Matematik

parameterfremstillinger

09. maj 2021 af MaryamS - Niveau: Universitet/Videregående

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Svar #1
09. maj 2021 af mathon

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Svar #2
09. maj 2021 af mathon

$\small \begin{array}{lllllll} \blacklozenge\\& \overrightarrow{v}_1=\begin{pmatrix} -\sin(u)\\ \cos(u) \end{pmatrix}\\\\& \overrightarrow{v}_2= \begin{pmatrix} \frac{2\cdot \left ( 1+t^2 \right )-2t\cdot 2t}{\left (1+t^2 \right )^2}\\\\ \frac{-2t-(1-t^2)\cdot 2t}{\left (1+t^2 \right )^2} \end{pmatrix}=\begin{pmatrix} \frac{2(1-t)^2}{(1+t^2)^2}\\\\ \frac{-4t}{(1+t^2)^2} \end{pmatrix}\\\\\\ \blacklozenge\\& v_1=\sqrt{(-\sin(u))^2+(\cos(u))^2}=1\\\\\\& v_2=\sqrt{\left ( \frac{2(1-t)^2}{(1+t^2)^2} \right )^2+\left ( \frac{-4t}{(1+t^2)^2} \right )^2}=\frac{\sqrt{4\cdot (1-t)^4+16t^2}}{1+t^2}=\frac{2\cdot \sqrt{(1-t)^4+4t^2}}{1+t^2} \end{array}$

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Svar #3
09. maj 2021 af mathon

$\small \begin{array}{lllllll} \blacklozenge\\& \textup{s\ae ttes }&t=\tan\left ( \frac{u}{2} \right )\\& \textup{haves:}\\\\&& \frac{2t}{1+t^2}=2\cdot \tan\left ( \frac{u}{2} \right )\cdot \frac{1}{1+\tan\left ( \frac{u}{2} \right )^2}=2\cdot \frac{\sin\left ( \frac{u}{2} \right )}{\cos\left ( \frac{u}{2} \right )}\cdot \cos^2\left ( \frac{u}{2} \right )=\sin\left (u \right )\\\\&& \left ( 1-t^2 \right )\cdot \cos^2\left ( \frac{u}{2} \right )=\left ( 1-\frac{\sin^2\left ( \frac{u}{2} \right )}{\cos^2\left ( \frac{u}{2} \right )} \right )\cdot \cos^2\left ( \frac{u}{2} \right )=\cos^2\left ( \frac{u}{2} \right )-\sin^2\left ( \frac{u}{2} \right )=\cos(u)\\& \textup{dvs}\\&& \overrightarrow{s}_1(u)=\begin{pmatrix} \cos(u)\\ \sin(u) \end{pmatrix}\\\\&& \overrightarrow{s}_2(u)=\begin{pmatrix} \sin(u)\\ \cos(u) \end{pmatrix}\\\\& \end{array}$

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Svar #4
09. maj 2021 af mathon

$\small \begin{array}{llllll} \blacklozenge\\& \textup{s\aa \ }\\&& \overrightarrow{s}_1\left ( \frac{\pi}{4} \right )=\begin{pmatrix} \cos\left ( \frac{\pi}{4} \right )\\ \sin\left ( \frac{\pi}{4} \right ) \end{pmatrix}=\begin{pmatrix} \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} \end{pmatrix}=\begin{pmatrix} \sin\left ( \frac{\pi}{4} \right )\\ \cos\left ( \frac{\pi}{4} \right ) \end{pmatrix}=\overrightarrow{s}_2\left ( \frac{\pi}{4} \right ) \end{array}$

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Svar #5
10. maj 2021 af mathon

$\small \begin{array}{llllll} \blacklozenge\\& \textup{Kurven er }\\&\textup{enhedscirklen:}\\&&x^2+y^2=1\\\\& \textup{hvis tangent-}\\& \textup{ligning i }(x_o,y_o)\\& \textup{er:}\\&& x_o x+y_oy=1\\\\& \textup{tangent i}\\& \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right )\textup{:}\\&& \frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y=1\\\\&& x+y=\sqrt{2}\\\\&\textup{Normalen i}\\& \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right )\textup{:}\\&& y=x+b\\\\&& \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}+b\\\\&& b=0\\& \textup{dvs}\\&& y=x\quad \textup{som g\aa r gennem }(0,0) \end{array}$

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Svar #6
10. maj 2021 af mathon

$\small \begin{array}{llllll} \blacklozenge\\& \textup{Krumningscirklen }\\& \textup{i }\left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right )\textup{:}\\&& \left (x-\frac{\sqrt{2}}{2} \right )^2+\left (y-\frac{\sqrt{2}}{2} \right )^2=1 \end{array}$

Svar #7
11. maj 2021 af MaryamS

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