Question
(a) Topic-3.2- Implicit Differentiation
(b) Topic-2.10- Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
(c) Topic-2.10- Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
(d) Topic- 5.7- Using the Second Derivative Test to Determine Extrema
Consider the function \(y = f (x)\) whose curve is given by the equation \(2y2 − 6 = y sin x\) for \(y > 0\).
(a) Show that \(\frac{dy}{dx}=\frac{ycos x}{4y-sin x}.\)
(b) Write an equation for the line tangent to the curve at the point (0, \(\sqrt{3}\) ).
(c) For \(0 \leq x \leq \pi\) and \(y > 0\), find the coordinates of the point where the line tangent to the curve is horizontal.
(d) Determine whether f has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.
▶️Answer/Explanation
5(a)
\(\frac{d}{dx} \left(2y^2 – 6\right) = \frac{d}{dx} \left(y \sin x\right)
\implies 4y \frac{dy}{dx} = \frac{dy}{dx} \sin x + y \cos x\)
\(\implies 4y \frac{dy}{dx} – \frac{dy}{dx} \sin x = y \cos x \implies \frac{dy}{dx} \left(4y – \sin x\right) = y \cos x\)
\(\implies \frac{dy}{dx} = \frac{y \cos x}{4y – \sin x}\)
5(b)
At the point \((0, \sqrt{3}), \quad \frac{dy}{dx} = \frac{\sqrt{3} \cos 0}{4\sqrt{3} – \sin 0} = \frac{1}{4}\).
An equation for the tangent line is \(y = \sqrt{3} + \frac{1}{4}x\).
5(c)
\(\frac{dy}{dx} = \frac{y \cos x}{4y – \sin x} = 0 \implies y \cos x = 0 \text{ and } 4y – \sin x \neq 0\)
\(y \cos x = 0 \text{ and } y > 0 \implies x = \frac{\pi}{2}\)
\(\text{When } x = \frac{\pi}{2}, \, y \sin x = 2y^2 – 6 \implies y \sin \frac{\pi}{2} = 2y^2 – 6\)
\(\implies y = 2y^2 – 6 \implies 2y^2 – y – 6 = 0\)
\(\implies (2y + 3)(y – 2) = 0 \implies y = 2\)
\(\text{When } x = \frac{\pi}{2} \text{ and } y = 2, \, 4y – \sin x = 8 – 1 \neq 0.\)
\(\text{Therefore, the line tangent to the curve is horizontal at the point } \left(\frac{\pi}{2}, 2\right).\)
5(d)
\(\frac{d^2y}{dx^2} = \frac{(4y – \sin x) \left(\frac{dy}{dx} \cos x – y \sin x\right) – (y \cos x) \left(4 \frac{dy}{dx} – \cos x\right)}{(4y – \sin x)^2} \)
\(\text{When } x = \frac{\pi}{2} \text{ and } y = 2, \)
\(\frac{d^2y}{dx^2} = \frac{(4 \cdot 2 – \sin \frac{\pi}{2}) \left(0 \cdot \cos \frac{\pi}{2} – 2 \cdot \sin \frac{\pi}{2}\right) – (2 \cos \frac{\pi}{2}) \left(4 \cdot 0 – \cos \frac{\pi}{2}\right)}{(4 \cdot 2 – \sin \frac{\pi}{2})^2} \)
\(= \frac{(7)(-2) – (0)(0)}{(7)^2} = \frac{-2}{7} < 0. \)
\(f \text{ has a relative maximum at the point } \left(\frac{\pi}{2}, 2\right) \text{ because } \frac{dy}{dx} = 0 \text{ and } \frac{d^2y}{dx^2} < 0. \)