Chapter 1

$x=2+\frac{3}{y+1},$ or $y=\mathrm{-1}+\frac{3}{x-2}.$ This equation describes a portion of a rectangular hyperbola centered at $\left(2,\mathrm{-1}\right).$

One possibility is $x\left(t\right)=t,y\left(t\right)=t^2+2t.$ Another possibility is $x\left(t\right)=2t-3,y\left(t\right)={\left(2t-3\right)}^2+2\left(2t-3\right)=4t^2-8t+3.$

There are, in fact, an infinite number of possibilities.

$x^{\prime }\left(t\right)=2t-4$ and $y^{\prime }\left(t\right)=6t^2-6,$ so $\frac{dy}{dx}=\frac{6t^2-6}{2t-4}=\frac{3t^2-3}{t-2}.$
This expression is undefined when $t=2$ and equal to zero when $t=\mathrm{\pm 1}.$

The equation of the tangent line is $y=24x+100.$

$\frac{d^{2}y}{d{x}^2}=\frac{3t^2-12t+3}{2{\left(t-2\right)}^3}.$ Critical points $\left(5,4\right),\left(\mathrm{-3},\mathrm{-4}\right),\text{and}\left(\mathrm{-4},4\right).$

$A=3\pi$ (Note that the integral formula actually yields a negative answer. This is due to the fact that $x\left(t\right)$ is a decreasing function over the interval $\left[0,2\pi \right];$ that is, the curve is traced from right to left.)

$s=2\left({10}^{3\text{/}2}-2^{3\text{/}2}\right)\approx 57.589$

$A=\frac{\pi \left(494\sqrt{13}+128\right)}{1215}$

$\left(8\sqrt{2},\frac{5\pi }{4}\right)$ and $\left(\mathrm{-2},2\sqrt{3}\right)$

The name of this shape is a cardioid, which we will study further later in this section.

$y=x^2,$ which is the equation of a parabola opening upward.

Symmetric with respect to the polar axis.

$A=3\pi \text{/}2$

$A=\frac{4\pi }{3}+2\sqrt{3}$

$s=3\pi$

$x=2{\left(y+3\right)}^2-2$

$\frac{{\left(x+1\right)}^2}{16}+\frac{{\left(y-2\right)}^2}{9}=1$

$\frac{{\left(y+2\right)}^2}{9}-\frac{{\left(x-1\right)}^2}{4}=1.$ This is a vertical hyperbola. Asymptotes $y=\mathrm{-2}\pm \frac{3}{2}\left(x-1\right).$

$e=\frac{c}{a}=\frac{\sqrt{74}}{7}\approx 1.229$

Here $e=0.8$ and $p=5.$ This conic section is an ellipse.

The conic is a hyperbola and the angle of rotation of the axes is $\theta =22.5\text{°}.$

orientation: bottom to top

orientation: left to right

$y=\frac{x^2}{4}+1$

Asymptotes are $y=x$ and $y=\text{−}x$

$y=\frac{\sqrt{x+1}}{2}$; domain: $x\in \left[\mathrm{–1},\mathrm{–\infty }\right).$

$\frac{x^2}{16}+\frac{y^2}{9}=1;$ domain $x\in \left[\mathrm{-4},4\right].$

$y=3x+2;$ domain: all real numbers.

${(x-1)}^2+{(y-3)}^2=1;$ domain: $x\in \left[0,2\right].$

$y=\sqrt{x^2-1};$ domain: $x\in \left(\mathrm{-\infty },-1\right].$

$y^2=\frac{1-x}{2};$ domain: $x\in [-1,1].$

$y=\text{ln}x;$ domain: $x\in [1,\infty ).$

$y=\text{ln}x;$ domain: $x\in (0,\infty ).$

$x^2+y^2=4;$ domain: $x\in [\mathrm{-2},2].$

line

parabola

circle

ellipse

hyperbola

The equations represent a cycloid.

22,092 meters at approximately 51 seconds.

0

$\frac{\mathrm{-3}}{5}$

$\text{Slope}=0;$ $y=8.$

Slope is undefined; $x=2.$

$$\begin{gathered} \tan t = -2 \\ (\frac{4}{\sqrt{5}},\frac{\mathrm{-8}}{\sqrt{5}}), (\frac{4}{\sqrt{5}},\frac{\mathrm{-8}}{\sqrt{5}}) \end{gathered}$$, $\left(\frac{-4}{\sqrt{5}},\frac{8}{\sqrt{5}}\right)$.

No points possible; undefined expression.

