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Функције - примена извода 3

Примена извода. Угао између кривих.

Задаци

Текст задатака објашњених у видео лекцији:

Пр.5)   Одредити угао под којим се секу криве ${x^2} + {y^2} - 4x = 1$

            и ${x^2} + {y^2} - 2y = 9$.

Пр.5)

\[\begin{gathered}
{x^2} + {y^2} - 4x = 1| \cdot \left( { - 1} \right) \hfill \\
\underline {{x^2} + {y^2} - 2y = 9} \hfill \\
4x - 2y = 8|:2 \hfill \\
2x - y = 4 \hfill \\
y = 2x - 4 \hfill \\
{x^2} + {\left( {2x - 4} \right)^2} - 4x = 1 \hfill \\
{x^2} + 4{x^2} - 16x + 16 - 4x = 1 \hfill \\
5{x^2} - 20x + 15 = 0 \hfill \\
{x^2} - 4x + 3 = 0 \hfill \\
{x_{1/2}} = \frac{{4 \pm \sqrt {16 - 2} }}{2} = \frac{{4 \pm 2}}{2} \hfill \\
\begin{array}{*{20}{c}}
{{x_1} = 1}&{}&{}&{{x_2} = 3} \\
{{y_1} = 2 - 4}&{}&{}&{{y_2} = 2 \cdot 3 - 4} \\
{{y_1} = - 2}&{}&{}&{{y_2} = 2} \\
{A\left( {1; - 2} \right)}&{}&{}&{B\left( {3;2} \right)}
\end{array} \hfill \\
\hfill \\
t:y - {y_0} = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
{f_1}:{x^2} + {y^2} - 4x = 1 \hfill \\
2x + 2y \cdot y' - 4 = 0 \hfill \\
2yy' = - 2x + 4 \hfill \\
y' = \frac{{ - 2x + 4}}{{2y}} = {f_1}\left( x \right) \hfill \\
{f_1}\left( {{x_0}} \right) = \frac{{ - 2 \cdot 3 + 4}}{{2 \cdot 2}} = - \frac{1}{2} = {k_1} \hfill \\
\end{gathered} &{}&{}&\begin{gathered}
{f_2}:{x^2} + {y^2} - 2y = 9 \hfill \\
2x + 2y \cdot y' - 2y' = 0 \hfill \\
y'\left( {2y - 2} \right) = - 2x \hfill \\
y' = \frac{{ - 2x}}{{2y - 2}} = {f_2}\left( x \right) \hfill \\
{f_2}\left( {{x_0}} \right) = \frac{{ - 2 \cdot 3}}{{2 \cdot 2 - 2}} = - 3 = {k_2} \hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
tg\varphi = \left| {\frac{{{k_1} - {k_2}}}{{1 + {k_1}{k_2}}}} \right| = \left| {\frac{{ - \frac{1}{2} + 3}}{{1 + \left( { - \frac{1}{2}} \right) \cdot \left( { - 3} \right)}}} \right| = \left| {\frac{{\frac{5}{2}}}{{\frac{5}{2}}}} \right| = 1 \hfill \\
\varphi = {45^ \circ } \hfill \\
\end{gathered} \]