Tangent Line Dbeaba

1. **State the problem:** We want to find the equation of the tangent line to the curve defined implicitly by $$\frac{1}{3} y^3 + 2xy - 100 = 0$$ at the point $(-2, 2)$. 2. **Recall the formula:** The tangent line at a point $(x_0,y_0)$ has the equation $$y - y_0 = m(x - x_0)$$ where $m$ is the slope of the tangent line at that point. 3. **Find the slope $m$:** Since the curve is given implicitly, we use implicit differentiation to find $\frac{dy}{dx}$. Differentiate both sides of $$\frac{1}{3} y^3 + 2xy - 100 = 0$$ with respect to $x$. 4. **Differentiate term-by-term:** - For $\frac{1}{3} y^3$, use the chain rule: $$\frac{d}{dx}\left(\frac{1}{3} y^3\right) = y^2 \frac{dy}{dx}$$ - For $2xy$, use the product rule: $$\frac{d}{dx}(2xy) = 2 \left(y + x \frac{dy}{dx}\right)$$ - The derivative of $-100$ is $0$. 5. **Write the differentiated equation:** $$y^2 \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} = 0$$ 6. **Group terms with $\frac{dy}{dx}$:** $$\left(y^2 + 2x\right) \frac{dy}{dx} = -2y$$ 7. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{-2y}{y^2 + 2x}$$ 8. **Evaluate the slope at the point $(-2, 2)$:** Calculate numerator: $$-2 \times 2 = -4$$ Calculate denominator: $$2^2 + 2 \times (-2) = 4 - 4 = 0$$ 9. **Interpretation:** The denominator is zero, so the slope is undefined at $(-2, 2)$. This means the tangent line is vertical. 10. **Equation of vertical tangent line:** Since the tangent line is vertical at $x = -2$, the equation is $$x = -2$$ **Final answer:** The equation of the tangent line to the curve at $(-2, 2)$ is $$x = -2$$.