Piece-wise defined functions - Revision history

Alexandk at 01:15, 4 April 2022

2022-04-04T01:15:03Z

2022-04-04T01:08:45Z

2022-04-04T01:05:36Z

2022-04-04T01:04:00Z

2022-04-04T00:57:03Z

2022-04-04T00:50:50Z

2022-04-04T00:45:52Z

2022-04-02T18:03:15Z

2022-04-02T18:01:08Z

2022-04-02T18:00:46Z

@@ Line 57: / Line 57: @@
 <math> y =
 \begin{cases}
-.5 - gx^2/18,  & 0\le x \le 3\sqrt{3/g}\\
+h - gx^2/18,  & 0\le x \le 3\sqrt{2h/g}\\
-.5 - g(x-6(k+1)\sqrt{2h/g})^2/18, & (3+6k)\sqrt{2h/g} \le x \le (9+6k)\sqrt{2h/g}\mbox{ for }k =0,1,2,\dots
+h - g(x-6(k+1)\sqrt{2h/g})^2/18, & (3+6k)\sqrt{2h/g} \le x \le (9+6k)\sqrt{2h/g}\mbox{ for }k =0,1,2,\dots
 \end{cases}
 </math>

@@ Line 53: / Line 53: @@
 </math>
-For subsequent computations in b), it will be useful to solve the above problem in more general setting of arbitrary initial height <math> h.</math>
+For subsequent computations in b), it will be useful to solve the above problem in more general setting of arbitrary initial height <math> h.</math> Namely we get <math> x = 3t </math> and  <math> y = h - gt^2/2 </math> for <math> 0\le t \le \sqrt{2h/g}. </math> Solving this in terms of <math> x,</math> we obtain <math> y = h - gx^2/18,</math> <math> 0\le x \le 3\sqrt{2h/g}, </math> and
 b) If 20% of energy is lost on each bounce, then the maximal height the ball reaches after the first bounce is 80% of the initial height, i.e. <math> 1.5\cdot 0.8. </math> Similarly, the height after the <math> k+ 1 </math> bounce will be <math> 1.5\cdot(0.8)^{k+1},</math> <math> k=0,1,2,\dots </math> Thus, for <math> 3\sqrt{3/g} \le x \le 9\sqrt{3/g} </math> we have <math> y = 1.5 - g(x-6)^2/18 </math>

@@ Line 52: / Line 52: @@
 \end{cases}
 </math>
 b) If 20% of energy is lost on each bounce, then the maximal height the ball reaches after the first bounce is 80% of the initial height, i.e. <math> 1.5\cdot 0.8. </math> Similarly, the height after the <math> k+ 1 </math> bounce will be <math> 1.5\cdot(0.8)^{k+1},</math> <math> k=0,1,2,\dots </math> Thus, for <math> 3\sqrt{3/g} \le x \le 9\sqrt{3/g} </math> we have <math> y = 1.5 - g(x-6)^2/18 </math>

@@ Line 49: / Line 49: @@
 \begin{cases}
 .5 - gx^2/18,  & 0\le x \le 3\sqrt{3/g}\\
-.5 - g(x-6(k+1))^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
+.5 - g(x-6(k+1)\sqrt{3/g})^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
 \end{cases}
 </math>

← Older revision		Revision as of 00:57, 4 April 2022
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	Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain		Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain

−	<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>	+	<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>
		+
		+	a) Assuming conservation of energy and using symmetry, we see that
		+
		+
		+	<math> y =
		+	\begin{cases}
		+	1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
		+	1.5 - g(x-6(k+1))^2/18, & (3+6k)\sqrt{3/g} \le x \le (9+6k)\sqrt{3/g}\mbox{ for }k =0,1,2,\dots
		+	\end{cases}
		+	</math>

← Older revision		Revision as of 00:50, 4 April 2022
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	A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.		A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.

−	Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 3 - gt^2/2 </math> for <math> 0\le t \le \sqrt{2h/g}. </math>	+	Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 1.5 - gt^2/2 </math> for <math> 0\le t \le \sqrt{3/g}. </math> Solving this in terms of <math> x,</math> we obtain
		+
		+	<math> y = 1.5 - gx^2/18,</math> <math> 0\le x \le 3\sqrt{3/g}. </math>

← Older revision		Revision as of 00:45, 4 April 2022
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	A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.		A person throws a ball forward at 3 m/s. The ball bounces off the ground, reaches some maximal height, then falls and bounces again, and so on. If the initial altitude of the ball is 1.5 m, find the trajectory of the ball if a) no energy is lost b) about 20% of energy is lost on each bounce.
		+
		+	Note that the ball follows the projectile trajectory until it hits the ground, i.e. <math> x = 3t </math> and <math> y = 3 - gt^2/2 </math> for <math> 0\le t \le \sqrt{2h/g}. </math>

@@ Line 2: / Line 2: @@
 (examples may vary).
-Below we try to give a bit more motivation to piece-wise functions (which is not just a nuance for students, but rather indispensable tool in mathematics and applications).
+Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).
 ''' Pool table '''

@@ Line 30: / Line 30: @@
 \begin{cases}
 \sqrt{2gx},  & 0\le x \le 10 \\
-\sqrt{2(20-x)g-10(x-10)^2}, 10 \le x \le 15.5
+\sqrt{2(20-x)g-10(x-10)^2}, & 10 \le x \le 15.5
 \end{cases}
 </math>

← Older revision		Revision as of 18:00, 2 April 2022
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	for <math> 10 \le x \le 10+(mg/k)+A.</math>		for <math> 10 \le x \le 10+(mg/k)+A.</math>

−	Therefore, again using <math> m= 60 </math>kg ~~and~~ <math> k = 600 N/m </math>, we obtain	+	Therefore, again using <math> m= 60 </math>kg, <math> k = 600 N/m </math>, <math> g\approx 9.8</math>m/s<sup>2</sup>, and <math> A \approx 4.5,</math> we obtain
−

		+	<math> v(x) =
		+	\begin{cases}
		+	\sqrt{2gx}, & 0\le x \le 10 \\
		+	\sqrt{2(20-x)g-10(x-10)^2}, 10 \le x \le 15.5
		+	\end{cases}
		+	</math>