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Rectangular Coordinates for Physics Help (page 2)

By — McGraw-Hill Professional
Updated on Sep 17, 2011

Finding Linear Equation Based On Graph

Now suppose that we are working in rectangular coordinates, and we know the exact values of two points P and Q . These two points define a straight line; this is one of the fundamental rules of geometry. Call the line L . Let’s give the coordinates of the points these names:

P = ( x p , y p )

Q = ( x q , y q )

The slope m of the line L is given by either of the following formulas:

m = ( y qy p )/( x qx p )

m = ( y py q )/( x px q )

 

The point-slope equation of L can be determined based on the known coordinates of P or Q . Therefore, either of the following formulas represent the line L :

yy p = m ( xx p )

yy q = m ( xx q )

Equation Of Parabola

The cartesian-coordinate graph of a quadratic equation where y is substituted for 0 in the standard form (recall this from Chapter 1) is a parabola :

y = ax 2 + bx + c

where a ≠ 0. (If a = 0, then the equation is linear, not quadratic.) To plot a graph of the preceding equation, first determine the coordinates of the point ( x 0 , y 0 ) where

x 0 = − b /(2 a )

y 0 = cb 2 /(4 a )

This point represents the base point of the parabola; that is, the point at which the curvature is sharpest and at which the slope of a line tangent to the curve is zero. Once this point is known, find four more points by “plugging in” values of x somewhat greater than and less than x 0 and determining the corresponding y values. These x values, call them x −2 , x −1 , x 1 , and x 2 , should be equally spaced on either side of x 0 , such that

x −2 < x −1 < x 0 < x 1 < x 2

x −1x −2 = x 0x −1 = x 1x 0 = x 2x 1

This will give you five points that lie along the parabola and that are symmetrical relative to the axis of the curve. The graph can then be inferred (this means that you can make a good educated guess), provided that the points are chosen judiciously. Some trial and error may be required. If a > 0, the parabola will open upward. If a < 0, the parabola will open downward.

Equation Of Parabola Practice Problems

Example 1

Consider the following formula:

y = x 2 + 2 x + 1

The base point is

x 0 = −2/2 = −1

y 0 = 1 − 4/4 = 1 − 1 = 0

Therefore, ( x 0 , y 0 ) = (−1,0)

This point is plotted first. Next, plot the following points:

x −2 = x 0 − 2 = −3

y −2 = (−3) 2 + 2 (−3) + 1 = 9−6 + 1 = 4

Therefore, ( x −2 , y −2 ) = (−3, 4)

x −1 = x 0 − 1 = −2

y −1 = (−2) 2 + 2 (−2) + 1 = 4 − 4 + 1 = 1

Therefore, ( x −1 , y −1 ) = (−2, 1)

x 1 = x 0 + 1 = 0

y 1 = (0) 2 + 2(0) + 1 = 0 + 0 + 1 = 1

Therefore, ( x 1 , y 1 ) = (0, 1)

x 2 = x 0 + 2 = 1

y 2 = (1) 2 + 2(1) + 1 = 1 + 2 + 1 = 4

Therefore, ( x 2 , y 2 ) = (1, 4)

The five known points are plotted as shown in Fig. 3-6. From these, the curve can be inferred.

Graphing Schemes Rectangular Coordinates Equation Of Parabola

Fig. 3-6 . Plot of the parabola y = x 2 + 2 x + 1.

Example 2

Consider the following formula:

y = −2 x 2 + 4 x − 5

The base point is

x 0 = −4/−4 = 1

y 0 = − 5 − 16/−8 = −5 + 2 = −3

Therefore, ( x 0 , y 0 ) = (1, −3)

This point is plotted first. Next, plot the following points:

x −2 = x 0 − 2 = −1

y −2 = −2 (−1) 2 + 4 (−1) −5 = −2 − 4 − 5 = −11

Therefore, ( x −2 , y −2 ) = (−1, −11)

x −1 = x 0 − 1 = 0

y −1 = −2(0) 2 + 4 (0) − 5 = −5

Therefore, ( x −1 , y −1 ) = (0, −5)

x 1 = x 0 + 1 = 2

y −1 = −2(2) 2 + 4(2) + 5 = −8 + 8 − 5 = −5

Therefore, ( x 1 , y 1 ) = (2, −5)

x 2 = x 0 + 2 = 3

y 2 = −2(3) 2 + 4 (3) + 5 = −18 + 12 − 5 = −11

Therefore, ( x 2 , y 2 ) = (3, −11)

The five known points are plotted as shown in Fig. 3-7. From these, the curve can be inferred.

 

Graphing Schemes Rectangular Coordinates Equation Of Parabola

Fig. 3-7 . Plot of the parabola y = −2 x 2 + 4 x − 5.

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