i know you're not gonna read this sir but truly informative lecture, everything explained conceptually and in an easy manner, no wonder why IIT alumni excel in their work
In last example the point does not lies on second surface. And moreover, if there are two surfaces touching at a point their normals at that point must be common. And even they will not have particular tangent line, instead they would have tangent plane.
In the last example it will be possible to get a tangent line with an extra condition only, if you take a vector parallel to line (x,y,z)-(a,b,c) where (a,b,c) Point of intersection of curves then ((x,y,z)-(a,b,c)).grad(f)=0......(1) at ( grad(f) at (a,b,c) ) (1) Will give locus of all tangent lines which is equal to the tangent plane.
@@nssjeecell-inorganicchemis9228 the question is right but sir wrote it wrong while solving Therefore the point is not satisfied It is satisfied in original question
i know you're not gonna read this sir but truly informative lecture, everything explained conceptually and in an easy manner, no wonder why IIT alumni excel in their work
In last example the point does not lies on second surface.
And moreover, if there are two surfaces touching at a point their normals at that point must be common. And even they will not have particular tangent line, instead they would have tangent plane.
Thanks
No this is not true, take some example and you will find normals are not same.
for ex take a curve in xz plane and another in yz plane with a common point on z axis u can easily observe.
Yes I was thinking the same
At 22:07 the equation for z should be z=t+2
exactly thats what i was thinking
thanks for pointing it out
better than IIT madras proffs...
very good explanation
you have made this very easy to understand sir
thank you sir
Very great😁🎉👍.. Lectures...
I will come to you for charan sparsh after my ETE😊 .Thank you so much sir.
Sir in the last question the point P0 doesn't satisfies the curve g
But we got the concept which is more imp. 👍👍
In the last example it will be possible to get a tangent line with an extra condition only, if you take a vector parallel to line (x,y,z)-(a,b,c) where (a,b,c) Point of intersection of curves then
((x,y,z)-(a,b,c)).grad(f)=0......(1)
at ( grad(f) at (a,b,c) )
(1) Will give locus of all tangent lines which is equal to the tangent plane.
22:07 z=t+2
Marvellous💯
Thank you giving lot of knowledge
Nice vedio sir great work of iit
thank you sir💖
Very good lecture sir 👌👌👌👌
bohot badiya
what a explanation sir jii
Perfect video sir......it's very useful
31:11 SIR THE 'g' SURFACE IS NOT PASSING THROUGH (1,1,1) !!!!!!!!!!
thankyou for awesome lecture sir
In the last example why won't we get a tangent plane??
Thanks Sir.🙏
awesome lecture sir
14:10
Tq sir
Last example is ¿¿¿¿ But the lecture is very helpful 🙏🙏🙏😊😊😊
Thank you sir
Love you sir not as a teacher 😩
Last example is wrong....
26:23 z=-1
How
last example is wrong, except that, a good video
yes because that point does not lie on second curve
@@nssjeecell-inorganicchemis9228 the question is right but sir wrote it wrong while solving
Therefore the point is not satisfied
It is satisfied in original question
Exactly what I thought because grad(f) and grad(g) were antiparallel and there cross product will be jero.
22:05 z = 2 + t