Skillnad mellan versioner av "2.5 Lösning 6"
Taifun (Diskussion | bidrag) m |
Taifun (Diskussion | bidrag) m |
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| Rad 1: | Rad 1: | ||
:<math> \begin{array}{rclcl} f(x) & = & C \cdot e\,^{k\,x} & & \\ | :<math> \begin{array}{rclcl} f(x) & = & C \cdot e\,^{k\,x} & & \\ | ||
\end{array}</math> | \end{array}</math> | ||
| + | |||
<math> \, f(0) = 50 </math> innebär: | <math> \, f(0) = 50 </math> innebär: | ||
| + | |||
| + | :<math> \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}</math> | ||
| + | |||
| + | :<math> \begin{array}{rclcl} f(x) & = & 50 \cdot e\,^{k\,x} & & \\ | ||
| + | f\,'(x) & = & 50 \cdot k \cdot e\,^{k\,x} & & \\ | ||
| + | \end{array}</math> | ||
| + | |||
| + | <math> \, f\,'(0) = 50 </math> innebär: | ||
:<math> \begin{array}{rclcl} f(0) & = & C \cdot e\,^{k\,\cdot\, 0} & = & 50 \\ | :<math> \begin{array}{rclcl} f(0) & = & C \cdot e\,^{k\,\cdot\, 0} & = & 50 \\ | ||
Versionen från 31 oktober 2014 kl. 12.04
\[ \begin{array}{rclcl} f(x) & = & C \cdot e\,^{k\,x} & & \\ \end{array}\]
\( \, f(0) = 50 \) innebär:
\[ \begin{array}{rclcl} f(0) & = & C \cdot e\,^{k\,\cdot\, 0} & = & 50 \\ & & C \cdot e\,^{0} & = & 50 \\ & & C \cdot 1 & = & 50 \\ & & C & = & 50 \end{array}\]
\[ \begin{array}{rclcl} f(x) & = & 50 \cdot e\,^{k\,x} & & \\ f\,'(x) & = & 50 \cdot k \cdot e\,^{k\,x} & & \\ \end{array}\]
\( \, f\,'(0) = 50 \) innebär:
\[ \begin{array}{rclcl} f(0) & = & C \cdot e\,^{k\,\cdot\, 0} & = & 50 \\ & & C \cdot e\,^{0} & = & 50 \\ & & C \cdot 1 & = & 50 \\ & & C & = & 50 \end{array}\]
