Chapter 10

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

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

$x=-\frac{1}{4}{\left(y-5\right)}^2+2$

$x=y^2$

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

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

$y=4x^2$

$x=-\frac{1}{8}{\left(y-3\right)}^2+2$

$\frac{{\left(x+1\right)}^2}{9}+\frac{{\left(y-2\right)}^2}{4}=1$

$\frac{{\left(x-1\right)}^2}{1/4}+\frac{y^2}{9}=1$

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

$\frac{x^2}{3}+\frac{y^2}{5}=1$

$\frac{x^2}{4}+\frac{{\left(y-3\right)}^2}{6}=1$

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

$x^2-\frac{y^2}{3}=1$

$y^2-\frac{x^2}{24}=1$

$\frac{{\left(y-3\right)}^2}{4}-\frac{{\left(x-1\right)}^2}{9}=1$

$\frac{{\left(x-1\right)}^2}{9}-\frac{{\left(y-3\right)}^2}{4}=1$

$\frac{x^2}{4}-\frac{y^2}{3}=1$

$\frac{x^2}{3}-\frac{{\left(y-1\right)}^2}{9}=1$

${\left(y-2\right)}^2-\frac{x^2}{10}=1$

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

T

$\sqrt{2}$

$4/3$

$109/2$

$12/5$

$-\ln \left(2-\sqrt{3}\right)\approx 1.31696$

${\int }_0^1\sqrt{1+4x^2}𝑑x$

${\int }_0^1\sqrt{1+\frac{1}{4x}}𝑑x$

${\int }_{-1}^1\sqrt{1+\frac{x^2}{1-x^2}}𝑑x$

${\int }_1^2\sqrt{1+\frac{1}{x^4}}𝑑x$

$1.4790$

Simpson’s Rule fails, as it requires one to divide by 0. However, recognize the answer should be the same as for $y=x^2$; why?

Simpson’s Rule fails.

$1.4058$

$2\pi {\int }_0^{1}2x\sqrt{5}𝑑x=2\pi \sqrt{5}$

$2\pi {\int }_0^1\sqrt{x}\sqrt{1+1/\left(4x\right)}𝑑x=\pi /6\left(5\sqrt{5}-1\right)$

T

rectangular

Possible Answer: $x=t$, $y=9-4t$

Possible Answer: $x=-9+7\cos t$, $y=4+7\sin t$

Possible Answer: $x=\frac{5}{4}t+\frac{11}{4}$, $y=t$, $[-3,1]$

Possible Answer: $x=t$, $y=t^2+2t$, $(-\mathrm{\infty },-1]$

$x=\left(t+11\right)/6$, $y=\left(t^2-97\right)/12$. At $t=1$, $x=2$, $y=-8$.

$y^{\prime }=6x-11$; when $x=2$, $y^{\prime }=1$.

$x={\cos }^{-1}t$, $y=\sqrt{1-t^2}$. At $t=1$, $x=0$, $y=0$.

$y^{\prime }=\cos x$; when $x=0$, $y^{\prime }=1$.

$x=4t$, $y=-16t^2+64t$

$x=10t$, $y=-16t^2+320t$

$x=3\cos \left(2\pi t\right)+1$, $y=3\sin \left(2\pi t\right)+1$; other answers possible

$x=5\cos t$, $y=\sqrt{24}\sin t$; other answers possible

$x=2\tan t$, $y=\pm 6\sec t$; other answers possible

$y=-1.5x+8.5$

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

$y=2x+3$

$y=e^{2x}-1$

$x^2-y^2=1$

$y=\frac{b}{a}\left(x-x_0\right)+y_0$; line through $(x_0,y_0)$ with slope $b/a$.

$\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$; ellipse centered at $(h,k)$ with horizontal axis of length $2a$ and vertical axis of length $2b$.

