## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 638: 4

#### Answer

$$c(t)= (2+t,1-2t)$$ (Other parametrizations are possible.)

#### Work Step by Step

Since the equation of the line is $y=5-2x$, then its slope is $m=-2$. Then we have $\frac{s}{r}=-2$, so take (for example) $r=1$ and $s=-2$. The line passes through one (sample) point $(0,5)$. Thus, the parametric equations are $$x=a+rt=t, \quad y= b+st=5-2t.$$ That is, $$c(t)=(x(t),y(t))=(t,5-2t).$$ Now, we define new parametrization $$c_0(s)=(s+h,5-2s-2h),$$ Since we need $c_0(0)=(2,1)=(h,5-2h)$, then $h=2$. Hence the required parametrization is $$c_0(s)= (2+s,1-2s)$$ We rewrite the result in the form of $c(t)$: $$c(t)= (2+t,1-2t)$$ (Other parametrizations are possible.)

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