R2 2014 vår LØSNING

DEL 1

Oppgave 1

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

$\displaystyle f'(x) = 3\cos(3x)$

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

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

Oppgave 2

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

La $\displaystyle u = x^2$

$\displaystyle \begin{align*} & \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \\ & \Rightarrow \mathrm{d}u = 2x \space \mathrm{d}x \end{align*}$

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

b) $\displaystyle \int_1^{e} x \cdot \ln x \, \mathrm{d}x$

La $\displaystyle u = \ln x$ og $\displaystyle v' = x$:

$\displaystyle \begin{align*} \int_1^{e} x \cdot \ln x \, \mathrm{d}x & = \left[ \ln x \cdot \frac{1}{2} x^2 - \int \frac{1}{x} \cdot \frac{1}{2} x^2 \right]_1^{e} \\ & = \left[ \frac{1}{2} x^2 \cdot \ln x - \frac{1}{2} \int x \, \mathrm{d}x \right]_1^{e} \\ & = \left[ \frac{1}{2} x^2 \cdot \ln x - \frac{1}{2} \cdot \frac{1}{2} x^2 \right]_1^{e} \\ & = \frac{1}{2} \left[ x^2 \cdot \ln x - \frac{1}{2} x^2 \right]_1^{e} \\ & = \frac{1}{2} \left( (e^2 \cdot \ln e - \frac{1}{2} \cdot e^2) - (1^2 \cdot \ln1 - \frac{1}{2} \cdot 1^2) \right) \\ & = \frac{1}{2} \left( (e^2 - \frac{1}{2} \cdot e^2) - (0 - \frac{1}{2}) \right) \\ & = \frac{1}{2} \left( \frac{e^2}{2} + \frac{1}{2} \right) \\ & = \frac{1}{2} \cdot \frac{e^2 + 1}{2} \\ & = \frac{e^2 + 1}{4} \end{align*}$

Oppgave 3

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

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

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

$\displaystyle \begin{align*} f ' ' (x) & = 0 \\ 4e^{2x} - 4e^x & = 0 \\ 4\left(e^x\right)^2 - 4e^x & = 0 \\ 4e^x\left(e^x - 1\right) & = 0 \\ e^x - 1 & = 0 \\ e^x & = 1 \\ x & = 0 \end{align*}$

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

Oppgave 4

$\displaystyle s(x) = 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ...$

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

b)

$\displaystyle \begin{align*} s(x) & = 3 \\ 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ... & = 3 \\ \frac{1}{1 - \left(1 - x\right)} & = 3 \\ \frac{1}{x} & = 3 \\ 1 & = 3x \\ x & = \frac{1}{3}\end{align*}$

$\displaystyle \begin{align*} s(x) & = \frac{1}{3} \\ 1 + \left(1 - x\right) + \left(1 - x\right)^2 + \left(1 - x\right)^3 + ... & = \frac{1}{3} \\ \frac{1}{x} & = \frac{1}{3} \end{align*}$

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

Oppgave 5

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

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

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

Hvilket skulle vises.

b) $\displaystyle l \perp \alpha \Leftrightarrow \vec{r}_{l} = \vec{n}_{\alpha}$

$\displaystyle\vec{r}_{l} = [2,1,-2]$

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

c)

$\displaystyle \begin{align*} 2\left( 3 + 2t \right) + \left(4 + t\right) - 2\left( 2 - 2t\right) + 3 & = 0 \\ 6 + 4t + 4 + t - 4 + 4t + 3 & = 0 \\ 9 + 9t & = 0 \\ 9t & = -9 \\ t & = -1\end{align*}$

Skjæringspunkt $\displaystyle = \left( 3 + 2\left( -1\right), 4 + \left( -1\right), 2 - 2\left(-1\right)\right) = \left(1,3,4\right)$

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

Oppgave 6

$\displaystyle f(x) = a\sin \left(c\space x + \varphi\right) + d$

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

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

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

$\displaystyle \varphi = \frac{c}{\frac{\left(2 - 0 \right)}{2}} = \frac{\frac{\pi}{2}}{1} = \frac{\pi}{2}$

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

Hvilket skulle vises.

b)

Oppgave 7

$\displaystyle y' - 3y = 2 \space , \space y(0) = \frac{1}{3}$

METODE 1

Differensiallikningen kan løses med en integrerende faktor.

$\displaystyle \begin{align*} y' - 3y & = 2 \space |\cdot e^{-3x} \\ y' \cdot e^{-3x} - 3y \cdot e^{-3x} & = 2\cdot e^{-3x} \\ \left(y \cdot e^{-3x}\right) & = 2e^{-3x} \\ y \cdot e^{-3x} & = \int 2e^{-3x} \, \mathrm{d}x \\ y \cdot e^{-3x} & = \frac{2}{-3} e^{-3x} + C \space |\cdot\frac{1}{e^{-3x}} \\ y & = -\frac{2}{3} + \frac{C}{e^{-3x}} \\ y & = Ce^{3x} - \frac{2}{3}\end{align*}$

METODE 2

Differensiallikningen er separabel.