$y=\text{−}\left(\frac{4}{e}\right)x+5$

$y=\mathrm{–2}x+3$

$\frac{\pi }{4},\frac{5\pi }{4},\frac{3\pi }{4},\frac{7\pi }{4}$

$\frac{dy}{dx}=\text{−}\text{tan}(t)$

$\frac{dy}{dx}=\frac{3}{4}$ and $\frac{d^{2}y}{d{x}^2}=0,$ so the curve is neither concave up nor concave down at $t=3.$ Therefore the graph is linear and has a constant slope but no concavity.

$\frac{dy}{dx}=4,\frac{d^{2}y}{d{x}^2}=\mathrm{-6}\sqrt{3};$ the curve is concave down at $\theta =\frac{\pi }{6}.$

No horizontal tangents. Vertical tangents at $(1,0),(\mathrm{-1},0).$

$\text{−}{\text{sec}}^3\left(\pi t\right)$

Horizontal $\left(0,\mathrm{-9}\right);$ vertical $\left(\text{±}2,\mathrm{-6}\right).$

1

0

4

Concave up on $t>0.$

$\frac{e^{\frac{\mathrm{\pi }}{2}}–1}{2}$

$\frac{3\pi }{2}$

$6\pi a^2$

$2\pi ab$

$\frac{1}{3}\left(2\sqrt{2}-1\right)$

$7.075$

$6a$

$6\sqrt{2}$

$\frac{2\pi \left(247\sqrt{13}+64\right)}{1215}$

59.101

$\frac{8\pi }{3}\left(17\sqrt{17}-1\right)$

$B\begin{array}{cc}\left(3,\frac{\text{−}\pi }{3}\right)\hfill & B\left(\mathrm{-3},\frac{2\pi }{3}\right)\hfill \end{array}$

$D\left(5,\frac{7\pi }{6}\right)D\left(\mathrm{-5},\frac{\pi }{6}\right)$

$(5, 5.356) (–5, 2.214)$

$(10, 2.214)(–10, 5.356)$

$\left(2\sqrt{3},\mathrm{-5.759}\right)\left(\mathrm{-2}\sqrt{3},2.618\right)$

$\left(\begin{array}{cc}\text{−}\sqrt{3},\hfill & \mathrm{-1}\hfill \end{array}\right)$

$\left(\begin{array}{cc}-\frac{\sqrt{3}}{2},\hfill & \frac{\mathrm{-1}}{2}\hfill \end{array}\right)$

$\left(\begin{array}{cc}0,\hfill & 0\hfill \end{array}\right)$

Symmetry with respect to the x-axis, y-axis, and origin.

Symmetric with respect to x-axis only.

Symmetry with respect to x-axis only.

Line $y=x$

$y=1$

Hyperbola; polar form $r^2\text{cos}(2\theta )=16$ or $r^2=16\text{sec}(2\theta ).$

$r=\frac{2}{3\text{cos}\theta -\text{sin}\theta }$

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

$x\text{tan}\sqrt{x^2+y^2}=y$

y-axis symmetry

y-axis symmetry

x- and y-axis symmetry and symmetry about the pole

x-axis symmetry

x- and y-axis symmetry and symmetry about the pole

no symmetry

a line

Answers vary. One possibility is the spiral lines become closer together and the total number of spirals increases.

$\frac{9}{2}{\displaystyle {\int }_0^{\pi }{\text{sin}}^2\theta d\theta }$

$32{\displaystyle {\int }_0^{\pi \text{/}2}{\text{sin}}^2(2\theta )d\theta }$

$\frac{1}{2}{\displaystyle {\int }_{\pi }^{2\pi }{\left(1-\text{sin}\theta \right)}^{2}d\theta }$

${\displaystyle {\int }_{{\text{sin}}^{\mathrm{-1}}\left(2\text{/}3\right)}^{\pi \text{/}2}{\left(2-3\text{sin}\theta \right)}^2}d\theta$

${\displaystyle {\int }_{\pi \text{/}3}^{\pi }{\left(1-2\text{cos}\theta \right)}^{2}d\theta -{\displaystyle {\int }_0^{\pi \text{/}3}{\left(1-2\text{cos}\theta \right)}^2}}d\theta$

$4{\displaystyle {\int }_0^{\pi \text{/}3}d\theta +16{\displaystyle {\int }_{\pi \text{/}3}^{\pi \text{/}2}\left({\text{cos}}^2\theta \right)}}d\theta$

