R2 2014 vår LØSNING

DEL 1

Oppgave 1

a) $f(x)=\sin (3x)$

$f^{\prime }(x)=3\cos (3x)$

b) $g(x)=e^{2x}\cdot \cos x$

$g^{\prime }(x)=2e^{2x}\cdot \cos x+e^{2x}\cdot (-\sin x)=e^{2x}(2\cos x-\sin x)$

Oppgave 2

a) $\int 2x\cdot \sin (x^2)\mathrm{d}x$

La $u=x^2$

$\begin{array}{rl}& \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}=2x\\ & \Rightarrow \mathrm{d}u=2x\mathrm{d}x\end{array}$

$\int 2x\cdot \sin (x^2)\mathrm{d}x=\int \sin u\mathrm{d}u=-\cos u+C=-\cos (x^2)+C$

b) ${\int }_1^{e}x\cdot \ln x\mathrm{d}x$

La $u=\ln x$ og $v^{\prime }=x$:

$\begin{array}{rl}{\int }_1^{e}x\cdot \ln x\mathrm{d}x& ={[\ln x\cdot \frac{1}{2}{x}^2-\int \frac{1}{x}\cdot \frac{1}{2}{x}^2]}_1^e\\ & ={[\frac{1}{2}{x}^2\cdot \ln x-\frac{1}{2}\int x\mathrm{d}x]}_1^e\\ & ={[\frac{1}{2}{x}^2\cdot \ln x-\frac{1}{2}\cdot \frac{1}{2}{x}^2]}_1^e\\ & =\frac{1}{2}{[x^2\cdot \ln x-\frac{1}{2}{x}^2]}_1^e\\ & =\frac{1}{2}((e^2\cdot \ln e-\frac{1}{2}\cdot e^2)-(1^2\cdot \ln 1-\frac{1}{2}\cdot 1^2))\\ & =\frac{1}{2}((e^2-\frac{1}{2}\cdot e^2)-(0-\frac{1}{2}))\\ & =\frac{1}{2}(\frac{e^2}{2}+\frac{1}{2})\\ & =\frac{1}{2}\cdot \frac{e^2+1}{2}\\ & =\frac{e^2+1}{4}\end{array}$

Oppgave 3

$f(x)=e^{2x}-4e^x,D_f=\text{R}$

$f^{\prime }(x)=2e^{2x}-4e^x$

$f^{″}(x)=4e^{2x}-4e^x$

$\begin{array}{rl}{f}^{″}(x)& =0\\ 4e^{2x}-4e^x& =0\\ 4{\left(e^x\right)}^2-4e^x& =0\\ 4e^x(e^x-1)& =0\\ e^x-1& =0\\ e^x& =1\\ x& =0\end{array}$

Vendepunkt: $(0,f(0))=(0,e^{2\cdot 0}-4e^0)=(0,1-3)=(0,-3)$

Oppgave 4

$s(x)=1+(1-x)+{(1-x)}^2+{(1-x)}^3+...$

a) $|k|<1\Rightarrow |1-x|<1\Rightarrow 0<x<2$

b)

$\begin{array}{rl}s(x)& =3\\ 1+(1-x)+{(1-x)}^2+{(1-x)}^3+...& =3\\ \frac{1}{1-(1-x)}& =3\\ \frac{1}{x}& =3\\ 1& =3x\\ x& =\frac{1}{3}\end{array}$

$\begin{array}{rl}s(x)& =\frac{1}{3}\\ 1+(1-x)+{(1-x)}^2+{(1-x)}^3+...& =\frac{1}{3}\\ \frac{1}{x}& =\frac{1}{3}\end{array}$

$x\ne 3$ ettersom denne verdien ligger utenfor rekkens konvergensområde. Likningen har ingen løsning.

Oppgave 5

$\alpha$: $2x+y-2z+3=0$

a) Punktet $P(3,4,2)$ ligger ikke i planet $\alpha$ kun dersom punktets koordinater ikke tilfredstiller likningen til planet.

$2\left(3\right)+\left(4\right)-2\left(2\right)+3=6+4-4=6\ne 0\iff$ punktet $P(3,4,2)$ ligger ikke i planet $\alpha$.

