RD sharma class 10 Chapter 11 some applications of trigonometry

Chapter 3 Linear Equations in Two Variables Download Pdf of RD sharma class 10 Chapter 11 some applications of trigonometry

Share:

Read More

Class 12 Chapter 6 (Application Of Derivative) Class Notes part II

MAXIMA AND MINIMA

 

1. MAXIMA AND MINIMA:

   • Let f be a function defined on an interval I. Then

     1. f is said to have a maximum value in I, if there exists a point c in I such that f (c) ≥ f (x) , for all x ∈ I.

   • The number f (c) is called the maximum value of f in I and the point c is called a point of maximum value of f in I.

     1. f is said to have a minimum value in I, if there exists a point c in I such that f (c) ≤ f (x), for all x ∈ I.

   • The number f (c), in this case, is called the minimum value of f in I and the point c, in this case, is called a point of minimum value of f in I.

   • f(x) is said to have an extreme value in I if there exists a point c in I such that f is either a maximum value or a minimum value of f in I.

   • The number f (c), in this case, is called an extreme value of f in I and the point c is called an extreme point.

     Absolute maxima and minima

   • Let f be a function defined on the interval I and c ∈ I. Then

     1. f(c) is absolute minimum if f(x) ≥ f(c) for all x ∈ I.

     2. f(c) is absolute maximum if f(x) ≤ f(c) for all x ∈ I.

     3. c ∈ I is called the critical point off if f ′(c) = 0

     4. Absolute maximum or minimum value of a continuous function f on [a, b] occurs at a or b or at critical points off (i.e. at the points where f ′is zero)

   • If c₁ ,c₂, … , c_n are the critical points lying in [a , b], then

     absolute maximum value of f = max{f(a), f(c₁), f(c₂), … , f(c_n), f(b)} and absolute minimum value of f = min{f(a), f(c₁), f(c₂), … , f(c_n), f(b)}.

     Local maxima and minima

   • (a)A function f is said to have a local maxima or simply a maximum value at x a if f(a ± h) ≤ f(a) for sufficiently small h

   • (b)A function f is said to have a local minima or simply a minimum value at x = a if f(a ± h) ≥ f(a).

   • First derivative test : A function f has a maximum at a point x = a if

     1. f ′(a) = 0, and

     2. f ′(x) changes sign from + ve to –ve in the neighborhood of ‘a’ (points taken from left to right).

     However, f has a minimum at x = a, if

     1. f ′(a) = 0, and

     2. f ′(x) changes sign from –ve to +ve in the neighborhood of ‘a’.

        If f ′(a) = 0 and f’(x) does not change sign, then f(x) has neither maximum nor minimum and the point ‘a’ is called point of inflection.

     
   • The points where f ′(x) = 0 are called stationary or critical points. The stationary points at which the function attains either maximum or minimum values are called extreme points.
   

   • Second derivative test :

     1. a function has a maxima at x= a, if f ′(x) =0 and f ′′ (a) <0

     2. a function has a minima at x = a, if f ′(x) = 0 and f ′′(a) > 0.

 

EXAMPLE

QUESTION: If length of three sides of a trapezium other than base is equal to 10cm each, then find the area of the trapezium when it is maximum.

SOLUTION: The required trapezium is as given in Fig below. Draw perpendiculars DP and CQ on AB

ΔAPD is congruent to ΔBQC

Let AP = BQ = x cm

\[ \Rightarrow DP = QC = \sqrt {100 - {x^2}} \]

Area of Trapezium = \( \Rightarrow \dfrac{1}{2}(2x + 10 + 10)\sqrt {100 - {x^2}}  = (x + 10)\sqrt {100 - {x^2}} \)

\[ \Rightarrow A'(x) = \dfrac{{ - 2{x^2} - 10x - 100}}{{\sqrt {100 - {x^2}} }}\]

\[ \Rightarrow A'(x) = 0 \Rightarrow x = 5\]

Now, \( \Rightarrow A''(x) = \dfrac{{2{x^2} - 300x - 100}}{{{{\left( {100 - {x^2}} \right)}^{\dfrac{2}{3}}}}}\)

and \( \Rightarrow {\left. {A''(x)} \right|_{x = 5}} = \dfrac{{ - 30}}{{{{\left( {75} \right)}^{\frac{2}{3}}}}} < 0\)

thus , the area of trapezium is maximum at x = 5

\[Area = 75\sqrt 3 \,c{m^2}\]

Share:

Read More

Class 10 Chapter 9 (some applications on trigonometry) class notes

Notes on Some Applications of Trigonometry

Trigonometry can be used in many ways in the things around us like we can use it for calculating the height and distance of some objects without calculating them actually.

Heights and Distances

To find the height of an object or to find the distance of an object to the other we must know the meaning of some points -

Line of Sight

When we look at some object then the line made by our vision to the object is called the Line of Sight.

Horizontal Line

A horizontal line is a distance between the observer and the object.

Angle of Elevation

An angle of elevation is the angle made by the line of sight to the top of the object and the horizontal line. It is above the horizontal line i.e. the angle of elevation is made when we look up to the object.

Angle of Depression

An angle of depression is made when the observer needs to look down to see the object.The angle between the horizontal line and the line of sight is the angle of depression when the horizontal line is above the angle.

Some Solved Examples

To solve the problems related to the angle of elevation and angle of depression we must remember trigonometric ratios, trigonometric table and the trigonometric identities.

Example: 1

Find the height of the flagpole if the angle of elevation is 30° and the distance of flag from the observer is 15 m.

Solution:

Let x be the height of the flagpole.

Horizontal line is 15m.

Now, to calculate x, we need to take the trigonometric ratio which includes perpendicular and base.

Hence, the flagpole is 9 m.

Example: 2

A child was playing at the top of the hill. He had thrown a stone in the lake from the hill, the distance covered by the stone was 150 m and the angle of depression was 30°, then what is the height of the hill?

Solution:

Let the height of the hill be h.

The hypotenuse is 150 m and the angle of depression is 30°.

Hence the height of the hill is 75 m.

Example: 3

A person standing at point A is looking up at the angle of elevation of 45° to the aeroplane which is at the height of 100 m. As the airplane is going upwards, after some, the person was looking at the angle of elevation of 60°. Then what will be the increase in the height of the aeroplane from the ground at the angle of 60°?

Solution:

Given

∠CAB = 45°, ∠DAB = 60°

Distance of the aero plane from the ground = x + 100 m

In ∆ABC

AB = BC = 100 m

In ∆ABD

BD = BC + CD

100√3m = 100 + x

x = 100(√3 - 1) m

Hence, the increase in the height of the aero plane is100 (√3 -1) m.

Example: 4

A girl who is 1.2 m tall is watching a ballon moving in a horizontal line at a height of 88.2 m from the ground.The angle of elevation from her eyes is changed from 60° to 30°. Calculate the distance travelled by balloon.

Solution:

In ΔACE

In ΔBCG

CG = 87√3 m

Distance travelled by ballon = EG = CG - CE

87√3 - 29√3 = 58√3 m

cbse class 10th Heights and distances

Share:

Read More