Matematik

mathon hjælp

04. december 2020 af Matfuckdk - Niveau: A-niveau

Hjælp

Vedhæftet fil: Udklip.PNG

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Svar #1
04. december 2020 af mathon

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Svar #2
04. december 2020 af mathon

$\small \small \begin{array}{llllll} 1.\\& \begin{array}{llllll} \frac{\mathrm{d} C}{\mathrm{d} t}=37.5\\\\ \int dC=\int 37.5\,\mathrm{d} t\\\\ C(t)=37.5t+k\\\\& C(0)=37.5\cdot 0+350\\\\& k=350\\\\ C(t)=37.5t+350\\\\ t(C)=\frac{C-350}{37.5}\\\\\\ \left \{ t_{1000},t_{2000} \right \}=\left \{ \frac{1000-350}{37.5} ,\frac{2000-350}{37.5}\right \}=\left \{ 17.3,44 \right \} \end{array} \end{array}$

Svar #3
04. december 2020 af Matfuckdk

#2
$\small \small \begin{array}{llllll} 1.\\& \begin{array}{llllll} \frac{\mathrm{d} C}{\mathrm{d} t}=37.5\\\\ \int dC=\int 37.5\,\mathrm{d} t\\\\ C(t)=37.5t+k\\\\& C(0)=37.5\cdot 0+350\\\\& k=350\\\\ C(t)=37.5t+350\\\\ t(C)=\frac{C-350}{37.5}\\\\\\ \left \{ t_{1000},t_{2000} \right \}=\left \{ \frac{1000-350}{37.5} ,\frac{2000-350}{37.5}\right \}=\left \{ 17.3,44 \right \} \end{array} \end{array}$

Tak, men hvordan med opgave 2 og 3

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Svar #4
04. december 2020 af mathon

$\small \small \small \begin{array}{llllll} 2.\\& \begin{array}{llllll} \frac{\mathrm{d} C}{\mathrm{d} t}=53.25-0.045\cdot C\\\\ C{\, }'+0.045C=53.25\qquad \textup{panserformlen}\\\\ C(t)=e^{-0.045t}\cdot \int 53.25\cdot e^{0.045t}\,\mathrm{d}t\\\\ C(t)=e^{-0.045t}\cdot\left ( \frac{53.25}{0.045} \cdot e^{0.045t}+k\right )\\\\ C(t)=k\cdot e^{-0.045t}+1183 \\\\& C(0)=k\cdot 1+1183=350\\\\& k=-833\\\\ C(t)=1183-833\cdot e^{-0.045t}\\\\ t(C)=\frac{\ln\left ( \frac{833}{1183-C} \right )}{0.045}\\\\\\ t(1000)=\frac{\ln\left ( \frac{833}{1183-1000} \right )}{0.045}=34\;(min) \end{array} \end{array}$

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Svar #5
04. december 2020 af mathon

$\small \small \small \small \begin{array}{llllll} 3.\\& \begin{array}{llllll} C(t)=1183-833\cdot e^{-0.045t}\\\\ \underset{\lim\, t \to \infty}{ C(t)}=1183-833\cdot e^{-0.045\cdot \infty}=1183-0=1183\;(\textup{ppm}) \end{array} \end{array}$

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