Solutions

1.
Plot 2D + Implicit: x2 + y2 = 1, x2 - y2 = 1, x + y2 = 0

(Take -2≤x≤2 and -2≤y≤2. Choose Equal Scaling Along Each Axis)BITMAPSETProbSolvHint0.2006in0.243in0inq1dtbpF3.9998in2.7492in0ptFigure

2.
Plot 2D + Implicit: $\left(\vphantom{ x-1}\right.$x - 1$\left.\vphantom{ x-1}\right)^{{2}}_{}$ + $\left(\vphantom{ y+2}\right.$y + 2$\left.\vphantom{ y+2}\right)^{{2}}_{}$ = 1,$\left(\vphantom{ x-1}\right.$x - 1$\left.\vphantom{ x-1}\right)^{{2}}_{}$ - $\left(\vphantom{ y+2}\right.$y + 2$\left.\vphantom{ y+2}\right)^{{2}}_{}$ = 1,$\left(\vphantom{ x-1}\right.$x - 1$\left.\vphantom{ x-1}\right)$ + $\left(\vphantom{ y+2}\right.$y + 2$\left.\vphantom{ y+2}\right)^{{2}}_{}$ = 0

(Take -1≤x≤3 and -4≤y≤ 0. Choose Equal Scaling Along Each Axis) BITMAPSETProbSolvHint0.2006in0.243in0inq2dtbpF3.7559in2.4811in0ptFigure

3.
Plot 2D + Implicit: x2 + y2 = 4, x2 - y2 = 1

(Take -5≤x≤5 and -5≤y≤5. Choose Equal Scaling Along Each Axis)dtbpF3.813in2.5253in0ptFigure

BITMAPSETProbSolvHint0.2006in0.243in0inq3

Solve + Numeric:
x2 + y2 = 4
x2 - y2 = 1
x$\left(\vphantom{ 1,2}\right.$1, 2$\left.\vphantom{ 1,2}\right)$
y$\left(\vphantom{ 1,2}\right.$1, 2$\left.\vphantom{ 1,2}\right)$
, Solution is : $\left\{\vphantom{ x=1.58113883,y=1.224744871}\right.$x = 1.58113883, y = 1.224744871$\left.\vphantom{ x=1.58113883,y=1.224744871}\right\}$

4.
Plot 2D + Implicit: $\left\vert\vphantom{ x}\right.$x$\left.\vphantom{ x}\right\vert^{{2/3}}_{}$ + $\left\vert\vphantom{ y}\right.$y$\left.\vphantom{ y}\right\vert^{{2/3}}_{}$ = 1

(Take -1≤x≤1 and -1≤y≤1.)dtbpF3.9998in2.6446in0ptFigure

Without the absolute values, Scientific Notebook returns only the first quadrant portion of the graph.BITMAPSETProbSolvHint0.2006in0.243in0inq4

5.
Plot 2D + Implicit: x3 + y3 = 6xy

(Take -5≤x≤5 and -5≤y≤5 and set the grid to 50 by 50.)dtbpF2.6507in2.6507in0ptFigure

Notice how the folium of Descartes shows up as a level curve on the surface z = x3 + y3 - 6xy.

Plot 3D + Rectangular: x3 + y3 - 6xy

(Use Patch & Contour and take -5≤x≤5, -5≤y≤5, Turn 16, and Tilt 34.)BITMAPSETProbSolvHint0.2006in0.243in0inq5dtbpF3in2.0003in0pt

6.
Plot 3D + Rectangular: sin xy

(Choose Patch and Contour and take -4≤x≤4, -4≤y≤4, Turn 108, and Tilt 17.) dtbpF3.8484in2.5694in0ptFigure

Plot 2D + Implicit: xy = π/2, xy = 5π/2, xy = 3π/2

(Take -4≤x≤4 and -4≤y≤4.)BITMAPSETProbSolvHint0.2006in0.243in0inq6dtbpF4.0482in2.6991in0ptFigure

7.
Plot 3D + Rectangular: [s, cos t, sin t],[cos t, s, sin t]

(Take -2≤s≤2 and 0≤t≤2π.) BITMAPSETProbSolvHint0.2006in0.243in0inq7dtbpF3.4489in2.2987in0ptFigure

8.
Plot 3D + Tube: $\left[\vphantom{
\begin{array}{c}
(2+\sin t)10\cos t \\
(2+\cos t)10\sin t \\
3\sin 3t
\end{array}
}\right.$$\begin{array}{c}
(2+\sin t)10\cos t \\
(2+\cos t)10\sin t \\
3\sin 3t
\end{array}$$\left.\vphantom{
\begin{array}{c}
(2+\sin t)10\cos t \\
(2+\cos t)10\sin t \\
3\sin 3t
\end{array}
}\right]$; drag $\left[\vphantom{
\begin{array}{c}
20\cos t \\
20\sin t \\
-3\sin 3t
\end{array}
}\right.$$\begin{array}{c}
20\cos t \\
20\sin t \\
-3\sin 3t
\end{array}$$\left.\vphantom{
\begin{array}{c}
20\cos t \\
20\sin t \\
-3\sin 3t
\end{array}
}\right]$ onto the plot, set 0≤t≤2π, and set the radius for both items to 1.BITMAPSETProbSolvHint0.2006in0.243in0inq8dtbpF3.4489in2.2987in0ptFigure

Solve + Exact:

(2 + sin t)10 cos t = 20 cos s
(2 + cos t)10 sin t = 20 sin s
3 sin 3t = - 3 sin 3s
, Solution is : $\left\{\vphantom{ t=0,s=0}\right.$t = 0, s = 0$\left.\vphantom{ t=0,s=0}\right\}$,$\left\{\vphantom{ t=\pi ,s=\pi }\right.$t = π, s = π$\left.\vphantom{ t=\pi ,s=\pi }\right\}$

9.
Plot 3D + Rectangular: $\sqrt{{\left(
-x^{2}-y^{2}+1\right) }}$, - $\sqrt{{\left(
-x^{2}-y^{2}+1\right) }}$,${\frac{{1}}{{2}%
}}$ - $\left(\vphantom{ x+y}\right.$x + y$\left.\vphantom{ x+y}\right)$

(Take -1≤x≤1, -1≤y≤1, -1≤z≤1, and style Patch & Contour.)

dtbpF3in2.0003in0pt

Simplify: x2 + y2 + $\left(\vphantom{ \frac{1}{2}-\left( x+y\right) }\right.$${\frac{{1}}{{2}}}$ - $\left(\vphantom{ x+y}\right.$x + y$\left.\vphantom{ x+y}\right)$$\left.\vphantom{ \frac{1}{2}-\left( x+y\right) }\right)^{{2}}_{}$ = 2x2 +2xy + 2y2 - x - y + ${\frac{{1}}{{4}}}$ = 1

$\bigtriangleup$  = B2 -4AC = 22 -4×2×2 = -12 < 0, implying the curve is an ellipse or circle.BITMAPSETProbSolvHint0.2006in0.243in0inq9

10.
Plot 3D + Rectangular: xy

(This is the default plot with settings -5≤x≤5, -5≤y≤5, and style WireFrame.) BITMAPSETProbSolvHint0.2006in0.243in0inq10dtbpF4.0482in2.6991in0ptFigure

(Take -1≤x≤1, -1≤y≤1, Turn 90, Tilt , and choose the style Patch & Contour.)dtbpF3.9885in2.6576in0ptFigure

(Take -1≤x≤1, -1≤y≤1, Turn 90, Tilt , and choose the style Patch & Contour.)dtbpF3.9885in2.6576in0ptFigure



Subsections