Velocity filed is given as....
V = 2 x 3 i ^ − 5 x 2 y j ^ + 4 t k ^ d V ( x , y , z , t ) = ( ∂ V ∂ x ) y , z , t d x + ( ∂ V ∂ y ) x , z , t d y + ( ∂ V ∂ z ) y , x , t d z + ( ∂ V ∂ t ) x , y , z d t d V ( x , y , z , t ) d t = ( ∂ V ∂ x ) y , z , t d x d t + ( ∂ V ∂ y ) x , z , t d y d t + ( ∂ V ∂ z ) y , x , t d z d t + ( ∂ V ∂ t ) x , y , z a ⃗ = d V ( x , y , z , t ) d t = ( ∂ V ∂ x ) y , z , t V x + ( ∂ V ∂ y ) x , z , t V y + ( ∂ V ∂ z ) y , x , t V z + ( ∂ V ∂ t ) x , y , z V=2x^3\hat{i}-5x^2y\hat{j}+4t\hat{k}\\
dV(x,y,z,t)=\bigg( \cfrac{\partial V}{\partial x} \bigg)_{y,z,t}dx+\bigg( \cfrac{\partial V}{\partial y} \bigg)_{x,z,t}dy+\bigg( \cfrac{\partial V}{\partial z} \bigg)_{y,x,t}dz+\bigg( \cfrac{\partial V}{\partial t} \bigg)_{x,y,z}dt\\
\cfrac{dV(x,y,z,t)}{dt}=\bigg( \cfrac{\partial V}{\partial x} \bigg)_{y,z,t}\cfrac{dx}{dt}+\bigg( \cfrac{\partial V}{\partial y} \bigg)_{x,z,t}\cfrac{dy}{dt}+\bigg( \cfrac{\partial V}{\partial z} \bigg)_{y,x,t}\frac{dz}{dt}+\bigg( \cfrac{\partial V}{\partial t} \bigg)_{x,y,z}\\
\vec{a}=\cfrac{dV(x,y,z,t)}{dt}=\bigg( \cfrac{\partial V}{\partial x} \bigg)_{y,z,t}V_x+\bigg( \cfrac{\partial V}{\partial y} \bigg)_{x,z,t}V_y+\bigg( \cfrac{\partial V}{\partial z} \bigg)_{y,x,t}V_z+\bigg( \cfrac{\partial V}{\partial t} \bigg)_{x,y,z}
\\ V = 2 x 3 i ^ − 5 x 2 y j ^ + 4 t k ^ d V ( x , y , z , t ) = ( ∂ x ∂ V ) y , z , t d x + ( ∂ y ∂ V ) x , z , t d y + ( ∂ z ∂ V ) y , x , t d z + ( ∂ t ∂ V ) x , y , z d t d t d V ( x , y , z , t ) = ( ∂ x ∂ V ) y , z , t d t d x + ( ∂ y ∂ V ) x , z , t d t d y + ( ∂ z ∂ V ) y , x , t d t d z + ( ∂ t ∂ V ) x , y , z a = d t d V ( x , y , z , t ) = ( ∂ x ∂ V ) y , z , t V x + ( ∂ y ∂ V ) x , z , t V y + ( ∂ z ∂ V ) y , x , t V z + ( ∂ t ∂ V ) x , y , z Evaluating all four terms in above equation.....
V x ∂ V ∂ x = 12 x 5 i ^ − 20 x 4 y j ^ . . . . . . . . E q [ 1 ] V y ∂ V ∂ y = 25 x 4 y j ^ . . . . . . . . E q [ 2 ] V z ∂ V ∂ z = 0........ E q [ 3 ] ∂ V ∂ t = 4 k ^ . . . . . . . . E q [ 4 ] V_x\cfrac{\partial V}{\partial x}=12x^5\hat{i}-20x^4y\hat{j}........Eq[1]\\
V_y\cfrac{\partial V}{\partial y}=25x^4y\hat{j}........Eq[2]\\
V_z\cfrac{\partial V}{\partial z}=0........Eq[3]\\
\cfrac{\partial V}{\partial t}=4\hat{k}........Eq[4] V x ∂ x ∂ V = 12 x 5 i ^ − 20 x 4 y j ^ ........ Eq [ 1 ] V y ∂ y ∂ V = 25 x 4 y j ^ ........ Eq [ 2 ] V z ∂ z ∂ V = 0........ Eq [ 3 ] ∂ t ∂ V = 4 k ^ ........ Eq [ 4 ] Using all above four equations in expression of acceleration, we have....
a ⃗ = 12 x 5 i ^ + 5 x 4 y j ^ + 4 k ^ \vec{a}=12x^5\hat{i}+5x^4y\hat{j}+4\hat{k} a = 12 x 5 i ^ + 5 x 4 y j ^ + 4 k ^ Now at point (x,y,z,t)=(1,2,3,1)...
V ⃗ = 2 i ^ − 10 j ^ + 4 k ^ a ⃗ = 12 i ^ + 10 j ^ + 4 k ^ . . . . . . A n s \vec V=2\hat{i}-10\hat{j}+4\hat{k} \\
\vec a=12\hat{i}+10\hat{j}+4\hat{k}......Ans V = 2 i ^ − 10 j ^ + 4 k ^ a = 12 i ^ + 10 j ^ + 4 k ^ ...... A n s
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