Question

consider the curve defined by ( 9x - 36y + y^{3}=0 ) (see figure)
a. find the coordinates of the y intercepts of the curve
b. find ( \frac{dy}{dx} )
c. find the slope of the curve at each point where ( x = 0 )
a. the coordinates of the y - intercepts are (type ordered pairs. use a comma to separate answers as needed.)
b. ( \frac{dy}{dx}=)
c. there are three points on the curve at which ( x = 0 ). at the topmost point, the slope of the curve is. at the middle point, the slope of the curve is. at the bottommost point, the slope of the curve is

Explanation:

Step1: Find y - intercepts

For y - intercepts, set \(x = 0\) in the equation \(9x-36y + y^{3}=0\).
We get \(y^{3}-36y=0\), factor out \(y\): \(y(y^{2}-36)=0\), then \(y(y - 6)(y + 6)=0\).
So \(y=0\), \(y = 6\), \(y=-6\). The coordinates of y - intercepts are \((0,0)\), \((0,6)\), \((0,-6)\).

Step2: Differentiate implicitly

Differentiate \(9x-36y + y^{3}=0\) with respect to \(x\).
Using the sum rule and chain rule: \(\frac{d}{dx}(9x)-\frac{d}{dx}(36y)+\frac{d}{dx}(y^{3})=\frac{d}{dx}(0)\).
\(9-36\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0\).
Group the terms with \(\frac{dy}{dx}\): \((3y^{2}-36)\frac{dy}{dx}=-9\).
Then \(\frac{dy}{dx}=\frac{-9}{3y^{2}-36}=\frac{3}{12 - y^{2}}\).

Step3: Find slopes when \(x = 0\)

When \(x = 0\), \(y=-6\), \(y = 0\), \(y = 6\).
For \(y = 6\): \(\frac{dy}{dx}=\frac{3}{12-36}=-\frac{1}{8}\).
For \(y = 0\): \(\frac{dy}{dx}=\frac{3}{12-0}=\frac{1}{4}\).
For \(y=-6\): \(\frac{dy}{dx}=\frac{3}{12 - 36}=-\frac{1}{8}\).

Answer:

a. \((0,0)\), \((0,6)\), \((0,-6)\)
b. \(\frac{3}{12 - y^{2}}\)
c. At the top - most point (\(y = 6\)): \(-\frac{1}{8}\); at the middle point (\(y = 0\)): \(\frac{1}{4}\); at the bottom - most point (\(y=-6\)): \(-\frac{1}{8}\)