent at a point is just

f'(x_0)

; the normal is perpendicular, so its slope is

-1/f'(x_0)

. With a point and a slope, the line equations follow immediately.

Definition

At $(x_0,y_0)$ on $y=f(x)$ : tangent slope $m=f'(x_0)$ , tangent $y-y_0=m(x-x_0)$ ; normal slope $-1/m$ , $y-y_0=-\tfrac1m(x-x_0)$ . The tangent makes angle $\theta=\tan^{-1}m$ with the x-axis. A tangent is horizontal where $f'=0$ , vertical where $f'$ is undefined; parallel tangents share the same $m$ .

Tangent & normal at a point

m_{\text{tangent}}=\left.\frac{dy}{dx}\right|_{(x_1,y_1)} \qquad m_{\text{normal}}=-\frac{1}{dy/dx} \qquad y-y_1=m(x-x_1)

Worked example

Find the slope of the tangent to

y=x^3-2x

x=1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2021 · Q100Moderate

Example 1 · Application of Derivatives · Tangents and Slopes

The tangent to the curve

x^{2}=y

(1,1)

makes an angle

\theta

with the positive direction of

x

-axis. Which one of the following is correct?

The normal slope is the NEGATIVE reciprocal

If the tangent slope is

m

, the normal slope is

-1/m

— not

1/m

and not

-m

. Drop either the minus sign or the reciprocal and the normal line is wrong.

Drill 2 more on tangent and normal to a curve

Concept 2 of 2: Rates of change and small-change approximation

A derivative is a rate:

\dfrac{dy}{dt}

tells how fast

y

changes in time, and related quantities chain together via

\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}

. For a small input change, the derivative gives a fast linear estimate of the output change:

\Delta y\approx f'(x)\,\Delta x

Definition

Related rates: differentiate the relation w.r.t. time and substitute known rates (e.g. radius growing → area's rate $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$ ).
Approximation (differentials): $\Delta y\approx \dfrac{dy}{dx}\,\Delta x$ ; use it to estimate $f(x+\Delta x)\approx f(x)+f'(x)\Delta x$ .
Exact vs approximate change: the total (exact) change is $\Delta y = f(x+\Delta x)-f(x)$ , computed directly; the derivative gives only the linear approximation $f'(x)\,\Delta x$ . A question that says "total change" or "actual change" wants the exact value — the two differ by the curvature term (for $y=3x^2+2$ , $x:10\to10.1$ : exact $6.03$ , approximate $6$ ).

Related rates & small-change approximation

\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt} \qquad \Delta y\approx f'(x)\,\Delta x \qquad dy=f'(x)\,dx

Worked example

The radius of a circle grows at

3

cm/s. How fast is the area changing when

r=5

cm?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q72Easy

Example 2 · Application of Derivatives · Tangents and Slopes

The radius of a circle is increasing at the rate of

0.7

cm/sec. What is the rate of increase of its circumference?

"Total change" is NOT the differential estimate

f'(x)\,\Delta x

is an approximation. When a question asks for the total or actual change, compute

f(x+\Delta x)-f(x)

exactly — for

y=3x^2+2

from

x=10

10.1

that is

6.03

, and the option

6

is the planted approximation. Use differentials only when the question says "approximately" or "estimate".

Related rates need the CHAIN RULE

To get a time rate, differentiate the relation with respect to

t

and chain through the variable:

\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dt}

. Differentiating

A=\pi r^2

as if

r

were the variable gives

2\pi r

— the rate is

2\pi r\,\dfrac{dr}{dt}

, not

2\pi r

Drill 1 more on rates of change and small-change approximation

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Tangent and normal to a curve
Tangent & normal at a point
$m_{\text{tangent}}=\left.\frac{dy}{dx}\right|_{(x_1,y_1)} \qquad m_{\text{normal}}=-\frac{1}{dy/dx} \qquad y-y_1=m(x-x_1)$
Rates of change and small-change approximation
Related rates & small-change approximation
$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt} \qquad \Delta y\approx f'(x)\,\Delta x \qquad dy=f'(x)\,dx$

Watch out for (3)

The normal slope is the NEGATIVE reciprocal
→ Tangent and normal to a curve
"Total change" is NOT the differential estimate
→ Rates of change and small-change approximation
Related rates need the CHAIN RULE
→ Rates of change and small-change approximation

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