Answer
(f)
Work Step by Step
Since
$$\mathbf{r}=\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k} $$
Then
$$y=t^{2}-1,\ z=2 t \Rightarrow y=\frac{z^{2}}{4}-1$$
Hence, the graph is (f).
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 16: Integrals and Vector Fields - Section 16.1 - Line Integrals - Exercises 16.1 - Page 943: 7
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