Acceleration and streamline curvature, u = Axy

Given: $u = Axy$, $A = 2\ \text{m}^{-1}\text{s}^{-1}$, 2-D incompressible, steady, no $w$.

Step 1: find v from continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0\;\Rightarrow\;\frac{\partial v}{\partial y} = -Ay\;\Rightarrow\; v = -\frac{Ay^2}{2}$$

(taking $v = 0$ at $y = 0$). At $(2, 1)$:

$$u = 2\times 2\times 1 = 4\ \text{m/s},\qquad v = -\frac{2\times 1}{2} = -1\ \text{m/s}$$

Step 2: acceleration (steady flow, so only the convective part)

$$\begin{aligned} a_x &= u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = u(Ay) + v(Ax)\\ &= 4(2) + (-1)(4) = 4\ \text{m/s}^2\\ a_y &= u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = 0 + (-1)(-Ay) = (-1)(-2) = 2\ \text{m/s}^2 \end{aligned}$$ $$\vec a = 4\hat i + 2\hat j,\qquad |a| = 4.472\ \text{m/s}^2\ \text{at } 26.6^\circ \text{ above +x}$$

Step 3: radius of curvature. The normal component of acceleration is $a_n = V^2/R$ and $a_n = |\vec V\times\vec a|/|V|$.

$$\begin{aligned} |V| &= \sqrt{4^2 + 1^2} = 4.123\ \text{m/s}\\ |\vec V\times\vec a| &= |u a_y - v a_x| = |4(2) - (-1)(4)| = 12\\ R &= \frac{|V|^3}{|\vec V\times\vec a|} = \frac{4.123^3}{12} = 5.84\ \text{m} \end{aligned}$$

Step 4: streamline.

$$\frac{dy}{dx} = \frac{v}{u} = \frac{-Ay^2/2}{Axy} = -\frac{y}{2x}\;\Rightarrow\; \frac{dy}{y} = -\frac{dx}{2x}\;\Rightarrow\; xy^2 = C$$

Through $(2,1)$: $C = 2$, so $xy^2 = 2\ \text{m}^3$.

 y
 3 |.
 2 |  .
 1.5| .
 1 |    .--->V(4,-1) at (2,1)
   |     ^ .
   |   a(4,2)   .   .
 0.5|            .      .       .
   +----------------------------- x
   0    1    2    3   ...   8

At (2,1) the velocity vector is directed down-right (slope $-0.25$, tangent to the streamline); the acceleration vector $(4, 2)$ points up-right, and its normal component points towards the centre of curvature (the concave side of the streamline).

Answer: $\vec a = (4\hat i + 2\hat j)$ m/s², $|a| = 4.47$ m/s²; $R = 5.84$ m; streamline $xy^2 = 2$.