Delhi Cantt, Delhi, India - 110010
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
12
Board
IGCSE, CBSE
CBSE Subjects taught
Mathematics, Physics
IGCSE Subjects taught
Physics
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
15
Board
CBSE
CBSE Subjects taught
Physics
Taught in School or College
No
Teaching Experience in detail in Class 12 Tuition
I am giving tuitions to class 12th grade from last 15 years. I am providing concept of subjects to the students . The concept of Physics and Mathematics raises the confidence of the student. And the confidence in subject matter is the key to score high quality performance in result as well as in carrier building.
5 out of 5 2 reviews
Amitabh Anal
"He is really a teacher. He have a capability to give the complete idea of every topic either in Mathematics or Physics. He just go into the concept of relevant topic. So I highly recommend him for Physics and Mathematics for class 11th and 12th. "
Vaibhav Mishra
Class 11 Tuition
"He is one of those rare teachers who give their students a liberty to think out of the box by themselves. He has the ability to develop interest regarding the toughest subjects in science. Sir is quite good at explaining topics in a detailed way. One of those legends who can help you to master your concepts regarding physics and mathematics, he can adjust himself according to a student's pace to clear his/her doubts. Sir teach in a very friendly way, explaining tips and tricks to memorize even the toughest theroms or formulas. His educational influence is so great that it actually enabled me to write this comment describing my own experience. Thank you so much sir for being my teacher "
Answered on 08/10/2019 Learn CBSE/Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3
It is given that x = acos3θ and y = asin3θ.
Therefore, the slope of the tangent at is given by,
Hence, the slope of the normal at
Answered on 08/10/2019 Learn Tuition
Answered on 07/10/2019 Learn CBSE/Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3
The equation of the given curve is ay2 = x3.
On differentiating with respect to x, we have:
The slope of a tangent to the curve at (x0, y0) is.
The slope of the tangent to the given curve at (am2, am3) is
∴ Slope of normal at (am2, am3) =
Hence, the equation of the normal at (am2, am3) is given by,
y − am3 =
Answered on 07/10/2019 Learn CBSE/Class 12/Mathematics/Continuity and Differentiability/NCERT Solutions/Exercise 5.7
Let
Then,
Answered on 07/10/2019 Learn CBSE/Class 12/Mathematics/Continuity and Differentiability/NCERT Solutions/Exercise 5.7
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