Math · Calculus I · Worked example

Find a tangent slope on x³ + y³ = 6xy

The curve $x^{3} + y^{3} = 6 x y$ , the folium of Descartes, passes through $(3, 3)$ . Find $d y / d x$ and the tangent line at that point.

x^{3} + y^{3} = 6 x y

Check the point

$3^{3} + 3^{3} = 54$ and $6 \cdot 3 \cdot 3 = 54$ , so $(3, 3)$ is on the curve.

3^{3} + 3^{3} = 54 = 6 (3) (3)

Differentiate both sides

The term $x^{3}$ gives $3 x^{2}$ . The term $y^{3}$ is a function of a function, so it gives $3 y^{2} \cdot d y / d x$ . The right side $6 x y$ is a product of $x$ and $y$ , and the product rule gives $6 y + 6 x \cdot d y / d x$ .

3 x^{2} + 3 y^{2} \frac{d y}{d x} = 6 y + 6 x \frac{d y}{d x}

\begin{aligned} 3 x^{2} + 3 y^{2} \frac{d y}{d x} \\ = 6 y + 6 x \frac{d y}{d x} \end{aligned}

Collect the $d y / d x$ terms

Unlike the circle, $d y / d x$ appears on both sides here. Move every term with $d y / d x$ to the left and every other term to the right, then factor out $d y / d x$ .

(3 y^{2} - 6 x) \frac{d y}{d x} = 6 y - 3 x^{2}

Solve for $d y / d x$

Divide by $3 y^{2} - 6 x$ , which needs it to be nonzero, and cancel the common factor $3$ .

\frac{d y}{d x} = \frac{2 y - x^{2}}{y^{2} - 2 x}

Evaluate at $(3, 3)$

The top is $2 \cdot 3 - 9 = - 3$ and the bottom is $9 - 6 = 3$ , which makes the slope $- 1$ . The nudge in the opening case gave about $- 1.03$ .

{\frac{d y}{d x} |}_{(3, 3)} = \frac{6 - 9}{9 - 6} = - 1

Write the tangent line

Point-slope form through $(3, 3)$ with slope $- 1$ gives $y - 3 = - (x - 3)$ , which is $y = - x + 6$ . The curve is symmetric about the line $y = x$ , and the tangent crosses that line at a right angle.

y = - x + 6

Result

$d y / d x$ $= (2 y - x^{2}) / (y^{2} - 2 x)$ , and the tangent at $(3, 3)$ is $y = - x + 6$ .

\frac{d y}{d x} = \frac{2 y - x^{2}}{y^{2} - 2 x}

Your turn

1. One like the example

Find $d y / d x$ for $x^{3} + y^{3} = 2 x y$ , and the slope at $(1, 1)$ .

Show the answer and explanation

$d y / d x$ $= (2 y - 3 x^{2}) / (3 y^{2} - 2 x)$ , and the slope at $(1, 1)$ is $- 1$ .

The point is on the curve, since $1 + 1 = 2$ . Differentiating gives $3 x^{2} + 3 y^{2} \cdot d y / d x$ $= 2 y + 2 x \cdot d y / d x$ . Collecting gives $(3 y^{2} - 2 x) d y / d x$ $= 2 y - 3 x^{2}$ , and dividing gives the formula. At $(1, 1)$ it is $(2 - 3) / (3 - 2) = - 1$ .

\begin{array}{l} x^{3} + y^{3} = 2 x y \\ 3 x^{2} + 3 y^{2} \frac{d y}{d x} = 2 y + 2 x \frac{d y}{d x} \\ \frac{d y}{d x} = \frac{2 y - 3 x^{2}}{3 y^{2} - 2 x} \end{array}

2. A variation

Find $d y / d x$ for $x^{2} y + y^{2} = 5$ , and the slope at $(2, 1)$ . Here the product is on the left.

Show the answer and explanation

$d y / d x = - 2 x y / (x^{2} + 2 y)$ , and the slope at $(2, 1)$ is $- 2 / 3$ .

The point is on the curve, since $4 + 1 = 5$ . The term $x^{2} y$ is a product, with derivative $2 x y + x^{2} \cdot d y / d x$ , and $y^{2}$ gives $2 y \cdot d y / d x$ . That makes $2 x y + (x^{2} + 2 y) d y / d x = 0$ , so $d y / d x = - 2 x y / (x^{2} + 2 y)$ . At $(2, 1)$ that is $- 4 / 6 = - 2 / 3$ .

\begin{array}{l} x^{2} y + y^{2} = 5 \\ 2 x y + x^{2} \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0 \\ \frac{d y}{d x} = - \frac{2 x y}{x^{2} + 2 y} \end{array}

3. Somewhere new

Where does the curve $x^{2} - x y + y^{2} = 3$ have a horizontal tangent?

Show the answer and explanation

At $(1, 2)$ and $(- 1, - 2)$ .

Differentiating gives $2 x - y - x \cdot d y / d x + 2 y \cdot d y / d x$ $= 0$ , so $d y / d x$ $= (y - 2 x) / (2 y - x)$ . A horizontal tangent needs $d y / d x = 0$ , which means $y = 2 x$ with $2 y - x \neq 0$ . Putting $y = 2 x$ into the curve gives $x^{2} - 2 x^{2} + 4 x^{2} = 3 x^{2} = 3$ , so $x = \pm 1$ . The points are $(1, 2)$ and $(- 1, - 2)$ , where $2 y - x$ is $3$ and $- 3$ .

\begin{array}{l} x^{2} - x y + y^{2} = 3 \\ 2 x - y - x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0 \\ \frac{d y}{d x} = \frac{y - 2 x}{2 y - x} \end{array}

Keep exploring

Open the steps in Math: the Checker marks every row ✓. Drop the $6 y$ from the second row and it reads ✗, because the product rule gives $6 x y$ two terms.

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Check the point

Differentiate both sides

Collect the d⁢y⁢/⁢d⁢xdy/dx terms

Solve for d⁢y⁢/⁢d⁢xdy/dx

Evaluate at (3,3)(3,3)

Write the tangent line

Result

Your turn

1. One like the example

d⁢y⁢/⁢d⁢xdy/dx =(2⁢y−3x2)⁢/⁢(3y2−2⁢x)=(2y−3x2)/(3y2−2x), and the slope at (1,1)(1,1) is −1−1.

2. A variation

d⁢y⁢/⁢d⁢x=−2⁢x⁢y⁢/⁢(x2+2⁢y)dy/dx=−2xy/(x2+2y), and the slope at (2,1)(2,1) is −2⁢/3−2/3.

3. Somewhere new

At (1,2)(1,2) and (−1,−2)(−1,−2).

Keep exploring

Study it, then try it

Collect the $d y / d x$ terms

Solve for $d y / d x$

Evaluate at $(3, 3)$

$d y / d x$ $= (2 y - 3 x^{2}) / (3 y^{2} - 2 x)$ , and the slope at $(1, 1)$ is $- 1$ .

$d y / d x = - 2 x y / (x^{2} + 2 y)$ , and the slope at $(2, 1)$ is $- 2 / 3$ .

At $(1, 2)$ and $(- 1, - 2)$ .