Derivatives of vector-valued functions Differentiate the following function. r(t)=\langle te^{-t},\ t\ln

3kofbe 2021-11-23 Answered
Derivatives of vector-valued functions Differentiate the following function.
\(\displaystyle{r}{\left({t}\right)}={\left\langle{t}{e}^{{-{t}}},\ {t}{\ln{{t}}},{t}{\cos{{t}}}\right\rangle}\)

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Expert Answer

Opeance1951
Answered 2021-11-24 Author has 7109 answers
Step 1
Given, \(\displaystyle{r}{\left({t}\right)}={\left\langle{t}{e}^{{-{t}}},\ {t}\ {e}{n}{t},\ {t}{\cos{{t}}}\right\rangle}\)
So, obtain differntiate individually
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{e}^{{-{t}}}\right)}={e}^{{-{t}}}+{t}{\left({e}^{{-{t}}}\right)}{\left(-{1}\right)}\)
\(\displaystyle={\left({1}-{t}\right)}{e}^{{-{t}}}\)
Step 2
Similarly,
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\ln{{t}}}\right)}={\frac{{{t}}}{{{t}}}}+{1}\times{\ln{{t}}}\)
\(\displaystyle={\ln{{t}}}+{1}\)
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({t}{\cos{{t}}}\right)}={\cos{{t}}}+{t}{\left(-{\sin{{t}}}\right)}\)
\(\displaystyle={\cos{{t}}}-{t}{\sin{{t}}}\)
\(\displaystyle\therefore{r}'{\left({t}\right)}={\left\langle{\left({1}-{t}\right)}{e}^{{-{t}}},\ {\left({1}+{\ln{{t}}}\right)},\ {\left({\cos{{t}}}-{t}{\sin{{t}}}\right)}\right\rangle}\)
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