## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 638: 7

#### Answer

$c\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right) = \left( {2t + 1,4t + 3} \right)$

#### Work Step by Step

We may write $x\left(t\right)=2t+1$, so that $y\left(t\right)=2\left(2t+1\right)+1=4t+3$. So, the parametrization is $c\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right) = \left( {2t + 1,4t + 3} \right)$ When $t=0$, we have $c\left(0\right)=\left({1,3}\right)$. When $t=1$, we have $c\left(1\right)=\left({3,7}\right)$. Thus, the path $c\left(t\right)$ traces the line $y=2x+1$ from $\left({1,3}\right)$ to $\left( {3,7} \right)$ for $0 \le t \le 1$.

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