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Vanga Goutham Reddy Class 12 Tuition trainer in Hyderabad

Vanga Goutham Reddy

experience 1 yrs of Exp
students 1 student
locationImg Begumpet, Hyderabad
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I am a engineering graduate student . I am good at all subjects. I am graduated from JNTUH. I want to teach the tuitions as I want to share my knowledge to other students.

Languages Spoken

Telugu Mother Tongue (Native)

English Proficient

Hindi Proficient

Education

Anurag group of institutions 2018

Bachelor of Technology (B.Tech.)

Address

Begumpet, Hyderabad, India - 500016

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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

1

Board

CBSE, International Baccalaureate, State, ISC/ICSE

Subjects taught

History, Biology, Political Science, Economics, Computer Science, Electronics, English, Engineering Graphics, Physics, Sanskrit, Geography, Mathematics, Chemistry, Telugu

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Economic Application, Sanskrit, English Literature, Science, History and Civics, Social Science, Telugu, Chemistry, Biology, Physics, Mathematics, Geography, English

Taught in School or College

No

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, State, ISC/ICSE

Subjects taught

Telugu, Geography, English, Electronics, Mathematics, Computer Science, History, Engineering Graphics, Physics, Biotechnology, Biology, Economics, Chemistry

Taught in School or College

No

BTech Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Electrical & Electronics subjects

High-Voltage Engineering, Switchgear And Protection, Circuit Theory, Control Systems, Electric Drives, (Dc- Ac) Direct Current-Alternatiing Current System Interaction, Embedded Processor Architecture, Direct Energy Conversion, Design Of Embedded Controllers, Applications of Digital Signal Processing (DSP), Electrical Circuit Analysis, High-Voltage Direct Current (Hvdc) Transmission, Power Electronics, Power System Protection, Regulation & Security, Electromagnetic Theory, Analog And Digital Communication, Data Structures & Algorithms, Alternative Energy Sources, Discrete Fourier Transforms And Digital Filter Design, Electric Energy Systems, Power System Engineering

BTech Branch

BTech 1st Year Engineering, BTech Electrical & Electronics

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

No

BTech 1st Year subjects

Engineering Chemistry, Engineering Mathematics (M1), Basic Electronics, Engineering Physics, Engineering Graphics, Advanced Mathematics (M2), Environmental Studies, Basic Electrical Technology, Computer science

Reviews

No Reviews yet!

Answers by Vanga Goutham Reddy

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

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Post a Lesson Post a Lesson

Slope of normal and slope of tangent are perpendicular to each other.e.g., slope of normal x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ=... ...more

Slope of normal and slope of tangent are perpendicular to each other.
e.g., slope of normal 

x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)

dividing equations (2) by (1),

dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθ
at θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b

Answer Answers 5 Comments Comments
Dislike Bookmark

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Determinants/NCERT Solutions/Miscellaneous Exercise 4

Ask a Question Ask a Question

Post a Lesson Post a Lesson

(x+a)× - x + x =0 (x+a) × - x + x = 0 (x+a) - ax² -ax² =0 2ax² + a²x + 2a²x + a³ -2ax² = 0 3a²x +a³ = 0 a² (3x + a) = 0 a != 0, ( 3x + a)= 0 3x = -a ...more

(x+a)×[(x+a)² - x²] - x [x(x+a) - x²] + x [ x²- x(x+a)]  =0

(x+a) ×[ x²+2ax+a² - x²] - x[ x² + ax -x²] + x [ x² - x² - ax] = 0

(x+a) [2ax + a²] - ax² -ax² =0

2ax² + a²x + 2a²x + a³ -2ax² = 0

3a²x +a³ = 0

a² (3x + a) = 0

a != 0, ( 3x + a)= 0

3x = -a

 

Answer Answers 3 Comments Comments
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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

1

Board

CBSE, International Baccalaureate, State, ISC/ICSE

Subjects taught

History, Biology, Political Science, Economics, Computer Science, Electronics, English, Engineering Graphics, Physics, Sanskrit, Geography, Mathematics, Chemistry, Telugu

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Economic Application, Sanskrit, English Literature, Science, History and Civics, Social Science, Telugu, Chemistry, Biology, Physics, Mathematics, Geography, English

Taught in School or College

No

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, State, ISC/ICSE

Subjects taught

Telugu, Geography, English, Electronics, Mathematics, Computer Science, History, Engineering Graphics, Physics, Biotechnology, Biology, Economics, Chemistry

Taught in School or College

No

BTech Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Electrical & Electronics subjects

High-Voltage Engineering, Switchgear And Protection, Circuit Theory, Control Systems, Electric Drives, (Dc- Ac) Direct Current-Alternatiing Current System Interaction, Embedded Processor Architecture, Direct Energy Conversion, Design Of Embedded Controllers, Applications of Digital Signal Processing (DSP), Electrical Circuit Analysis, High-Voltage Direct Current (Hvdc) Transmission, Power Electronics, Power System Protection, Regulation & Security, Electromagnetic Theory, Analog And Digital Communication, Data Structures & Algorithms, Alternative Energy Sources, Discrete Fourier Transforms And Digital Filter Design, Electric Energy Systems, Power System Engineering

BTech Branch

BTech 1st Year Engineering, BTech Electrical & Electronics

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

No

BTech 1st Year subjects

Engineering Chemistry, Engineering Mathematics (M1), Basic Electronics, Engineering Physics, Engineering Graphics, Advanced Mathematics (M2), Environmental Studies, Basic Electrical Technology, Computer science

No Reviews yet!

Answers by Vanga Goutham Reddy

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

Ask a Question Ask a Question

Post a Lesson Post a Lesson

Slope of normal and slope of tangent are perpendicular to each other.e.g., slope of normal x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ=... ...more

Slope of normal and slope of tangent are perpendicular to each other.
e.g., slope of normal 

x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)

dividing equations (2) by (1),

dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθ
at θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b

Answer Answers 5 Comments Comments
Dislike Bookmark

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Determinants/NCERT Solutions/Miscellaneous Exercise 4

Ask a Question Ask a Question

Post a Lesson Post a Lesson

(x+a)× - x + x =0 (x+a) × - x + x = 0 (x+a) - ax² -ax² =0 2ax² + a²x + 2a²x + a³ -2ax² = 0 3a²x +a³ = 0 a² (3x + a) = 0 a != 0, ( 3x + a)= 0 3x = -a ...more

(x+a)×[(x+a)² - x²] - x [x(x+a) - x²] + x [ x²- x(x+a)]  =0

(x+a) ×[ x²+2ax+a² - x²] - x[ x² + ax -x²] + x [ x² - x² - ax] = 0

(x+a) [2ax + a²] - ax² -ax² =0

2ax² + a²x + 2a²x + a³ -2ax² = 0

3a²x +a³ = 0

a² (3x + a) = 0

a != 0, ( 3x + a)= 0

3x = -a

 

Answer Answers 3 Comments Comments
Dislike Bookmark
x

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