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Find a vector parametrization of the curve x=-4z^2 in the xz-plane.

Question:

Find a vector parametrization of the curve {eq}x=-4z^2 {/eq} in the xz-plane.

Parametric Equations:

Consider a curve in the form of {eq}x {/eq} and {eq}y {/eq}, defining both {eq}x {/eq} and {eq}y {/eq} in terms of third variable say {eq}t {/eq} {eq}x = f\left( t \right) \ , y = g\left( t \right) {/eq} This is defined as parametric equations.

Answer and Explanation:

Given curve {eq}x= -4z^2 \\ {/eq}

Let {eq}z=t \\ {/eq}

Then {eq}x = -4t^2 \\ {/eq} The parametric equations are {eq}(-4t^2,0,t)\\ {/eq} In vector form {eq}\vec{r(t)} = -4t^2 \vec{i} + t \vec{k} {/eq}


Learn more about this topic:

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Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5
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