$\displaystyle \begin{align*} y' - 3y & = 2 \\ y' & = 3y + 2 \space |\cdot\frac{1}{3y + 2} \\ y' \cdot \frac{1}{3y + 2} & = 1 \\ \frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{1}{3y + 2} & = 1 \\ \frac{\mathrm{d}y}{3y + 2} & = \mathrm{d}x \\ \int \frac{\mathrm{d}y}{3y + 2} & = \int \mathrm{d}x \\ \frac{1}{3}\ln|3y + 2| + C_1 & = x + C_2 \\ \frac{1}{3}\ln|3y + 2| & = x + C_2 - C_1 \\ C_2 - C_1 = C_3 \Rightarrow \ln|3y + 2| & = 3x + C_3 \\ 3y + 2 & = e^{3x + C_3} \\ 3y + 2 & = e^{3x} \cdot e^{C_3} \\ e^{C_3} = C_4 \Rightarrow 3y + 2 & = C_4e^{3x} \\ 3y & = C_4e^{3x} - 2 \\ \frac{C_4}{3} = C \Rightarrow y & = Ce^{3x} - \frac{2}{3}\end{align*}$

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

DEL 2

Oppgave 1

a) Setningen forteller at punktene $\displaystyle A(4,3,1)$, $\displaystyle B(2,2,0)$ og $\displaystyle C(1,2,-2)$ bestemmer entydig et plan $\displaystyle \alpha$ kun hvis punktene ikke ligger langs en rett linje.

$\displaystyle \vec{AB} ≠ k\cdot \vec{AC}\space k\in \R

\vec{AB} = [2-4,2-3,0-1] = [-2,-1,-1] ≠ k\cdot\vec{AC} = k\cdot [1-2,2-2,-2-0] = k\cdot [-1,0,-2]$

Hvilket skulle vises.

b) $\displaystyle \begin{align*} \vec{n}_{\alpha} & = \vec{AB} \times \vec{AC} \\ & = [(-1)(-2) - (-1)0, - (-2)(-2) - (-1)(-1), (-2)0 - (-1)(-1)] \\ & = [2,-3,-1]\end{align*}$

$\displaystyle\begin{align*} \alpha: 2(x-4) -3(y-3) - (z-1) & = 0 \\ 2x - 8 - 3y + 9 - z + 1 & = 0 \\ 2x - 3y - z + 2 & = 0 \end{align*}$

c)

$\displaystyle \begin{align*} V_{ABCT} & = 3 \\ \frac{1}{6}|(\vec{AB} \times \vec{AC}) \cdot \vec{AT}| & = 3 \\ \frac{1}{6}|[2,-3,-1]\cdot[2-4,5-3,(4t+1)-1]| & = 3 \\ \frac{1}{6}|2(-2) + (-3)2 + (-1)4t| & = 3 \\ \frac{1}{6}|-10-4t| & = 3 \\ |-10-4t| & = 18 \end{align*}$

Oppgave 4

d)

Påstanden \[P(n):\quad1+\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+\cdots+\frac{n}{2^{n-1}}=4-\frac{n+2}{2^{n-1}}\quad,\quad n\in\mathbb{N}\] er på formen \[a_1+a_2+a_3+\cdots+a_n=f(n)\quad,\quad n\in\mathbb{N}\] der \[a_n=\frac{n}{2^{n-1}}\qquad\textrm{og}\qquad f(n)=4-\frac{n+2}{2^{n-1}}\] For å vise $P(n)$ ved hjelp av induksjon, vises først at $P(1)$ stemmer, så at $P(n)\Rightarrow P(n+1)$.

Steg 1: $P(1)$ stemmer hvis $a_1=f(1)$. Dette er ekvivalent med at $a_1-f(1)=0$. Sjekker om det er tilfelle: \[a_1-f(1)=\frac{1}{2^{1-1}}-\left(4-\frac{1+2}{2^{1-1}}\right)=1-4+3=0\] Så $P(1)$ stemmer.

Steg 2: Antar at $P(n)$ stemmer. Dvs at

\[a_1+a_2+a_3+\cdots+a_n=f(n)\] Legger til $a_{n+1}$ på begge sider: \[a_1+a_2+a_3+\cdots+a_n+a_{n+1}=f(n)+a_{n+1}\] Venstre side i likningen over er lik venstre side i likningen til $P(n+1)$. Høyre side i likningen over er lik høyre side i likningen til $P(n+1)$ hvis $f(n)+a_{n+1}=f(n+1)$. Dette er ekvivalent med at $f(n)+a_{n+1}-f(n+1)=0$. Sjekker om det er tilfelle: \begin{align*} f(n)+a_{n+1}-f(n+1) & = 4-\frac{n+2}{2^{n-1}}+\frac{n+1}{2^{n+1-1}}-\left(4-\frac{n+1+2}{2^{n+1-1}}\right) \\ & = 4-\frac{n+2}{2^{n-1}}+\frac{n+1}{2^n}-4+\frac{n+3}{2^n} \\ & = -\frac{n+2}{2^{n-1}}\cdot\frac{2^1}{2^1}+\frac{n+1}{2^n}+\frac{n+3}{2^n} \\ & = -\frac{2n+4}{2^n}+\frac{n+1}{2^n}+\frac{n+3}{2^n} \\ & = 2^{-n}(-(2n+4)+n+1+n+3)\\ & = 0 \end{align*} Så $P(n)\Rightarrow P(n+1)$ og dermed stemmer $P(n)$ for alle $n$.