Quadratics in Real Life

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Quadratics in Real Life
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• Quadratic functions are more than algebraic
  curiositiesthey are widely used in science,
  business, and engineering.
• The U-shape of a parabola can describe the
  trajectories of water jets in a fountain and a
  bouncing ball, or be incorporated into structures
  like the parabolic reflectors that form the base
  of satellite dishes and car headlights.
• Quadratic functions help forecast business profit
  and loss, plot the course of moving objects, and
  assist in determining minimum and maximum values.
  
• Most of the objects we use every day, from cars
  to clocks, would not exist if someone, somewhere
  hadn't applied quadratic functions to their
  design.

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We commonly use quadratic equations in situations
where two things are multiplied together and they
both depend of the same variable. For example,
when working with area, if both dimensions are
written in terms of the same variable, we use a
quadratic equation. Because the quantity of a
product sold often depends on the price, we
sometimes use a quadratic equation to represent
revenue as a product of the price and the
quantity sold. Quadratic equations are also
used when gravity is involved, such as the path
of a ball or the shape of cables in a suspension
bridge.
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Balls, Arrows, Missiles and Stones

• If you throw a ball (or shoot an arrow, fire a
  missile or throw a stone) it will go up into the
  air, slowing down as it goes, then come down
  again ...
• ... and a Quadratic Equation tells you where it
  will be!

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Example Throwing a Ball

• A ball is thrown straight up, from 3 m above the
  ground, with a velocity of 14 m/s. When does it
  hit the ground?
• Ignoring air resistance, we can work out its
  height by adding up these three things
• The height starts at 3 m   3
• It travels upwards at 14 meters per second (14
  m/s)   14t
• Gravity pulls it down, changing its speed by
  about 5 m/s per second (5 m/s2)   -5t2  

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Add them up and the height h at any time t is

• h-5t2 14t 3
• And the ball will hit the ground when the height
  is zero
• -5t2 14t 3 0
• It will be easier to solve if we factor out the
  negative 1
• -1(5t2 - 14t - 3 ) 0
• Divide both sides by -1 and we have

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5t2 - 14t - 3 0

• 5t2 - 14t 3 (5t 1)(t 3)
• So our two solutions are
• t -0.2 and t 3
• The "t -0.2" is a negative time, impossible in
  our case.
• The "t 3" is the answer we want
• The ball hits the ground after 3 seconds!

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What does it look like?

• Here is the graph of the Parabola h -5t2 14t
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• It shows you the height of the ball vs time
• Some interesting points
• (0,3) When t0 (at the start) the ball is at 3 m
• (-0.2,0) Says that -0.2 seconds BEFORE we threw
  the ball it was at ground level ... this never
  happened, so our common sense says to ignore it!
• (3,0) Says that at 3 seconds the ball is at
  ground level.
• Note also that the ball reaches nearly 13 meters
  high.

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What about that maximum?

• You can find exactly where the top point is!
• Find where (along the horizontal axis) the top
  occurs using -b/2a
• t -b/2a -(-14)/(2 5) 14/10 1.4 seconds
• Then find the height using that value (1.4)
• h -5t2 14t 3 -5(1.4)2 14 1.4 3
  12.8 meters
• So the ball reaches the highest point of 12.8
  meters after 1.4 seconds.

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Lets take a look at some more examples

• Profit