Naraina, Delhi Details verified of Vinay Kumar Aggarwal✕
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English Basic
IGNOU Pursuing
Master of Arts (M.A.)
Naraina, Delhi, India - 110028
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
3
Board
CBSE, State
Subjects taught
Economics, Accountancy, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
3
Board
CBSE, State
Subjects taught
Economics, Accountancy, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in BA Tuition
3
Field tutored for
Mathematics, Economics
BA Mathematics Subjects
Calculus
Type of class
Regular Classes
Class strength catered to
One on one/ Private Tutions, Group Classes
Taught in School or College
Yes
BA Economics Subjects
Statistical Methods for Economics, Macroeconomics, Microeconomics
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in BCom Tuition
1
BCom Subject
Accounting Information Systems, Economics of Regulation of Domestic and Foreign Exchange Markets, Indian Economy
Class strength catered to
One on one/ Private Tutions, Group Classes
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
4
Board
CBSE
Subjects taught
Mathematics, Hindi, English, Science, EVS
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
Subjects taught
Mathematics
Taught in School or College
No
Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3
Ask a Question
Post a Lesson
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 sinθ/a
at θ = π/2 , dy/dx = 2bsinπ/2/a = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b
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Answers 5 
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
3
Board
CBSE, State
Subjects taught
Economics, Accountancy, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
3
Board
CBSE, State
Subjects taught
Economics, Accountancy, Mathematics
Taught in School or College
Yes
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in BA Tuition
3
Field tutored for
Mathematics, Economics
BA Mathematics Subjects
Calculus
Type of class
Regular Classes
Class strength catered to
One on one/ Private Tutions, Group Classes
Taught in School or College
Yes
BA Economics Subjects
Statistical Methods for Economics, Macroeconomics, Microeconomics
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in BCom Tuition
1
BCom Subject
Accounting Information Systems, Economics of Regulation of Domestic and Foreign Exchange Markets, Indian Economy
Class strength catered to
One on one/ Private Tutions, Group Classes
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
4
Board
CBSE
Subjects taught
Mathematics, Hindi, English, Science, EVS
Taught in School or College
No
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
2
Board
CBSE
Subjects taught
Mathematics
Taught in School or College
No
Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3
Ask a Question
Post a Lesson
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 sinθ/a
at θ = π/2 , dy/dx = 2bsinπ/2/a = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b
Like 0 Answers 5 Comments Ask a Question
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