Find the positive value for which the vector
r
(
t
)
⃗
=
⟨
4
t
,
6
t
2
,
6
t
2
−
5
⟩
\displaystyle \vec{{{r}{\left({t}\right)}}}={\left\langle{4}{t},{6}{t}^{{2}},{6}{t}^{{2}}-{5}\right\rangle}
r
(
t
)
=
⟨
4
t
,
6
t
2
,
6
t
2
−
5
⟩
is perpendicular to
r
′
(
t
)
⃗
\displaystyle \vec{{{r}'{\left({t}\right)}}}
r
′
(
t
)
.
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\displaystyle
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter DNE for Does Not Exist, oo for Infinity