This commit is contained in:
baiobelfer
commit
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import matplotlib.pyplot as plt
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import numpy as np
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# Parametry ruchu
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v = 5 # prędkość w m/s
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s0 = 2 # początkowe położenie w m
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# Generowanie czasu i położenia
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t = np.arange(0, 9, 1) # czas od 0 do 8 sekund
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s = v * t + s0 # położenie w funkcji czasu
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# Rysowanie wykresu
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plt.figure(figsize=(8, 6))
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plt.plot(t, s, marker='o', color='b', label=f's(t) = {v}*t + {s0}')
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plt.title("Wykres zależności położenia s od czasu t")
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plt.xlabel("Czas (s)")
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plt.ylabel("Położenie s (m)")
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plt.legend()
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plt.grid(True)
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plt.show()
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import matplotlib.pyplot as plt
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import numpy as np
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# Parametry ruchu liniowego
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v = 5 # prędkość w m/s
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s0 = 2 # początkowe położenie w m
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# Funkcja liniowa: s(t) = v * t + s0
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t = np.linspace(0, 8, 100) # dokładniejszy przedział czasu do rysowania wykresu
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s_linear = v * t + s0 # położenie dla funkcji liniowej
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# Funkcja kwadratowa: s_quad(t) = a*t^2 + b*t + c
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a, b, c = 1, -5, 5 # dobór współczynników do uzyskania 2 punktów przecięcia
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s_quad = a * t**2 + b * t + c # położenie dla funkcji kwadratowej
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# Znajdowanie punktów przecięcia
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intersection_points = []
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for i in range(len(t)):
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if np.isclose(s_linear[i], s_quad[i], atol=0.1): # tolerancja na różnice wartości
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intersection_points.append((t[i], s_linear[i]))
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# Rysowanie wykresów
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plt.figure(figsize=(10, 6))
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plt.plot(t, s_linear, label=f'Liniowa: s(t) = {v}*t + {s0}', color='b')
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plt.plot(t, s_quad, label=f'Kwadratowa: s(t) = {a}*t^2 + ({b})*t + {c}', color='r')
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plt.xlabel("Czas (s)")
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plt.ylabel("Położenie s (m)")
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plt.title("Wykres zależności położenia s od czasu t")
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# Oznaczanie punktów przecięcia
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for x, y in intersection_points:
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plt.plot(x, y, 'go') # zielony punkt dla przecięć
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plt.text(x, y, f'({x:.2f}, {y:.2f})', fontsize=9, ha='right')
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plt.legend()
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plt.grid(True)
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plt.show()
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\relax
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l.13 \maketitle
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...
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l.17 ...zadania jest znalezienie punktów przecię
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cia dwóch funkcji: funkcj...
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...
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l.17 ...funkcji kwadratowej. Obie funkcje opisują
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zależność położenia ...
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l.23 \section*{Rozwiązanie}
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...
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l.25 Aby znaleźć punkty przecię
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cia, przyrównujemy obie funkcje do siebie:
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?
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l.30 Przenoszą
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l.30 Przenosząc wszystkie wyrazy na jedną
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l.30 ...nosząc wszystkie wyrazy na jedną stronę
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...
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l.35 Równanie to można rozwią
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zać przy użyciu wzoru kwadratowego:
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...
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l.42 Dla potrzeb zadania przyję
|
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liśmy następujące wartości parametrów:
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l.42 Dla potrzeb zadania przyjęliśmy nastę
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pujące wartości parametrów:
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l.42 Dla potrzeb zadania przyjęliśmy następują
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ce wartości parametrów:
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?
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l.44 \item Prę
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dkość \( v = 5 \, \text{m/s} \)
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...
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l.45 \item Począ
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tkowe położenie \( s_0 = 2 \, \text{m} \)
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l.49 \section*{Punkty przecięcia}
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l.51 Rozwią
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zując równanie kwadratowe, otrzymaliśmy następujące wartoś...
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l.51 Rozwiązują
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c równanie kwadratowe, otrzymaliśmy następujące wartoś...
