Solution 18.8b

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(New page: <math>\begin{align} & \mathbf{v}=\frac{d\mathbf{r}}{dt}=\left( 2-\frac{t}{5} \right)\mathbf{i}+2\mathbf{j} \\ & v=\sqrt{{{\left( 2-\frac{t}{5} \right)}^{2}}+{{2}^{2}}}=\sqrt{4-\frac{4t}{5...)
Current revision (17:17, 27 March 2011) (edit) (undo)
 
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<math>\begin{align}
<math>\begin{align}
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& \mathbf{v}=\frac{d\mathbf{r}}{dt}=\left( 2-\frac{t}{5} \right)\mathbf{i}+2\mathbf{j} \\ & v=\sqrt{{{\left( 2-\frac{t}{5} \right)}^{2}}+{{2}^{2}}}=\sqrt{4-\frac{4t}{5}+\frac{{{t}^{2}}}{25}} \\
+
& \mathbf{v}=\frac{d\mathbf{r}}{dt}=\left( 2-\frac{t}{5} \right)\mathbf{i}+2\mathbf{j} \\ & v=\sqrt{{{\left( 2-\frac{t}{5} \right)}^{2}}+{{2}^{2}}}=\sqrt{8-\frac{4t}{5}+\frac{{{t}^{2}}}{25}} \\
\end{align}</math>
\end{align}</math>

Current revision

\displaystyle \begin{align} & \mathbf{v}=\frac{d\mathbf{r}}{dt}=\left( 2-\frac{t}{5} \right)\mathbf{i}+2\mathbf{j} \\ & v=\sqrt{{{\left( 2-\frac{t}{5} \right)}^{2}}+{{2}^{2}}}=\sqrt{8-\frac{4t}{5}+\frac{{{t}^{2}}}{25}} \\ \end{align}

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