Implicit differentiation and inverse trigonometric functions, question 28

Topic: Implicit differentiation and inverse trigonometric functions
Difficulty: Challenging
Suggested time: 6 min
Marks: 4
Theory: UBC Math 100 §4.5 CLP-1 §2.11Key results

The curve $x^{2} + xy + y^{2} = 3$ has exactly two points at which its tangent line is horizontal. Find the $y$ -coordinate of the higher of those two points.

Your answer

A number.

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Theory UBC Math 100 §4.5 CLP-1 §2.11Key results

Worked solution

Strategy

Implicit differentiation gives y′ as a fraction; horizontal tangents are where its numerator vanishes. That condition relates x and y, and substituting it into the original equation locates the points, whose y-coordinates can then be compared.

Working

$2x + y + x\,y' + 2y\,y' = 0 \quad\Longrightarrow\quad y' = -\frac{2x + y}{x + 2y}$
Product rule on xy; collect the y′ terms.
$y' = 0 \iff 2x + y = 0 \iff y = -2x \qquad (\text{with } x + 2y \neq 0)$
Horizontal tangent: the numerator is zero and the denominator is not.
$x^{2} + x(-2x) + (-2x)^{2} = 3 \quad\Longrightarrow\quad 3x^{2} = 3 \quad\Longrightarrow\quad x = \pm 1$
Substitute y = −2x into the curve.
$x = 1 \Rightarrow (1, -2), \qquad x = -1 \Rightarrow (-1, 2)$
Each y from y = −2x. At both points x + 2y = −3x ≠ 0, so the tangents really are horizontal.
$\text{higher point } (-1, 2):\quad y = 2$
Check on the curve: 1 − 2 + 4 = 3. This is the highest point of the whole loop.

Answer

y = 2, \text{ at the point } (-1, 2)

Check

As a quadratic in x, x² + yx + y² − 3 = 0 has real roots only if its discriminant 12 − 3y² ≥ 0, so y ≤ 2 everywhere on the curve.

Where this goes wrong

Setting the denominator x + 2y to zero instead of the numerator. That finds the points with vertical tangents, (2, −1) and (−2, 1), and gives 1; a horizontal tangent needs the numerator 2x + y to vanish.

Leans on

The product rule, from Differentiation rules

Try a similar question

The point

(1, 2)

lies on the curve below. Find

\dfrac{dy}{dx}

at that point.

x^{2} + xy + y^{2} = 7

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Implicit differentiation and inverse trigonometric functions practice question 28