Hvilket skulle vises.

b) $l\perp \alpha \iff {\overrightarrow{r}}_l={\overrightarrow{n}}_{\alpha }$

${\overrightarrow{r}}_l=[2,1,-2]$

$\Rightarrow l$: $\begin{array}{rl}x& =3+2t\\ y& =4+t\\ z& =2-2t\end{array}$

c)

$\begin{array}{rl}2(3+2t)+(4+t)-2(2-2t)+3& =0\\ 6+4t+4+t-4+4t+3& =0\\ 9+9t& =0\\ 9t& =-9\\ t& =-1\end{array}$

Skjæringspunkt $=(3+2(-1),4+(-1),2-2(-1))=(1,3,4)$

d) $D=\frac{|2\cdot 3+1\cdot 4-2\cdot 2+3|}{\sqrt{2^2+1^2+{(-2)}^2}}=\frac{|6+4-4+3|}{\sqrt{4+1+4}}=\frac{9}{\sqrt{9}}=\frac{9}{3}=3$

Oppgave 6

$f(x)=a\sin (cx+\phi )+d$

$a=\frac{7-3}{2}=2$

$d=3+a=3+2=5$

$c=\frac{2\pi }{p}=\frac{2\pi }{2(2-0)}=\frac{2\pi }{4}=\frac{\pi }{2}$

$\phi =\frac{c}{\frac{(2-0)}{2}}=\frac{\frac{\pi }{2}}{1}=\frac{\pi }{2}$

$\Rightarrow f(x)=2\sin (\frac{\pi }{2}x+\frac{\pi }{2})+5$.

Hvilket skulle vises.

b)

Oppgave 7

$y^{\prime }-3y=2,y(0)=\frac{1}{3}$

METODE 1

Differensiallikningen kan løses med en integrerende faktor.

$\begin{array}{rl}{y}^{\prime }-3y& =2|\cdot e^{-3x}\\ y^{\prime }\cdot e^{-3x}-3y\cdot e^{-3x}& =2\cdot e^{-3x}\\ (y\cdot e^{-3x})& =2e^{-3x}\\ y\cdot e^{-3x}& =\int 2e^{-3x}\mathrm{d}x\\ y\cdot e^{-3x}& =\frac{2}{-3}{e}^{-3x}+C|\cdot \frac{1}{e^{-3x}}\\ y& =-\frac{2}{3}+\frac{C}{e^{-3x}}\\ y& =C{e}^{3x}-\frac{2}{3}\end{array}$

METODE 2

Differensiallikningen er separabel.

$\begin{array}{rl}{y}^{\prime }-3y& =2\\ y^{\prime }& =3y+2|\cdot \frac{1}{3y+2}\\ y^{\prime }\cdot \frac{1}{3y+2}& =1\\ \frac{\mathrm{d}y}{\mathrm{d}x}\cdot \frac{1}{3y+2}& =1\\ \frac{\mathrm{d}y}{3y+2}& =\mathrm{d}x\\ \int \frac{\mathrm{d}y}{3y+2}& =\int \mathrm{d}x\\ \frac{1}{3}\ln |3y+2|+C_1& =x+C_2\\ \frac{1}{3}\ln |3y+2|& =x+C_2-C_1\\ C_2-C_1=C_3\Rightarrow \ln |3y+2|& =3x+C_3\\ 3y+2& =e^{3x+C_3}\\ 3y+2& =e^{3x}\cdot e^{C_3}\\ e^{C_3}=C_4\Rightarrow 3y+2& =C_4{e}^{3x}\\ 3y& =C_4{e}^{3x}-2\\ \frac{C_4}{3}=C\Rightarrow y& =C{e}^{3x}-\frac{2}{3}\end{array}$

$\begin{array}{rl}y(0)=\frac{1}{3}& \Rightarrow C{e}^{3\cdot 0}-\frac{2}{3}=\frac{1}{3}\\ & \Rightarrow C=\frac{1}{3}+\frac{2}{3}\\ & \Rightarrow C=1\\ & \Rightarrow y=e^{3x}-\frac{2}{3}\end{array}$