$9\pi$

$\frac{9\pi }{4}$

$\frac{9\pi }{8}$

$\frac{18\pi -27\sqrt{3}}{2}$

$\frac{4}{3}\left(4\pi -3\sqrt{3}\right)$

$\frac{3}{2}\left(4\pi -3\sqrt{3}\right)$

$2\pi -4$

${\displaystyle {\int }_0^{2\pi }\sqrt{{\left(1+\text{sin}\theta \right)}^2+{\text{cos}}^2\theta }}d\theta$

$\sqrt{2}{\displaystyle {\int }_0^1{e}^{\theta }d\theta }$

$\frac{\sqrt{10}}{3}\left(e^6-1\right)$

32

6.238

2

4.39

$A=\pi {\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{\pi }{2}\text{and}\frac{1}{2}{\displaystyle {\int }_0^{\pi }\left(1+2\text{sin}\theta \text{cos}\theta \right)}d\theta =\frac{\pi }{2}$

$C=2\pi \left(\frac{3}{2}\right)=3\pi \text{and}{\displaystyle {\int }_0^{\pi }3}d\theta =3\pi$

$C=2\pi \left(5\right)=10\pi \text{and}{\displaystyle {\int }_0^{\pi }10}d\theta =10\pi$

$\frac{dy}{dx}=\frac{f^{\prime }(\theta )\text{sin}\theta +f\left(\theta \right)\text{cos}\theta }{f^{\prime }\left(\theta \right)\text{cos}\theta -f(\theta )\text{sin}\theta }$

The slope is $\frac{1}{\sqrt{3}}.$

The slope is 0.

At $\left(4,0\right),$ the slope is undefined. At $\left(\mathrm{-4},\frac{\pi }{2}\right),$ the slope is 0.

The slope is undefined at $\theta =\frac{\pi }{4}.$

Slope = −1.

Slope is $\frac{\mathrm{-2}}{\pi }.$

Calculator answer: −0.836.

Horizontal tangent at $\left(\text{±}\sqrt{2},\frac{\pi }{6}\right),$ $\left(\text{±}\sqrt{2},-\frac{\pi }{6}\right).$

Horizontal tangents at $\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}.$ Vertical tangents at $\frac{\pi }{6},\frac{5\pi }{6}$ and also at the pole $(0,0).$

$y^2=16x$

$x^2=2y$

$x^2=\mathrm{-4}\left(y-3\right)$

${\left(x+3\right)}^2=8\left(y-3\right)$

$\frac{x^2}{16}+\frac{y^2}{12}=1$

$\frac{x^2}{13}+\frac{y^2}{4}=1$

$\frac{{\left(y-1\right)}^2}{16}+\frac{{\left(x+3\right)}^2}{12}=1$

$\frac{x^2}{16}+\frac{y^2}{12}=1$

$\frac{x^2}{25}-\frac{y^2}{11}=1$

$\frac{x^2}{7}-\frac{y^2}{9}=1$

$\frac{{\left(y+2\right)}^2}{4}-\frac{{\left(x+2\right)}^2}{32}=1$

$\frac{x^2}{4}-\frac{y^2}{32}=1$

$e=1,$ parabola

$e=\frac{1}{2},$ ellipse

$e=3,$ hyperbola

$r=\frac{4}{5+\text{cos}\theta }$

$r=\frac{4}{1+2\text{sin}\theta }$

Hyperbola

Ellipse

Ellipse

At the point 2.25 feet above the vertex.

0.5625 feet

Length is 96 feet and height is approximately 26.53 feet.

$r=\frac{2.616}{1+0.995\text{cos}\theta }$

$r=\frac{5.192}{1+0.0484\text{cos}\theta }$

True.

False. Imagine $y=t+1,$ $x=\text{−}t+1.$

$y=1-x^3$

$\frac{x^2}{16}+{(y-1)}^2=1$

Symmetric about polar axis

$r^2=\frac{4}{{\text{sin}}^2\theta -{\text{cos}}^2\theta }$

$y=\frac{3\sqrt{2}}{2}+\frac{1}{5}\left(x+\frac{3\sqrt{2}}{2}\right)$

$\frac{e^2}{2}$

$9\sqrt{10}$

${\left(y+5\right)}^2=\mathrm{-8}x+32$

$\frac{{\left(y+1\right)}^2}{16}-\frac{{\left(x+2\right)}^2}{9}=1$

$e=\frac{2}{3},$ ellipse

$\frac{y^2}{{19.03}^2}+\frac{x^2}{{19.63}^2}=1,$ $e=0.2447$