Answer to Question #149098 in Calculus for Usman

Question

The velocity vector in a fluid flow is given by V = 2x^3 i - 5x^2 y j + 4t k. Find the velocity and acceleration of a fluid particle at (1, 2, 3) at time , t = 1.

Answer 1

Velocity filed is given as....

$$V=2x^3\hat{i}-5x^2y\hat{j}+4t\hat{k}\\ dV(x,y,z,t)=\bigg( \cfrac{\partial V}{\partial x} \bigg)_{y,z,t}dx+\bigg( \cfrac{\partial V}{\partial y} \bigg)_{x,z,t}dy+\bigg( \cfrac{\partial V}{\partial z} \bigg)_{y,x,t}dz+\bigg( \cfrac{\partial V}{\partial t} \bigg)_{x,y,z}dt\\ \cfrac{dV(x,y,z,t)}{dt}=\bigg( \cfrac{\partial V}{\partial x} \bigg)_{y,z,t}\cfrac{dx}{dt}+\bigg( \cfrac{\partial V}{\partial y} \bigg)_{x,z,t}\cfrac{dy}{dt}+\bigg( \cfrac{\partial V}{\partial z} \bigg)_{y,x,t}\frac{dz}{dt}+\bigg( \cfrac{\partial V}{\partial t} \bigg)_{x,y,z}\\ \vec{a}=\cfrac{dV(x,y,z,t)}{dt}=\bigg( \cfrac{\partial V}{\partial x} \bigg)_{y,z,t}V_x+\bigg( \cfrac{\partial V}{\partial y} \bigg)_{x,z,t}V_y+\bigg( \cfrac{\partial V}{\partial z} \bigg)_{y,x,t}V_z+\bigg( \cfrac{\partial V}{\partial t} \bigg)_{x,y,z} \\$$

Evaluating all four terms in above equation.....

$$V_x\cfrac{\partial V}{\partial x}=12x^5\hat{i}-20x^4y\hat{j}........Eq[1]\\ V_y\cfrac{\partial V}{\partial y}=25x^4y\hat{j}........Eq[2]\\ V_z\cfrac{\partial V}{\partial z}=0........Eq[3]\\ \cfrac{\partial V}{\partial t}=4\hat{k}........Eq[4]$$

Using all above four equations in expression of acceleration, we have....

$$\vec{a}=12x^5\hat{i}+5x^4y\hat{j}+4\hat{k}$$

Now at point (x,y,z,t)=(1,2,3,1)...

$\vec V=2\hat{i}-10\hat{j}+4\hat{k} \\ \vec a=12\hat{i}+10\hat{j}+4\hat{k}......Ans$