Question

Math

Posted 5 months ago

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Consider the curve given by the equation $x y^{2}-x^{3} y=8$. It can be shown that $\frac{d y}{d x}=\frac{3 x^{2} y-y^{2}}{2 x y-x^{3}}$.
Write the equation of the vertical line that is tangent to the curve.
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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

To find the vertical tangent, we need to identify where the denominator of $\frac{dy}{dx}$ is zero, since vertical tangents occur when the slope of the tangent is undefined (which happens when the denominator is zero)

step 2

Set the denominator of $\frac{dy}{dx}$ equal to zero: $2xy - x^3 = 0$

step 3

Factor out the common term $x$: $x(2y - x^2) = 0$

step 4

Solve for $y$ when $x \neq 0$: $2y - x^2 = 0 \Rightarrow y = \frac{x^2}{2}$

step 5

Substitute $y = \frac{x^2}{2}$ into the original equation $xy^2 - x^3y = 8$ to find the $x$-values where the vertical tangent occurs

step 6

After substitution, we get $x\left(\frac{x^2}{2}\right)^2 - x^3\left(\frac{x^2}{2}\right) = 8$. Simplify this to find the $x$-values

step 7

Simplifying gives $x\frac{x^4}{4} - \frac{x^5}{2} = 8$. This simplifies to $\frac{x^5}{4} - \frac{2x^5}{4} = 8$, which further simplifies to $-\frac{x^5}{4} = 8$

step 8

Solve for $x$: $x^5 = -32$

step 9

Take the fifth root of both sides to find $x$: $x = \sqrt[5]{-32}$

step 10

The fifth root of $-32$ is $-2$, so $x = -2$

step 11

The equation of the vertical line is $x = -2$

Answer

The equation of the vertical line that is tangent to the curve is $x = -2$.

Key Concept

Vertical Tangents and Slope

Explanation

Vertical tangents occur where the slope of the curve is undefined, which corresponds to the points where the denominator of the derivative $\frac{dy}{dx}$ is zero.

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