Question

The position vector r(t) =<<n t, 1/ t^2, t^4>> describes the path of an object moving in space. (a) Find the velocity vector, speed, and acceleration vector of the object. (b) Evaluate the velocity vector and acceleration vector of the object at the given value of t = sqrt3

Vectors and spaces
ANSWERED
asked 2021-03-08

The position vector \(\displaystyle{r}{\left({t}\right)}={\left\langle{\ln}\ {t},\frac{{1}}{{t}^{{2}}},{t}^{{4}}\right\rangle}\) describes the path of an object moving in space.
(a) Find the velocity vector, speed, and acceleration vector of the object.
(b) Evaluate the velocity vector and acceleration vector of the object at the given value of \(\displaystyle{t}=\sqrt{{3}}\)

Expert Answers (1)

2021-03-09

a) Velocity vector is the derivative of position vector.
So to find velocity vector, differentiate position vector with respect to t. We get \(\displaystyle{v}{\left({t}\right)}={\left\langle\frac{{1}}{{5}},-\frac{{2}}{{t}^{{3}}},{4}{t}^{{3}}\right\rangle}\)
Now speed is the magnitude of velocity vector. That is,
Speed \(\displaystyle={\left|{\left|{v}\right|}\right|}\)
\(\displaystyle=\sqrt{{{\left(\frac{{1}}{{t}}\right)}^{{2}}+{\left(-\frac{{2}}{{t}^{{3}}}\right)}^{{2}}+{\left({4}{t}^{{3}}\right)}^{{2}}}}\)
\(\displaystyle=\sqrt{{\frac{{1}}{{t}^{{2}}}+\frac{{4}}{{t}^{{6}}}+{16}{t}^{{6}}}}\)
\(\displaystyle=\sqrt{{\frac{{{t}^{{4}}+{4}+{16}{t}^{{12}}}}{{t}^{{6}}}}}\)
\(\displaystyle=\frac{\sqrt{{{\left({16}{t}^{{12}}+{t}^{{4}}+{4}\right)}}}}{{t}^{{3}}}\)
Acceleration vector is the derivative of the velocity vector.
So to find acceleration vector, differentiate velocity vector with respect to t. We get,
\(a(t)=\left\langle-\frac{1}{t^{2}},\ -2(-\frac{3}{t^{4}}),\ 4(3t^{2}) \right\rangle\)

\(=\left\langle -\frac{1}{t^{2}},\ \frac{6}{t^{4}},\ 12t^{2}\right\rangle\)
b) Put \(\displaystyle{t}=\sqrt{{3}}\) in the velocity vector, we get
\(\displaystyle{v}{\left(\sqrt{{3}}\right)}={\left\langle\frac{{1}}{\sqrt{{3}}},-\frac{{2}}{{\left(\sqrt{{3}}\right)}^{{3}}},{4}{\left(\sqrt{{3}}\right)}^{{3}}\right\rangle}\)
\(\displaystyle={\left\langle\frac{{1}}{\sqrt{{3}}},-\frac{{2}}{{{3}\sqrt{{3}}}},{12}\sqrt{{3}}\right\rangle}\)
Put \(\displaystyle{t}=\sqrt{{3}}\) in the acceleration vector, we get
\(\displaystyle{a}{\left(\sqrt{{3}}\right)}={\left\langle-\frac{{1}}{{\left(\sqrt{{3}}\right)}^{{2}}},\frac{{6}}{{\left(\sqrt{{3}}\right)}^{{4}}},{12}{\left(\sqrt{{3}}\right)}^{{2}}\right\rangle}\)
\(\displaystyle={\left\langle-\frac{{1}}{{3}},\frac{{6}}{{9}},{12}{\left({3}\right)}\right\rangle}\)
\(\displaystyle={\left\langle-\frac{{1}}{{3}},\frac{{2}}{{3}},{36}\right\rangle}\)

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