Skillnad mellan versioner av "2.5 Lösning 6"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
||
| Rad 16: | Rad 16: | ||
<math> f\,'(0) = 5 </math> innebär: | <math> f\,'(0) = 5 </math> innebär: | ||
| − | :<math> \begin{array}{rcrcl} | + | :<math> \begin{array}{rcrcl} f\,'(0) & = & 50 \cdot k \cdot e\,^{k\,\cdot\, 0} & = & 5 \\ |
| − | + | & & 50 \cdot k \cdot e\,^{0} & = & 5 \\ | |
| − | + | & & 50 \cdot k \cdot 1 & = & 5 \\ | |
| − | + | & & k & = & {5 \over 50} \\ | |
| − | + | & & k & = & 0,1 | |
| − | + | ||
\end{array}</math> | \end{array}</math> | ||
Versionen från 31 oktober 2014 kl. 12.16
\[ \begin{array}{rclcl} f(x) & = & C \cdot e\,^{k\,x} & & \\ \end{array}\]
\( \, f(0) = 50 \) innebär:
\[ \begin{array}{rcrcl} f(0) & = & C \cdot e\,^{k\,\cdot\, 0} & = & 50 \\ & & C \cdot e\,^{0} & = & 50 \\ & & C \cdot 1 & = & 50 \\ & & C & = & 50 \end{array}\]
\[ \begin{array}{rclcl} f(x) & = & 50 \cdot e\,^{k\,x} & & \\ f\,'(x) & = & 50 \cdot k \cdot e\,^{k\,x} & & \\ \end{array}\]
\( f\,'(0) = 5 \) innebär:
\[ \begin{array}{rcrcl} f\,'(0) & = & 50 \cdot k \cdot e\,^{k\,\cdot\, 0} & = & 5 \\ & & 50 \cdot k \cdot e\,^{0} & = & 5 \\ & & 50 \cdot k \cdot 1 & = & 5 \\ & & k & = & {5 \over 50} \\ & & k & = & 0,1 \end{array}\]
