Problem

Find the slope of the tangent line to the curve \[ x^{2}-3 x y-2 y^{3}=20 \] at the point $(3,-1)$.

Solution

Step 1 :First, we need to find the derivative of the given equation. The derivative of a function gives us the slope of the tangent line at any point. So, we differentiate the given equation implicitly with respect to \(x\).

Step 2 :Taking the derivative of \(x^{2}-3 x y-2 y^{3}=20\) with respect to \(x\), we get \(2x - 3y - 3xy' - 6y^{2}y' = 0\).

Step 3 :Now, we solve this equation for \(y'\), the derivative of \(y\) with respect to \(x\).

Step 4 :Rearranging the terms, we get \(3xy' + 6y^{2}y' = 2x - 3y\).

Step 5 :Factoring out \(y'\), we get \(y'(3x + 6y^{2}) = 2x - 3y\).

Step 6 :So, \(y' = \frac{2x - 3y}{3x + 6y^{2}}\). This is the derivative of \(y\) with respect to \(x\), which gives us the slope of the tangent line at any point \((x, y)\) on the curve.

Step 7 :Now, we substitute the given point \((3, -1)\) into the derivative to find the slope of the tangent line at this point.

Step 8 :Substituting \(x = 3\) and \(y = -1\) into \(y' = \frac{2x - 3y}{3x + 6y^{2}}\), we get \(y' = \frac{2(3) - 3(-1)}{3(3) + 6(-1)^{2}} = \frac{6 + 3}{9 + 6} = \frac{9}{15}\).

Step 9 :Simplifying \(\frac{9}{15}\) gives us \(\frac{3}{5}\).

Step 10 :So, the slope of the tangent line to the curve at the point \((3, -1)\) is \(\boxed{\frac{3}{5}}\).

From Solvely APP
Source: https://solvelyapp.com/problems/8125/

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