Difference between revisions of "Piece-wise defined functions"
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(examples may vary). | (examples may vary). | ||
| − | Below we try to give a bit more motivation to piece-wise functions (which | + | Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications). |
''' Pool table ''' | ''' Pool table ''' | ||
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Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get | Substituting <math> \triangle x = x - (10+ mg/k) \</math> into the formula for <math> v, </math> and simplifying, we get | ||
| − | <math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2} | + | <math> v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2}</math> |
| + | for <math> 10 \le x \le 10+(mg/k)+A.</math> | ||
| + | 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> | ||
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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 = 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> | ||
| + | |||
| + | 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)\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> | ||
| + | |||
| + | 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 | ||
| + | |||
| + | <math> y = | ||
| + | \begin{cases} | ||
| + | h - gx^2/18, & 0\le x \le 3\sqrt{2h/g}\\ | ||
| + | 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> | ||
Latest revision as of 21:15, 3 April 2022
A typical introduction of piece-wise defined functions in a calculus textbook looks something like this: there are functions that take on different formulas depending on the intervals (pieces) of their domains, for example... (examples may vary).
Below we try to give a bit more motivation to piece-wise functions (which are rather indispensable tools in mathematics and applications).
Pool table
A pool table has dimensions 10 ft by 5 ft. A ball is hit from a corner towards the longer side at the angle 30 degrees (with respect to the shorter side). Describe the trajectory of the ball before it hits the side second time.
Even numbers
Determine how many positive even numbers are there thad do not exceed a given positive integer.
Bungee jumping
Alice is doing bungee jumping and jumps from a tall bridge. She uses a bungee cord that is 10 meters long (non-stretched). The spring constant of the cord is 600 N/m. Alice weighs 60 kg. Find how Alices velocity changes as a function of distance from the bridge as she descents form the bridge to the lowest point. You can make necessary simplifying assumption (no air resistance, weight of the cord is negligible, Alice jumps directly downwards, the cord is attached at the centre of mass, etc.).
Choose the x-axis (in meters) pointing downwards and with the origin coinciding with Alice's initial position on the bridge. For the first part of her journey, i.e. when [math]\displaystyle{ 0\le x \le 10 }[/math], the only acting force is gravity, so the speed is [math]\displaystyle{ gt }[/math], or, equivalently, [math]\displaystyle{ v(x) = \sqrt{2gx} }[/math]. After the point where [math]\displaystyle{ x = 10 }[/math], a restoring force comes into play. Velocity at that point is [math]\displaystyle{ v(10) = \sqrt{2g\cdot 10} = 2\sqrt{5g} }[/math]. The system is in the equilibrium when the restoring force is equal to the gravity, i.e. when [math]\displaystyle{ x = 10+ mg/k. }[/math] If [math]\displaystyle{ \triangle x }[/math] denotes the displacement from the equilibrium, the velocity is [math]\displaystyle{ v(\triangle x) = \sqrt{k/m} \sqrt{A^2 - (\triangle x)^2} }[/math], where [math]\displaystyle{ A }[/math] is the amplitude and can be computed by substituting [math]\displaystyle{ \triangle x = - mg/k }[/math] and therefore (using [math]\displaystyle{ m= 60 }[/math]kg, [math]\displaystyle{ k = 600 N/m }[/math], and [math]\displaystyle{ g\approx 9.8 }[/math]m/s2) is [math]\displaystyle{ A = \sqrt{(mg/k)^2 +20mg/k}\approx 4.5. }[/math]
Substituting [math]\displaystyle{ \triangle x = x - (10+ mg/k) \ }[/math] into the formula for [math]\displaystyle{ v, }[/math] and simplifying, we get
[math]\displaystyle{ v(x) = \sqrt{20(20-x)g - (k/m) (x-10)^2} }[/math]
for [math]\displaystyle{ 10 \le x \le 10+(mg/k)+A. }[/math]
Therefore, again using [math]\displaystyle{ m= 60 }[/math]kg, [math]\displaystyle{ k = 600 N/m }[/math], [math]\displaystyle{ g\approx 9.8 }[/math]m/s2, and [math]\displaystyle{ A \approx 4.5, }[/math] we obtain
[math]\displaystyle{ 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]
Ball trajectory
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]\displaystyle{ x = 3t }[/math] and [math]\displaystyle{ y = 1.5 - gt^2/2 }[/math] for [math]\displaystyle{ 0\le t \le \sqrt{3/g}. }[/math] Solving this in terms of [math]\displaystyle{ x, }[/math] we obtain
[math]\displaystyle{ y = 1.5 - gx^2/18, }[/math] [math]\displaystyle{ 0\le x \le 3\sqrt{3/g}. }[/math]
a) Assuming conservation of energy and using symmetry, we see that
[math]\displaystyle{ y =
\begin{cases}
1.5 - gx^2/18, & 0\le x \le 3\sqrt{3/g}\\
1.5 - 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]
For subsequent computations in b), it will be useful to solve the above problem in more general setting of arbitrary initial height [math]\displaystyle{ h. }[/math] Namely we get [math]\displaystyle{ x = 3t }[/math] and [math]\displaystyle{ y = h - gt^2/2 }[/math] for [math]\displaystyle{ 0\le t \le \sqrt{2h/g}. }[/math] Solving this in terms of [math]\displaystyle{ x, }[/math] we obtain [math]\displaystyle{ y = h - gx^2/18, }[/math] [math]\displaystyle{ 0\le x \le 3\sqrt{2h/g}, }[/math] and
[math]\displaystyle{ y = \begin{cases} h - gx^2/18, & 0\le x \le 3\sqrt{2h/g}\\ 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]