**PARAMETRIC CURVES**

**Assignment 10**

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**By
Gloria L. Jones**

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**A
parametric curve in the plane is a pair of functions:**

** x=f(t)**

** y=g(t)**

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**where
the two continuous functions define ordered pairs (x, y). In this assignment, we will graph a
line by using a parametric equation of a line through (7,5) with a slope of
3. Exploring sets of curves for
x=a+t and y=b+kt, various linear graphs were discovered. **

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**As the
values of k change, one notices a change in the slope of the line though
despite the movement, it is easy to notice a point that remains stationary
throughout. At a closer glance,
one notices this fixed point to be (3,5) from the parametric equations
above. Looking specifically at the
parametric equations, one notices that not only does the point (3,5) fall on
the graph at t=0 (approximation) but that the slope of the line appears to be **

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**y = 4x -7**

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**K=4. Based on this assumption, the
parametric equations for the line through (7,5) with a slope of 3 should
be
**

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**Algebraically
we know that the equation of this line is y-5=3(x-7) which yields y=3x-16 with
slope of 3 and y-intercept of -16.
In general, x=a+bt and y=c+dt are the parametric equations of a line
(both b and d0). Through
substitutions we have x=a+bt and then solving for t obtain t=(x-a)/b. Replacing t in the y=c+dt equation, we
have y=c+d((x-a)/b) which implies y=c+dx/b-da/b and thus y=(d/b)x+((bc-ad)/b)
so d/b is the slope and (bc-ad)/b is the y-intercept of the line. Therefore, in our example, since the
slope is 3, then d/b = 3/1 so d=3 and b=1. Since x=a+bt and y=c+dt, we can now say x=a+1t and
y=c+3t. When t=0, x=a and y=c so
(a,c) must be a point on the line.
Le (a,c) + (7,5) so x=7+t and y=5+3t are the parametric equations for
the line. A parametric sketch
using GSP confirms our calculation.**

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**y = 3x -16**

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**Notice
the two graphs coincide, pass through the point (7, 5), have a slope of 3 and a
y-intercept of -16.**

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