6) Find the equation of a tangent line to the curve \( x^{2} y+y^{3}=75 \) at the point \( (4,3) \)

To find the equation of the tangent line to the curve defined by the equation \( x^{2} y + y^{3} = 75 \) at the point \( (4, 3) \), we will follow these steps: 1. **Differentiate the equation implicitly** to find the slope of the tangent line. 2. **Evaluate the derivative at the point** \( (4, 3) \) to find the slope at that point. 3. **Use the point-slope form of the line** to write the equation of the tangent line. ### Step 1: Differentiate the equation implicitly We start with the equation: \[ x^{2} y + y^{3} = 75 \] Differentiating both sides with respect to \( x \): \[ \frac{d}{dx}(x^{2} y) + \frac{d}{dx}(y^{3}) = \frac{d}{dx}(75) \] Using the product rule on \( x^{2} y \) and the chain rule on \( y^{3} \): \[ 2xy + x^{2} \frac{dy}{dx} + 3y^{2} \frac{dy}{dx} = 0 \] Now, we can factor out \( \frac{dy}{dx} \): \[ x^{2} \frac{dy}{dx} + 3y^{2} \frac{dy}{dx} = -2xy \] \[ \frac{dy}{dx}(x^{2} + 3y^{2}) = -2xy \] Now, we can solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-2xy}{x^{2} + 3y^{2}} \] ### Step 2: Evaluate the derivative at the point \( (4, 3) \) Substituting \( x = 4 \) and \( y = 3 \) into the derivative: \[ \frac{dy}{dx} = \frac{-2(4)(3)}{(4^{2}) + 3(3^{2})} \] Calculating the numerator: \[ -2(4)(3) = -24 \] Calculating the denominator: \[ 4^{2} = 16 \quad \text{and} \quad 3(3^{2}) = 27 \quad \Rightarrow \quad 16 + 27 = 43 \] Thus, we have: \[ \frac{dy}{dx} = \frac{-24}{43} \] ### Step 3: Use the point-slope form of the line The point-slope form of the line is given by: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. Substituting \( m = \frac{-24}{43} \), \( x_1 = 4 \), and \( y_1 = 3 \): \[ y - 3 = \frac{-24}{43}(x - 4) \] Now, we can rearrange this to find the equation of the tangent line: \[ y - 3 = \frac{-24}{43}x + \frac{96}{43} \] \[ y = \frac{-24}{43}x + \frac{96}{43} + 3 \] \[ y = \frac{-24}{43}x + \frac{96}{43} + \frac{129}{43} \] \[ y = \frac{-24}{43}x + \frac{225}{43} \] Thus, the equation of the tangent line to the curve at the point \( (4, 3) \) is: \[ y = \frac{-24}{43}x + \frac{225}{43} \]