44. Given that a curve has a gradient function \( \frac{1}{(3-x)^{2}} \) and it passes through the point \( \left(2,1 \frac{1}{2}\right) \), find the equation of the curve.
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To find the equation of the curve, we first need to integrate the gradient function. The gradient function given is \( \frac{1}{(3-x)^{2}} \). Integrating this, we have: \[ \int \frac{1}{(3-x)^{2}} \, dx = -\frac{1}{3-x} + C \] where \( C \) is the constant of integration. Now, we need to find \( C \) using the fact that the curve passes through the point \( (2, 1.5) \): Substituting \( x = 2 \) into the equation: \[ y = -\frac{1}{3-2} + C = -1 + C \] Given that \( y = 1.5 \) when \( x = 2 \): \[ 1.5 = -1 + C \Rightarrow C = 2.5 \] Thus, the equation of the curve is: \[ y = -\frac{1}{3-x} + 2.5 \] Or, rearranging gives us: \[ y = 2.5 - \frac{1}{3-x} \] This is the equation of the curve that satisfies the given conditions.