Let's go through the step-by-step solution to go from the potential function, \[ \phi = Ux + \frac{q}{2\pi} \ln(r) \] to the velocity component, \[ u = \frac{\partial \phi}{\partial x} = U + \frac{q}{2\pi} \frac{\cos\theta}{r} \] where \( r \) and \( \theta \) are polar coordinates. --- ### **Step 1: Write Down the Potential Function** Given: \[ \phi = Ux + \frac{q}{2\pi} \ln(r) \] where: - \( U \) is a constant (could be a uniform flow speed) - \( q \) is a constant (could be a source strength) - \( x \) is the Cartesian coordinate - \( r \) is the radial (polar) coordinate, \( r = \sqrt{x^2 + y^2} \) --- ### **Step 2: Take the Partial Derivative with Respect to \( x \)** \[ u = \frac{\partial \phi}{\partial x} \] Let's differentiate term by term: - \( \frac{\partial}{\partial x}(Ux) = U \) - \( \frac{\partial}{\partial x} \left(\frac{q}{2\pi} \ln(r)\right) = \frac{q}{2\pi} \cdot \frac{\partial}{\partial x}(\ln(r)) \) Recall that: \[ \frac{\partial}{\partial x} \ln(r) = \frac{1}{r} \frac{\partial r}{\partial x} \] where \( r = \sqrt{x^2 + y^2} \). So, \[ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} \] Therefore, \[ \frac{\partial}{\partial x} \ln(r) = \frac{1}{r} \cdot \frac{x}{r} = \frac{x}{r^2} \] So, \[ u = U + \frac{q}{2\pi} \cdot \frac{x}{r^2} \] --- ### **Step 3: Express \( x \) in Terms of Polar Coordinates** Recall: - \( x = r \cos\theta \) - \( r^2 = x^2 + y^2 \) Substitute \( x \): \[ u = U + \frac{q}{2\pi} \cdot \frac{r\cos\theta}{r^2} = U + \frac{q}{2\pi} \cdot \frac{\cos\theta}{r} \] --- ### **Final Answer** \[ \boxed{ u = U + \frac{q}{2\pi} \frac{\cos\theta}{r} } \] **Explanation:** - Take the derivative of \( \phi \) with respect to \( x \). - Substitute the polar form for \( x \). - Simplify. Let me know if you need the same for the \( v \) (y-component)!