$t=\pm 1$

$t=\pi /2,3\pi /2$

$t=-1$

$t=k\pi$ for integer values of $k$

F

F

horizontal: $t=0$; vertical: none

horizontal: $t=-1/2$; vertical: $t=1/2$

horizontal: none; vertical: $t=0$

The solution is non-trivial; use identities $\sin \left(2t\right)=2\sin t\cos t$ and $\cos \left(2t\right)={\cos }^{2}t-{\sin }^{2}t=1-2{\sin }^{2}t$ to rewrite $dy/dt=2\sin t\left(2{\cos }^{2}t-{\sin }^{2}t\right)$ and $dx/dt=2\cos t\left(1-3{\sin }^{2}t\right)$. Horizontal: $\sin t=0$ when $t=0,\pi ,2\pi$, and $2{\cos }^{2}t-{\sin }^{2}t=0$ when $t={\tan }^{-1}\left(\sqrt{2}\right),\pi \pm {\tan }^{-1}\left(\sqrt{2}\right),\mathrm{\hspace{0.25em}2}\pi -{\tan }^{-1}\left(\sqrt{2}\right)$. Vertical: $\cos t=0$ when $t=\pi /2,3\pi /2$, and $1-3{\sin }^{2}t=0$ when $t={\sin }^{-1}\left(1/\sqrt{3}\right),\pi -{\sin }^{-1}\left(1/\sqrt{3}\right)$.

$t_0=0$; ${lim}_{t\to 0}\frac{dy}{dx}=0$.

$t_0=1$; ${lim}_{t\to 1}\frac{dy}{dx}=\mathrm{\infty }$.

$\frac{d^{2}y}{d{x}^2}=2$; always concave up

$\frac{d^{2}y}{d{x}^2}=-\frac{4}{{\left(2t-1\right)}^3}$; concave up on $(-\mathrm{\infty },1/2)$; concave down on $(1/2,\mathrm{\infty })$.

$\frac{d^{2}y}{d{x}^2}=-{\cot }^{3}t$; concave up on $(-\pi /2,0)$; concave down on $(0,\pi /2)$.

$\frac{d^{2}y}{d{x}^2}=\frac{4\left(13+3\cos \left(4t\right)\right)}{{\left(\cos t+3\cos \left(3t\right)\right)}^3}$, obtained with a computer algebra system; concave up on $(-{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right),{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right))$, concave down on $(-\frac{\pi }{2},-{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right))$; $({\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right),\frac{\pi }{2})$

$L=6\pi$

$L=2\sqrt{34}$

$2\pi$

$-{\displaystyle \frac{\sqrt{10}}{3}}+\ln \left(3+\sqrt{10}\right)+\sqrt{2}-\ln \left(1+\sqrt{2}\right)$

$L\approx 2.4416$ (actual value: $L=2.42211$)

$L\approx 4.19216$ (actual value: $L=4.18308$)

The answer is $16\pi$ for both (of course), but the integrals are different.

$\frac{6\pi a^2}{5}$

$SA\approx 8.50101$ (actual value $SA=8.02851$

Answers will vary.

T

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

$x^2+{\left(y-\frac{3}{2}\right)}^2=\frac{9}{4}$

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

$x=3$

$x^4+x^2{y}^2-y^2=0$

$x^2+y^2=4$

$\theta =\pi /4$

$r=5\sec \theta$

$r=\cos \theta /{\sin }^2\theta$

$r=\sqrt{7}$

$P(\sqrt{3}/2,\pi /6)$, $P(0,\pi /2)$, $P(-\sqrt{3}/2,5\pi /6)$

$P(0,0)=P(0,\pi /2)$, $P(\sqrt{2},\pi /4)$

$P(\sqrt{2}/2,\pi /12)$, $P(-\sqrt{2}/2,5\pi /12)$, $P(\sqrt{2}/2,3\pi /4)$, and the origin.

Answers will vary. If $m$ and $n$ do not have any common factors, then an interval of $2n\pi$ is needed to sketch the entire graph.

Using $x=r\cos \theta$ and $y=r\sin \theta$, we can write $x=f\left(\theta \right)\cos \theta$, $y=f\left(\theta \right)\sin \theta$.