Question:
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 ≤ x ≤ π 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
Ans:
(a)
2y2-6 = y sin x
4y y’ = y cos x + y’ sin x
4y y’ – y’ sin x = y cos x
y'(4y-sinx) = y cos x
\(y’ = \frac{ycos x}{4y-sinx}\)
\(\frac{dy}{dx} = \frac{ycos x}{4y-sinx}\)
(b)
\(\frac{dy}{dx}At (0,\sqrt{3})=\frac{\sqrt{3}cos(0)}{4\sqrt{3}-sin(0)}=\frac{\sqrt{3}(1)}{4\sqrt{3}-(0)}=\frac{\sqrt{3}}{4\sqrt{3}}=\frac{1}{4}\)
\(y-\sqrt{3}=\frac{1}{4}(x-0)\)
(c)
2y2 – 6 = y sin x
\(\frac{dy}{dx}=\frac{ycos x}{4y – sinx}\) y ≠0 : -6 = 0 sinx
\(0=\frac{ycos x}{4y – sinx}\)
\(x = \frac{\pi }{2}: 2y^{2}-6=y sin(\frac{\pi }{2})\)
\(2y^{2}-6=y (x)\)
0 = y cos x 2y2 – y-6 =0
y ≠ 0 x ≠ 0 (2y + 3) (y – 2) = 0
Since
0 = y cos x cos (0) = 1 y = 2 2y + 3 = 0
x = π/2 and 2y = -3
0 ≠ 1
f(x) has a horizontal tangent at \(\left ( \frac{\pi }{2},2 \right )\)
(d)
Since \(\frac{d^{2}y}{dx^{2}}<0\) at \(\left ( \frac{\pi }{2},2 \right )\), f has a relative maximum at that point.
\(\frac{dy}{dx}=\frac{y cos x}{4y-sin x}\)
\(\frac{d^{2}y}{dx^{2}} = \frac{(4y-sinx)\left [ y(-sinx)+y’cos x \right ]-(y cos x)(4y’-cosx)}{(4y-sin x)^{2}}\)
At \(\left ( \frac{\pi }{2},2 \right )\) : \(\frac{d^{2}y}{dx^{2}} = \frac{(4.2-1)\left [ 2(-1)+0 cos x \right ]-(2(0))(4(0)-0)}{(4(2)-1)^{2}}\)
\(\frac{d^{2}y}{dx^{2}} = \frac{(8-1)(-2)}{(8-1)^{2}}=\frac{7(-2)}{7}=\frac{-2}{7}\)
concave down