## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 14 - Calculus of Vector-Valued Functions - 14.1 Vector-Valued Functions - Exercises - Page 711: 41

#### Answer

$$r(t)=\lt \frac{\sqrt 3}{2}\cos t, \frac{1}{2}, \frac{\sqrt 3}{2}\sin t \gt.$$

#### Work Step by Step

The intersection of the plane $y = 1 /2$ with the sphere $x^2 + y^2 + z^2 = 1$ is given by $$x^2 + (1/4)^2 +z^2 = 1\Longrightarrow x^2 + z^2 = 3/4.$$ To find a parametrization, we put $x=\frac{\sqrt 3}{2}\cos t$ and $z=\frac{\sqrt 3}{2}\sin t$; that is, we have the parametrization $$r(t)=\lt \frac{\sqrt 3}{2}\cos t, \frac{1}{2}, \frac{\sqrt 3}{2}\sin t \gt.$$

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