## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 745: 2

#### Answer

$y=\dfrac{1}{x}-1$; $v(\dfrac{-1}{2})=4i-4j$ and $a(\dfrac{-1}{2})=-16i-16j$

#### Work Step by Step

We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$ In the given problem, we have $x=\dfrac{t}{t+1}; y=\dfrac{1}{t}$ Re-write as follows: $y=(\dfrac{1}{x})-1$ $v(t)=r'(t)=\dfrac{1}{(t+1)^2}i-\dfrac{1}{t^2}j \\ \implies v(\dfrac{-1}{2})=4i-4j$ Next, we have $a(t)=v'(t)=\dfrac{-2}{(t+1)^3}i+\dfrac{2}{t^3}j \\ \implies a(\dfrac{-1}{2})=-16i-16j$

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