Question
(a)-Topic-7.2 Verifying Solutions for Differential Equations
(b)-Topic-7.3 Sketching Slope Fields
(c)-Topic-7.4 Reasoning Using Slope Fields
(d)-Topic-7.4 Reasoning Using Slope Fields
6. Consider the curve given by the equation \(6xy=2+y^{3}\) .
(a) Show that \(\frac{dy}{dx}=\frac{2y}{y^{2}-2x}\).
(b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
(c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
(d) A particle is moving along the curve. At the instant when the particle is at the point\(\left ( \frac{1}{2′}-2 \right )\),its horizontal position is increasing at a rate of \(\frac{dx}{dt}=\frac{2}{3}\) unit per second. What is the value of \(\frac{dy}{dt}\) , the rate of change of the particle’s vertical position, at that instant?
▶️Answer/Explanation
6(a) \(\frac{dy}{dx} = \frac{2y}{y^{2} – 2x}.\)
\(\frac{d}{dx}(6xy) = \frac{d}{dx}(2+y^{3}) \Rightarrow 6y +6x\frac{dy}{dx} = 3y^{2}\frac{dy}{dx}\)
\(\Rightarrow 2y = \frac{dy}{dx}(y^{2} -2x) \Rightarrow \frac{dy}{dx} = \frac{2y}{y^{2}- 2x}\)
6(b) Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
For the line tangent to the curve to be horizontal, it is necessary that 2 0 y = (so y = 0 ) and that \(y^{2} – 2x \neq 0\)
Substituting y = 0 into \(6xy= 2-y^{3}\) yields the equation \(6x .0, x ⋅ = 2\) which has no solution.
Therefore, there is no point on the curve at which the line tangent to the curve is horizontal.
6(c) Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
For a line tangent to this curve to be vertical, it is necessary that \(2y\neq 0\) and that \(y^{2} – 2x = 0 (so x = \frac{y^{2}}{2})\).
Substituting \(x = \frac{y^{2}}{2}\) into \(6xy= 2-y^{3}\) yields the equation \(3y^{2} .y = 2 + y^{3} \Rightarrow 2y^{3} = 2\Rightarrow y = 1.\)
Substituting y = 1 in \(6xy= 2-y^{3}\) yields 6x = 2+1, or x \(\frac{1}{2}\). The tangent line to the curve is vertical at the point \((\frac{1}{2}, 1)\).
6(d) A particle is moving along the curve. At the instant when the particle is at the point \((\frac{1}{2}, -2)\), its horizontal position is increasing at a rate of \(\frac{dx}{dt}\) = \(\frac{2}{3}\) unit per second. What is the value of \(\frac{dy}{dt}\), the rate of change of the particle’s vertical position, at that instant?
\(6y\frac{dx}{dt} + 6x\frac{dy}{dt} = 0 + 3y^{2}\frac{dy}{dt}\)
At the point (x, y) = \((\frac{1}{2}, -2)\),
\(6(-2)(\frac{2}{3}) + 6(\frac{1}{2})\frac{dy}{dt} = 3(-2)^{2} \frac{dy}{dt}\)
\(\Rightarrow -8 + 3\frac{dy}{dt} = 12\frac{dy}{dt}\)
\(\Rightarrow \frac{dy}{dt} = -\frac{8}{9}\) unit per second