Question 5

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A curve is defined by $x^3 + 2y^2 + 2y = 48$ . It can be shown that $\dfrac{dy}{dx} = \dfrac{-3x^2}{4y + 2}$ .

$\text{(a)}$
A point on the curve near $(2, 4)$ has $x$ -coordinate $3$ . Use the tangent line at $(2, 4)$ to approximate its $y$ -coordinate.
$\text{(b)}$
Is the horizontal line $y = 1$ tangent to the curve? Give a reason.
$\text{(c)}$
The curve meets the positive $x$ -axis at $\left(\sqrt[3]{48},\, 0\right)$ . Is the tangent line there vertical? Give a reason.
$\text{(d)}$
For $t \ge 0$ , a particle moves along a different curve, $y^3 + 2xy = 24$ . At the moment it is at $(4, 2)$ , its $y$ -coordinate is decreasing at $2$ units per second. At that moment, how fast is its $x$ -coordinate changing?

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Understanding the Question

This question is about a curve defined implicitly, and tangent lines to it

Part 1 of 5

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The question

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A curve is defined by $x^3 + 2y^2 + 2y = 48$ . It can be shown that $\dfrac{dy}{dx} = \dfrac{-3x^2}{4y + 2}$ .

$\text{(a)}$
A point on the curve near $(2, 4)$ has $x$ -coordinate $3$ . Use the tangent line at $(2, 4)$ to approximate its $y$ -coordinate.
$\text{(b)}$
Is the horizontal line $y = 1$ tangent to the curve? Give a reason.
$\text{(c)}$
The curve meets the positive $x$ -axis at $\left(\sqrt[3]{48},\, 0\right)$ . Is the tangent line there vertical? Give a reason.
$\text{(d)}$
For $t \ge 0$ , a particle moves along a different curve, $y^3 + 2xy = 24$ . At the moment it is at $(4, 2)$ , its $y$ -coordinate is decreasing at $2$ units per second. At that moment, how fast is its $x$ -coordinate changing?

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This question is about a curve defined implicitly, and tangent lines to it

Part 1 of 5