Skillnad mellan versioner av "2.5 Lösning 6"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 14: | Rad 14: | ||
\end{array}</math> | \end{array}</math> | ||
| − | <math> | + | <math> f\,'(0) = 5 </math> innebär: |
| − | :<math> \begin{array}{rclcl} f(0) & = & | + | :<math> \begin{array}{rclcl} f\,'(0) & = & 50 \cdot k \cdot e\,^{k\,\cdot\, 0} & = & 5 \\ |
| − | + | & & 50 \cdot k \cdot e\,^{0} & = & 5 \\ | |
| − | + | & & 50 \cdot k \cdot 1 & = & 5 \\ | |
| − | + | & & 50 \cdot k & = & 5 \\ | |
| + | & & k & = & {5\over 50} \\ | ||
| + | & & k & = & 0,1 | ||
\end{array}</math> | \end{array}</math> | ||
Versionen från 31 oktober 2014 kl. 12.10
\[ \begin{array}{rclcl} f(x) & = & C \cdot e\,^{k\,x} & & \\ \end{array}\]
\( \, f(0) = 50 \) innebär:
\[ \begin{array}{rclcl} f(0) & = & C \cdot e\,^{k\,\cdot\, 0} & = & 50 \\ & & C \cdot e\,^{0} & = & 50 \\ & & C \cdot 1 & = & 50 \\ & & C & = & 50 \end{array}\]
\[ \begin{array}{rclcl} f(x) & = & 50 \cdot e\,^{k\,x} & & \\ f\,'(x) & = & 50 \cdot k \cdot e\,^{k\,x} & & \\ \end{array}\]
\( f\,'(0) = 5 \) innebär:
\[ \begin{array}{rclcl} f\,'(0) & = & 50 \cdot k \cdot e\,^{k\,\cdot\, 0} & = & 5 \\ & & 50 \cdot k \cdot e\,^{0} & = & 5 \\ & & 50 \cdot k \cdot 1 & = & 5 \\ & & 50 \cdot k & = & 5 \\ & & k & = & {5\over 50} \\ & & k & = & 0,1 \end{array}\]
