If →V is a differentiable vector function and f is a sufficient differentiable scalar function, then curl (f→V) is equal to
A
(gradf)×(→V)+f(curl→V)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
fcurl(→V)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(gradf)×(→V)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A(gradf)×(→V)+f(curl→V) Let →V=u^i+v^j+w^k curl(f→v)=∣∣
∣
∣
∣∣ijk∂∂x∂∂y∂∂zufvfwf∣∣
∣
∣
∣∣ =[∂∂y(wf)−∂∂z(vf)]^i+[∂∂z(uf)−∂∂x(wf)]^j+[∂∂x(vf)−∂∂y(uf)]^k =f[(∂w∂y−∂v∂z)^i+(∂u∂z−∂w∂x)^j+(∂v∂x−∂u∂y)^k] +[(w∂f∂y−∂f∂z)^i+(u∂f∂z−∂f∂x)^j+(v∂f∂x−∂f∂y)^k] curl(f→V)=f(curl→V)+(∂f∂x^i+∂f∂y^j+∂f∂z^k)×(u^i+v^j+w^k) curl(f→V)=f(curl→V)+(gradf)×→V
Alternative Solution: Curl(f→V)=▽×(f→V) =(▽f)×→V+f(▽×→V) f(curl→V)=(gradf)×→V+f(curl→V)
It is a vector identity just learn it without paying attention on its proof.