CLASS 11 PHYSICS IMPORTANT NOTES
IMPORTANT NOTES
✤ TWO LAND TECHNIQUE
Differentiation: It is method by which we can find change in One
physical Quantity w.r.t another quantity.
y2-y1 / x2-x1 =dy/dx
differentiation of y w.r.t. x
d/dx is called differential operator.
(1) d/dx un = n• xn-1 (2) d/dx sinx = cosx
(3) d/dx x5 = 5•x5-1 (4) d/dx tanx=sec2x
= 5x4 (5) d / dx cosx = -sinx
(6) d/dx constant = 0
ALGEBRA OF DIFFRENTIATION
(1) ADDITION RULE
d/dx (u+v) = d/dx u + d/dx v
(2) SUBSTRACTION RULE
d / dx (u-v) = d /dx u -d/dx v
(3) SCALAR FORM
d/dx [k][f(x)] = k • d/dx (f(x))
Ex:
d/dx (5 • x4) = 5 • d/dx x4
= 5 • 4x3
= 20x3
(4) PRODUCT RULE
d / dx (uxv) = u dv/ dx + v du/dx
(5) DIVISION RULE
d/dx(u/v) = [v(du/dx) – u(dv/dx)] / v2
❈ Velocity: It is defined as the rate of change of displacement
v= dx / dt SI unit : m/s
Q) If displacement x=t3+4t2+5
t = 2s
Ans) v= dx/dt
v=d / dt (t3+4t2+5)
v= d/dt t3 + d/dt 4t2 +d/dt 5
v= 3t2 + 8t
At t = 2s
= 3(2)2+8(2)
= 12 + 16
= 28 m/s
❈ ACCELERATION: Rate of change of velocity is called Acceleration
A= dv/dt
SI unit= m/s2
* 3 Markers Questions Asked
Q) If Displacement of a car at any constant t is given by
x = 5t4+6t2+9t+4
(i) Displacement at t =1
x = 5(1)4+6(1)2+9(1)+4
= 24m
(ii) Initial Velocity (t = 0)
v = dx/dt = 20t3+12t+9
= 9 m/s
(iii) Velocity at (t = 3s)
v = dx/dt = 20(3)3+12(3)+9
= 213 m/s
(iv) Acceleration = dv/dt
A = 60t2+12
A = 540+12
= 552 m/s2
* * * * * * * * * * * * *