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...
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l.51 ...równanie kwadratowe, otrzymaliśmy nastę
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pujące wartości \( t \) ...
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...
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l.51 ...anie kwadratowe, otrzymaliśmy następują
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ce wartości \( t \) oraz ...
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...
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l.51 ...ujące wartości \( t \) oraz odpowiadają
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ce im wartości położeni...
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...
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l.54 \item Punkt przecię
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cia 1: \( t_1 = \frac{-B + \sqrt{B^2 - 4 \cdot ...
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...
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l.55 \item Punkt przecię
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cia 2: \( t_2 = \frac{-B - \sqrt{B^2 - 4 \cdot ...
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l.56 \end{itemize}
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l.56 \end{itemize}
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l.56 \end{itemize}
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...
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l.60 ...towej wraz z oznaczeniem punktów przecię
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cia.
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l.63 ...idth=0.8\textwidth]{intersection_plot.png}
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{geometry}
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\geometry{a4paper, margin=1in}
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\title{Analiza Punktów Przecięcia Funkcji Liniowej i Kwadratowej}
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\author{}
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\date{\today}
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\begin{document}
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\maketitle
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\section*{Opis zadania}
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Celem zadania jest znalezienie punktów przecięcia dwóch funkcji: funkcji liniowej oraz funkcji kwadratowej. Obie funkcje opisują zależność położenia \( s \) od czasu \( t \), gdzie:
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\begin{itemize}
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\item Funkcja liniowa: \( s_{\text{linear}}(t) = v \cdot t + s_0 \)
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\item Funkcja kwadratowa: \( s_{\text{quad}}(t) = a \cdot t^2 + b \cdot t + c \)
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\end{itemize}
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\section*{Rozwiązanie}
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Aby znaleźć punkty przecięcia, przyrównujemy obie funkcje do siebie:
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\begin{equation}
|
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|
v \cdot t + s_0 = a \cdot t^2 + b \cdot t + c
|
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|
\end{equation}
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Przenosząc wszystkie wyrazy na jedną stronę, otrzymujemy równanie kwadratowe w postaci:
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\begin{equation}
|
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|
a \cdot t^2 + (b - v) \cdot t + (c - s_0) = 0
|
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|
\end{equation}
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Równanie to można rozwiązać przy użyciu wzoru kwadratowego:
|
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|
\begin{equation}
|
||||||
|
t = \frac{-(b - v) \pm \sqrt{(b - v)^2 - 4 \cdot a \cdot (c - s_0)}}{2 \cdot a}
|
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|
\end{equation}
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\section*{Parametry}
|
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Dla potrzeb zadania przyjęliśmy następujące wartości parametrów:
|
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|
\begin{itemize}
|
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|
\item Prędkość \( v = 5 \, \text{m/s} \)
|
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\item Początkowe położenie \( s_0 = 2 \, \text{m} \)
|
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\item Współczynniki funkcji kwadratowej: \( a = 1 \), \( b = -5 \), \( c = 5 \)
|
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\end{itemize}
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\section*{Punkty przecięcia}
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Rozwiązując równanie kwadratowe, otrzymaliśmy następujące wartości \( t \) oraz odpowiadające im wartości położenia \( s \):
|
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|
\begin{itemize}
|
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|
\item Punkt przecięcia 1: \( t_1 = \frac{-B + \sqrt{B^2 - 4 \cdot A \cdot C}}{2 \cdot A} \)
|
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|
\item Punkt przecięcia 2: \( t_2 = \frac{-B - \sqrt{B^2 - 4 \cdot A \cdot C}}{2 \cdot A}
|
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\end{itemize}
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\section*{Wykres}
|
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Poniżej przedstawiono wykres funkcji liniowej oraz kwadratowej wraz z oznaczeniem punktów przecięcia.
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\begin{center}
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\includegraphics[width=0.8\textwidth]{intersection_plot.png}
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\end{center}
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|
\end{document}
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|
||||||
Loading…
Reference in New Issue