Question

Find \(r(t)\)if \({r^'}(t) = 2ti + 3{t^2}j + \sqrt t k\) and \(r(1) = i + j\)

Short answer

Therefore, the solution is \({\bf{r}}(t) = {t^2}{\bf{i}} + {t^3}{\bf{j}} + \left( {\frac{2}{3}{t^{3/2}} - \frac{2}{3}} \right){\bf{k}}\)

Step-by-step solution

Step 1: Given information

The given value is:

\({r^\prime }(t) = 2ti + 3{t^2}j + \sqrt t k\)

Evaluate the integral

\({\bf{r}}(t) = \int {\left( {2t{\bf{i}} + 3{t^2}{\bf{j}} + \sqrt t {\bf{k}}} \right)} dt\)

\( = \left( {2 \cdot \frac{1}{2}{t^2} + {C_1}} \right){\bf{i}} + \left( {3 \cdot \frac{1}{3}{t^3} + {C_2}} \right){\bf{j}} + \left( {\frac{1}{{3/2}}{t^{3/2}} + {C_3}} \right){\bf{k}}\)

\( = \left( {{t^2} + {C_1}} \right){\bf{i}} + \left( {{t^3} + {C_2}} \right){\bf{j}} + \left( {\frac{2}{3}{t^{3/2}} + {C_3}} \right){\bf{k}}\)

Step 3: Find the constants

Use the given\({\bf{r}}(1) = {\bf{i}} + {\bf{j}}\)(which is vector\( < 1,1,0 > )\)to find the constants.

\(\begin{array}{l}1 = {t^2} + {C_1}\\1 = {(1)^2} + {C_1}\\{C_1} = 0\end{array}\)

\(\begin{array}{l}1 = {t^3} + {C_2}\\1 = {(1)^3} + {C_2}\\{C_2} = 0\end{array}\)

\(\begin{array}{l}0 = \frac{2}{3}{t^{3/2}} + {C_3}\\0 = \frac{2}{3}{(1)^{3/2}} + {C_3}\\{C_3} = - \frac{2}{3}\end{array}\)

\({\bf{r}}(t) = {t^2}{\bf{i}} + {t^3}{\bf{j}} + \left( {\frac{2}{3}{t^{3/2}} - \frac{2}{3}} \right){\bf{k}}\)

This implies,

\({\bf{i}}\int t {{\rm{e}}^{2t}}dt = t \cdot \frac{{{{\rm{e}}^{2t}}}}{2}{\bf{i}} - \int 1 \cdot \frac{{{{\rm{e}}^{2t}}}}{2} \cdot dt \cdot {\bf{i}} = t \cdot \frac{{{{\rm{e}}^{2t}}}}{2}{\bf{i}} - \frac{{{{\rm{e}}^{2t}}}}{4}{\bf{i}} = \frac{{{{\rm{e}}^{2t}}(2t - 1)}}{4}{\bf{i}}\)

Step 4: Simplify the function

Now, subtract and add 1 to simplify the function and use table of the elementary integrals,

\({\bf{j}}\int {\frac{t}{{1 - t}}} dt = {\bf{j}}\int {\frac{{(t - 1) + 1}}{{1 - t}}} dt\)

\( = {\bf{j}}\int {\left( {\frac{{t - 1}}{{1 - t}} + \frac{1}{{1 - t}}} \right)} dt\)

\( = {\bf{j}}\int {\left( { - 1 + \frac{1}{{1 - t}}} \right)} dt\)

\( = {\bf{j}}\int {\left( { - 1 - \frac{1}{{t - 1}}} \right)} dt\)

\( = {\bf{j}}\left( { - t - \ln |t - 1| + {c_2}} \right)\)

The next component is,

\({\bf{k}}\int {\frac{1}{{\sqrt {1 - {t^2}} }}} dt = {\bf{k}}\left( {{{\sin }^{ - 1}}t + {c_3}} \right)\)

Combining all three components together, we get,

\(\vec C = \left( {{c_1},{c_2},{c_3}} \right) = i{c_1} + j{c_2} + k{c_3}\)

\(\int {\left( {t{{\rm{e}}^{2t}}{\bf{i}} + \frac{t}{{1 - t}}{\bf{j}} + \frac{1}{{\sqrt {1 - {t^2}} }}{\bf{k}}} \right)} dt = {\bf{i}}\frac{{{{\rm{e}}^{2t}}(2t - 1)}}{4} + {\bf{j}}( - t - \ln |t - 1|) + {\bf{k}}{\sin ^{ - 1}}t + \vec C\)