## Calculus 8th Edition

Published by Cengage

# Chapter 13 - Vector Functions - 13.1 Vector Functions and Space Curves - 13.1 Exercises - Page 895: 44

#### Answer

$r(t)=ti+t^2j+4t^2+t^4k$ ; or, $r(t) =\lt t, t^2, 4t^2+t^4 \gt$

#### Work Step by Step

The parametric equations of a circle whose radius is $r$ are given as follows: $x=r \cos t ; y =r \sin t$ From the given problem, we have $z=4x^2+y^2$ and $y=x^2$ Also, $x^2+y^2=1+2y+y^2$ This implies that $y=\dfrac{x^2-1}{2}$ When we plug $x=t$, then we get $y=(x)^2=t^2$ This gives: , $z=4x^2+y^2$ or, $z=4t^2+(t^2)^2=4t^2+t^4$ Hence, we have the parametric equation in the vector form as follows: $r(t)=ti+t^2j+4t^2+t^4k$ ; or, $r(t) =\lt t, t^2, 4t^2+t^4 \gt$

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