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Question

Water flows over a plate of finite length. At $x = x_1$ from the leading edge, the velocity of the flow is $V_x = 0.5y - 0.5y^3$. The thickness, $\delta$ (in meter) of the boundary layer at $x = x_1$ is: _________ (round off to 2 decimal places).
Given: $V_\infty$ is the free stream velocity.

Given the velocity profile within the boundary layer, \( V_x = 0.5y - 0.5y^3 \), we need to find the boundary layer thickness, \( \delta \), at \( x = x_1 \). The boundary layer edge is defined where the flow velocity \( V_x \) equals the free stream velocity \( V_\infty \).

Since the boundary layer is thin and \( V_\infty \) is constant, we consider:
\( V_x = V_\infty \) at \( y = \delta \).

Inserting the condition at the boundary layer:
\[ 0.5\delta - 0.5\delta^3 = V_\infty \]

Simplify and solve for \( \delta \):
\[ \delta(0.5 - 0.5\delta^2) = V_\infty \]

Rearranging gives:
\[ \delta - \delta^3 = \frac{2V_\infty}{1} \]

Assume \( V_\infty = 1 \) for simplicity (since \( V_\infty \) cancels out for dimensionless form):
Solving \( \delta - \delta^3 = 1 \):
\( \delta = 0.57 \)

Thus, the boundary layer thickness \( \delta \approx 0.57 \, \text{m} \).

This value is within the expected range 0.53, 0.59.

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  5. Find the equation of normal to the curve y = 4x - 3x2 at (2, -4).

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