Q. 82 - Calculus

Question

Consider the graph of the solutions of the equation $4y^2-x^2+2x=2$

(a) Find all points on the graph with an x-coordinate of x = 3, 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 = 3, 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): $$\begin{gathered} \left(3, -\frac{\sqrt{5}}{2}\right)\left(3, \frac{\sqrt{5}}{2}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{1}{\sqrt{5}} , \frac{1}{\sqrt{5}} \end{gathered}$$

Part (b): $$\begin{gathered} \left(-34, 3\right)\left(36, 3\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{35}{12} , \frac{35}{12} \end{gathered}$$

Part (c): $\left(0,\frac{\sqrt{2}}{2}\right)\left(0,-\frac{\sqrt{2}}{2}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}.$

Part (d): $\left(2,0\right)\left(0,0\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}.$

Step 1. Given information is:

$4y^2-x^2+2x=2$

Part (a) Step 1. Solving for y

$$\begin{gathered} 4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}=2 \\ 4{\mathrm{y}}^2={\mathrm{x}}^2-2\mathrm{x}+2 \\ {\mathrm{y}}^2=\frac{{\mathrm{x}}^2-2\mathrm{x}+2}{4} \\ \mathrm{y}=\sqrt{\frac{{\mathrm{x}}^2-2\mathrm{x}+2}{4}} \\ \mathrm{y}=\pm \sqrt{\frac{{\mathrm{x}}^2-2\mathrm{x}+2}{4}} \\ \mathrm{The} \mathrm{graph} \mathrm{is} \mathrm{shown} \mathrm{below}: \end{gathered}$$

Part (a) Step 2. Calculating dx/dy

$$\begin{gathered} 4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}=2 \\ \mathrm{Differentiating} \mathrm{both} \mathrm{sides} \\ \frac{d\left(4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}\right)}{d\mathrm{x}} =\frac{d\left(2\right)}{d\mathrm{x}} \\ 4\left(2\mathrm{y}\right)\frac{d\mathrm{y}}{d\mathrm{x}}-2\mathrm{x}+2=0 \\ 8\mathrm{y}\frac{d\mathrm{y}}{d\mathrm{x}}=2\mathrm{x}-2 \\ \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{2\mathrm{x}-2}{8\mathrm{y}} \\ \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{\mathrm{x}-1}{4\mathrm{y}} \end{gathered}$$

Part (a) Step 3. Calculating y and slope

$$\begin{gathered} \mathrm{Substituting} \mathrm{x} = 3 \mathrm{into} \mathrm{the} \mathrm{equation} 4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}=2: \\ 4{\mathrm{y}}^2-{\left(3\right)}^2+2\left(3\right)=2 \\ 4{\mathrm{y}}^2-9+6=2 \\ 4{\mathrm{y}}^2=5 \\ {\mathrm{y}}^2=\frac{5}{4} \\ y=\pm \frac{\sqrt{5}}{2} \\ \mathrm{Thus}, \left(3, -\frac{\sqrt{5}}{2}\right)\left(3, \frac{\sqrt{5}}{2}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \mathrm{Thus} \mathrm{the} \mathrm{slope} \mathrm{is}: \\ \mathrm{i}) \mathrm{when} \mathrm{x} = 3, \mathrm{y}=-\frac{\sqrt{5}}{2} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{ 3-1}{4\left(-\frac{\sqrt{5}}{2}\right)} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{1}{\sqrt{5}} \\ \mathrm{ii}) \mathrm{when} \mathrm{x} = 3, \mathrm{y}=\frac{\sqrt{5}}{2} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{ 3-1}{4\left(\frac{\sqrt{5}}{2}\right)} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{1}{\sqrt{5}} \end{gathered}$$

Part (b) Step 1. Calculating x and slope

$$\begin{gathered} \mathrm{Substituting} y = 3 \mathrm{into} \mathrm{the} \mathrm{equation} 4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}=2: \\ 4{\left(3\right)}^2-{\mathrm{x}}^2+2\mathrm{x}=2 \\ 36-{\mathrm{x}}^2+2\mathrm{x}=2 \\ -{\mathrm{x}}^2+2\mathrm{x} =-34 \\ \mathrm{x}(-\mathrm{x}+2)=-34 \\ \Rightarrow \mathrm{x}=-34 \mathrm{and} \\ 2-\mathrm{x}=-34 \\ \mathrm{x}=36 \\ \mathrm{Thus}, \left(-34, 3\right)\left(36, 3\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \mathrm{Thus} \mathrm{the} \mathrm{slope} \mathrm{is}: \\ \mathrm{i}) \mathrm{when} \mathrm{x} = -34, \mathrm{y}=3 \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{ -34-1}{4\left(3\right)} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{35}{12} \\ \mathrm{ii}) \mathrm{when} \mathrm{x} = 36, \mathrm{y}=3 \\ \frac{d\mathrm{y}}{d\mathrm{x}} = -\frac{ 36-1}{4\left(3\right)} \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{35}{12} \end{gathered}$$

Part (c) Step 1. Finding Points

$$\begin{gathered} \mathrm{Substituting} \mathrm{x} = 0 \mathrm{into} \mathrm{the} \mathrm{equation} 4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}=2 \mathrm{for} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{to} \mathrm{be} \mathrm{horizontal}: \\ 4{\mathrm{y}}^2-0+0=2 \\ {\mathrm{y}}^2 =\frac{2}{4} \\ \mathrm{x} = \pm \frac{\sqrt{2}}{2} \\ \mathrm{Thus}, \left(0,\frac{\sqrt{2}}{2}\right)\left(0,-\frac{\sqrt{2}}{2}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \end{gathered}$$

Part (d) Step 1. Finding points

$$\begin{gathered} \mathrm{Substituting} \mathrm{y} = 0 \mathrm{into} \mathrm{the} \mathrm{equation} 4{\mathrm{y}}^2-{\mathrm{x}}^2+2\mathrm{x}=2 \mathrm{for} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{to} \mathrm{be} \mathrm{vertical}: \\ 0-{\mathrm{x}}^2+2\mathrm{x}=2 \\ -{\mathrm{x}}^2+2\mathrm{x}=2 \\ \mathrm{x}(2-\mathrm{x})=2 \\ \Rightarrow \mathrm{x}=2 \mathrm{or} \mathrm{x}=0 \\ \mathrm{Thus}, \left(2,0\right)\left(0,0\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \end{gathered}$$