Q. 83 - Calculus

Question

Consider the graph of the solutions of the equation $y^3+xy+2=0$

(a) Find all points on the graph with an x-coordinate of x = 1, and then find the slope of the tangent line at each of these points.

(b) Find all points on the graph with a y-coordinate of y = 1, and then find the slope of the tangent line at each of these points.

(c) Find all points where the graph has a horizontal tangent line.

(d) Find all points where the graph has a vertical tangent line.

Step-by-Step Solution

Answer

Part (a): $$\begin{gathered} \left(1, -2\right) \mathrm{is} \mathrm{the} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{2}{13} \end{gathered}$$

Part (b): $$\begin{gathered} \left(-3,1\right) \mathrm{is} \mathrm{the} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}} =-\infty \end{gathered}$$

Part (c): $\mathrm{There} \mathrm{is} \mathrm{no} \mathrm{point} \mathrm{where} \mathrm{tangent} \mathrm{line} \mathrm{is} \mathrm{horizontal}.$

Part (d): $\mathrm{There} \mathrm{is} \mathrm{no} \mathrm{point} \mathrm{where} \mathrm{tangent} \mathrm{line} \mathrm{is} \mathrm{vertical}.$

Step 1. Given information is:

$y^3+xy+2=0$

Part (a) Step 1. Calculating dy/dx

$$\begin{gathered} \mathrm{Here}, \\ {y}^3+xy+2=0 \\ \frac{d\left({\mathrm{y}}^3+\mathrm{xy}+2 \right)}{d\mathrm{x}} = 0 (\mathrm{Differentiating} \mathrm{both} \mathrm{sides}) \\ 3{\mathrm{y}}^2\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{x}\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}\frac{d\mathrm{x}}{d\mathrm{x}}+0= 0 \\ 3{\mathrm{y}}^2\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{x}\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}= 0 \\ (3{\mathrm{y}}^2+\mathrm{x})\frac{d\mathrm{y}}{d\mathrm{x}}+\mathrm{y}= 0 \\ \frac{d\mathrm{y}}{d\mathrm{x}}=-\frac{y}{3{\mathrm{y}}^2+\mathrm{x}} \end{gathered}$$

Part (a) Step 2. Calculating y and slope

$$\begin{gathered} \mathrm{Substituting} \mathrm{x} = 1 \mathrm{into} \mathrm{the} \mathrm{equation} {\mathrm{y}}^3+\mathrm{xy}+2=0: \\ {\mathrm{y}}^3+\left(1\right)\mathrm{y}+2=0 \\ {\mathrm{y}}^3+\mathrm{y}=-2 \\ \mathrm{y}({\mathrm{y}}^2+1)=-2 \\ \mathrm{y}=-2 \\ \mathrm{Thus}, \left(1, -2\right) \mathrm{is} \mathrm{the} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \mathrm{Thus} \mathrm{the} \mathrm{slope} \mathrm{is}: \\ \mathrm{x} = 1, \mathrm{y}=-2 \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{(-2)}{3{(-2)}^2+1} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{2}{3\left(4\right)+1} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{2}{13} \end{gathered}$$

Part (b) Step 1. Calculating x and slope

$$\begin{gathered} \mathrm{Substituting} \mathrm{y} = 1 \mathrm{into} \mathrm{the} \mathrm{equation} {\mathrm{y}}^3+\mathrm{xy}+2=0: \\ {1}^3+\mathrm{x}+2=0 \\ \mathrm{x}=-3 \\ \mathrm{Thus}, \left(-3,1\right) \mathrm{is} \mathrm{the} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \mathrm{Thus} \mathrm{the} \mathrm{slope} \mathrm{is}: \\ \mathrm{when} \mathrm{x} = -3, \mathrm{y}=1 \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{ 1}{3{\left(1\right)}^2+(-3)} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{-1}{0}=-\infty \end{gathered}$$

Part (c) Step 1. Finding points

$$\begin{gathered} \mathrm{Substituting} \mathrm{x} = 0 \mathrm{into} \mathrm{the} \mathrm{equation} {\mathrm{y}}^3+\mathrm{xy}+2=0 \mathrm{for} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{to} \mathrm{be} \mathrm{horizontal}: \\ {\mathrm{y}}^3+\left(0\right)\mathrm{y}+2=0 \\ {\mathrm{y}}^3=-2 \\ \mathrm{Thus}, \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{point} \mathrm{where} \mathrm{tangent} \mathrm{line} \mathrm{is} \mathrm{horizontal}. \end{gathered}$$

Part (d) Step 1. Finding points

$$\begin{gathered} \mathrm{Substituting} \mathrm{y} = 0 \mathrm{into} \mathrm{the} \mathrm{equation} {\mathrm{y}}^3+\mathrm{xy}+2=0 \mathrm{for} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{to} \mathrm{be} \mathrm{vertical}: \\ {0}^3+\mathrm{x}\left(0\right)+2=0 \\ 2=0 \\ \mathrm{Thus}, \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{point} \mathrm{where} \mathrm{tangent} \mathrm{line} \mathrm{is} \mathrm{vertical}. \end{gathered}$$