Q. 81 - Calculus

Question

Consider the circle of radius 1 centered at the origin, that is, the solutions of the equation

$x^2+y^2 = 1$

(a) Find all points on the graph with an x-coordinate of x = 1/2, 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 $y=\frac{\sqrt{2}}{2}$ , and then find the slope of the tangent line at each of these points.

(c) Find all points on the graph where the tangent line is vertical.

(d) Find all points on the graph where the tangent line has a slope of −1.

Step-by-Step Solution

Answer

Part (a): $$\begin{gathered} \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{1}{\sqrt{3}} \& -\frac{1}{\sqrt{3}} \end{gathered}$$

Part (b): $$\begin{gathered} \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \\ \frac{d\mathrm{y}}{d\mathrm{x}} = 1 \& -1 \end{gathered}$$

Part (c): $\left(1,0\right)\left(-1,0\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}.$

Part (d): $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}.$

Step 1. Given information is:

$$\begin{gathered} \mathrm{Function} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{is}: \\ {\mathrm{x}}^2+{\mathrm{y}}^2 = 1 \end{gathered}$$

Part (a) Step 1. Calculating dy/dx

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

Part (a) Step 2. Calculating y and slope

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

Part (b) Step 1. Calculating x and slope

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

Part (c) Step 1. Finding points

$$\begin{gathered} \mathrm{Substituting} \mathrm{y} = 0 \mathrm{into} \mathrm{the} \mathrm{equation} {\mathrm{x}}^2 +{\mathrm{y}}^2 =1 \mathrm{for} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{to} \mathrm{be} \mathrm{vertical}: \\ {\mathrm{x}}^2 +{\left(0\right)}^2 =1 \\ {\mathrm{x}}^2 =1 \\ \mathrm{x} = \pm 1 \\ \mathrm{Thus}, \left(1,0\right)\left(-1,0\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \end{gathered}$$

Part (d) Step 1. Finding points

$$\begin{gathered} \mathrm{Here}, \\ -\frac{\mathrm{x}}{\mathrm{y}} =-1 \\ \mathrm{x}=\mathrm{y} \\ \mathrm{Substituting} \mathrm{the} \mathrm{above} \mathrm{relation} \mathrm{in} \mathrm{given} \mathrm{equation}: \\ {\mathrm{y}}^2 +{\mathrm{y}}^2 =1 \\ 2{\mathrm{y}}^2 =1 \\ {\mathrm{y}}^2 =\frac{1}{2} \\ \mathrm{y} = \pm \frac{1}{\sqrt{2}} \\ \mathrm{Thus}, \\ {\mathrm{x}}^2 +{\left(\frac{1}{\sqrt{2}}\right)}^2 =1 \\ {\mathrm{x}}^2 +\frac{1}{2} = 1 \\ \mathrm{x} = \pm \frac{1}{\sqrt{2}} \\ \mathrm{Thus}, \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph}. \end{gathered}$$