MHT-CET · Chapter

Applications of Derivative

The concepts MHT-CET needs, one connected idea at a time.

0 / 11

concepts mastered

Tangents and rates0/5
Mean value theorems0/2
Increasing functions and extremes0/3
Angle between curves0/1

Ready for you

Up nextTangent and normal at a pointThe tangent at $(x_1, y_1)$ has slope $m = \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}$ :Start
Derivative as a rate of changeStart $\dfrac{dy}{dx}$ is the rate at which $y$ changes with $x$ ;
Rolle's theoremStartIf $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ and $f(a) = f(b)$ ,
Increasing and decreasing functionsStart $f$ increases where $f'(x) > 0$ and decreases where $f'(x) < 0$ :

Whole chapter

How the ideas build on each other. Each one opens once you've mastered what it needs.

Tangents and rates0 / 5

Tangent and normal at a pointThe tangent at $(x_1, y_1)$ has slope $m = \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}$ :

y - y_1 = m(x - x_1)

;

the normal has slope

-\dfrac{1}{m}

for motion

s(t)

,

velocity

v = \dfrac{ds}{dt}

and acceleration

a = \dfrac{dv}{dt}

then

f'(c) = 0

for some

c

in

(a, b)

:

a horizontal tangent

find where

f' = 0

and test the sign of

f'

between those points

y - y_1 = m(x - x_1)

;

the normal has slope

-\dfrac{1}{m}

Start

Tangents with a given slopeSet

\dfrac{dy}{dx}

equal to the slope wanted (parallel: equal; perpendicular:

-\dfrac{1}{m}

) and solve for the points of contactLocked

Derivative as a rate of change

\dfrac{dy}{dx}

is the rate at which

y

changes with

x

; for motion

s(t)

, velocity

v = \dfrac{ds}{dt}

and acceleration

a = \dfrac{dv}{dt}

Start

Related ratesWrite the relation between the quantities, differentiate both sides with respect to

t

, then substitute the known rates and valuesLocked

ApproximationsNear

a

, the curve hugs its tangent:

f(a + h) \approx f(a) + h\,f'(a)

, as in

\sqrt{8.95} \approx 3 - \dfrac{0.05}{6}

Locked

Mean value theorems0 / 2

Rolle's theoremIf $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ and $f(a) = f(b)$ , then $f'(c) = 0$ for some $c$ in $(a, b)$ : a horizontal tangentStart
Lagrange's mean value theoremIf $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , some $c$

Increasing functions and extremes0 / 3

Increasing and decreasing functions $f$ increases where $f'(x) > 0$ and decreases where $f'(x) < 0$ : find where $f' = 0$ and test the sign of $f'$ between those pointsStart
Local maxima and minimaAt a critical point $f'(c) = 0$ : a maximum if $f''(c) < 0$ ,
Greatest and least values in problemsWrite the quantity to optimise in one variable using the constraint, find its critical point, and check it is the maximum or minimum asked forLocked

Angle between curves0 / 1

Angle between two curvesAt a common point, the angle between two curves is the angle between their tangents; they cut at right angles when $m_1m_2 = -1$ Locked

in

(a, b)

has

f'(c) = \dfrac{f(b) - f(a)}{b - a}

:

a tangent parallel to the chord

Locked

a minimum if

f''(c) > 0

;

or watch

f'

change sign from

+

to

-

(maximum) or

-

to

+

Locked