آهنگ تغییرات

\(\begin{array}{l}v(x) = {x^3}\\\\\frac{{v}}{{x}} = \frac{{v(5) - v(2)}}{{5 - 2}} = \frac{{125 - 8}}{3} = 39\end{array}\)

\(\begin{array}{l}s(r) = \pi {r^2} \to {s^`}(r) = 2\pi r\\\\{s^`}(5) = 2\pi r\left( 5 \right) = 10\pi \end{array}\)

\(\begin{array}{l}f(t) = 30 + 10{t^2}\\\\f(2) = 30 + 10{(2)^2} = 70\\\\f(7) = 30 + 10{(7)^2} = 520\\\\\frac{{f}}{{x}} = \frac{{f(7) - f(2)}}{{7 - 2}} = \frac{{520 - 70}}{5} = 90\end{array}\)

\(\begin{array}{l}l = (1) + (2) - 2(1)(2)\cos \alpha \to l(\alpha ) = \sqrt {5 - 4\cos \alpha } \\\\{l^`}(\alpha ) = \frac{{4\cos \alpha }}{{2\sqrt {5 - 4\cos \alpha } }} = \frac{{2\sin \alpha }}{{\sqrt {5 - 4\cos \alpha } }}\\\\{l^`}(\frac{\pi }{4}) = \frac{{2\sin (\frac{\pi }{4})}}{{\sqrt {5 - 4\cos (\frac{\pi }{4})} }} = \frac{{2\sqrt {\frac{2}{2}} }}{{\sqrt {5 - 4(\frac{{\sqrt 2 }}{2})} }} = \frac{{\sqrt 2 }}{{\sqrt {5 - 2\sqrt 2 } }}\end{array}\)

\(\begin{array}{l}s(r) = \pi {r^2}s(p) = \pi {(\frac{p}{{2\pi }})^2} = {\frac{p}{{4\pi }}^2}\\\\s(p) = {\frac{p}{{4\pi }}^2} \to {s^`}(p) = \frac{{2p}}{{4\pi }} = \frac{p}{{2\pi }}\\\\s(5\pi ) = \frac{{5\pi }}{2} = 2/5\end{array}\)

تابع تخلیه آب

\(\begin{array}{l}v(t) = 60 - 60{(1 - \frac{t}{{200}})^2} = 60(1 - (1 - \frac{t}{{100}} + \frac{{{t^2}}}{{4000}}))\\\\ = 60(\frac{t}{{100}} - \frac{{{t^2}}}{{4000}})60 \times \frac{{400t - {t^2}}}{{4000}} = \frac{3}{{2000}}(400t - {t^2})\end{array}\)

الف

\(\begin{array}{l}v(t) = \frac{3}{{2000}}(400t - {t^2})\\\\v(0) = \frac{3}{{2000}}(400(0) - {(0)^2}) = 0\\\\v(60) = \frac{3}{{2000}}(400(60) - {(60)^2}) = 30/6\\\\\frac{{v}}{{t}} = \frac{{v(60) - v(0)}}{{60 - 0}} = \frac{{30/6 - 0}}{{60}} = 0/51\end{array}\)

ب

\(\begin{array}{l}v(t) = \frac{3}{{2000}}(400t - {t^2}) \to v(t) = \frac{3}{{2000}}(400 - 2t) = \frac{3}{{1000}} = (200 - t)\\\\{v^`}(100) = \frac{3}{{1000}}(200 - 100) = \frac{3}{{10}} = 0/3\end{array}\)

\(x(t) = {t^2} - t + 6\)

الف

\(\begin{array}{l}x(3) = {(3)^2} - 5(3) + 6 = 9 - 15 + 6 = 0\\\\x(5) = {(5)^2}\_5(5) + 6 = 25 - 25 + 6 = 6\\\\v = \frac{{x}}{{t}} = \frac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \frac{{6 - 0}}{{5 - 3}} = 3\frac{m}{s}\end{array}\)

ب

\({x^`}(t) = 2t - 5 \to {x^`}(2) = 2(2) - 5 = - 1\frac{m}{s}\)

1 معادله ی داده شده یک سهمی و چون در آن \(a = - 5\)پس نمودار سهمی رو به پایین بوده و دارای نقطه یmax است.

2 چون بعد از 4 ثانیه توپ مجددا به زمین بر میگردد .لذا دامنه ی تابع \(D = \left[ {0,4} \right]\)می شود.

3 

\(\begin{array}{l}y(t) = - 5{t^2} + 20t\\\\y(0) = - 5{(0)^2} + 20(0) = 0\\\\y(2) = - 5{(2)^2} + 20(2) = - 20 + 4\\\\y(t) = - 5{t^2} + 20t\\\\\frac{{y}}{{t}} = \frac{{y(2) - y(0)}}{{2 - 0}} = \frac{{20 + 0}}{2} = 10\end{array}\)

4 

\(\begin{array}{l}y(t) = - 5{t^2} + 20t \to {y^`}(t) = - 10t + 20t\\\\{y^`}(1) = - 10(1) + 20 = 10\frac{m}{s}\end{array}\)

5 

\({y^`}(4) = - 10(4) + 20 = - 20\frac{m}{s}\)

6  بالاترین ارتفاع توپ زمانی است که\(t = 2\) باشد.لذا

\({y^`}(2) = - 10(2) + 20 = 0\frac{m}{s}\)

یعنی سرعت لحظه ای توپ در این لحظه برابر صفر است.(ایست لحظه ای )