Question

Find \(r(t)\) if \(r'(t) = ti + {e^t}j + t{e^t}k\) and\(r(0) = i + j + k\).

Short answer

So the final answer : \(\left( {\frac{1}{2}{t^2} + 1} \right)i + {e^t}j + (t{e^t} - {e^t} + 2)k\)

Step-by-step solution

Step 1: Integrate each component

\(r'(t) = ti + {e^t}j + t{e^t}k\)

\(r(0) = i + j + k\)

Integrate each component of r’(t) separately. For the first component:

\(\int {t\;dt = \frac{1}{2}{t^2} + {c_1}} \)

The 1^st component of r(0) is 1 so we can solve for \({c_1}\)

\(\begin{array}{l}\frac{1}{2}{(0)^2} + {c_1} = 1\\{c_1} = 1\\r(t) = \left( {\frac{1}{2}{t^2} + 1} \right)i\end{array}\)

Step 2: Integration for second component

Integrate each component of r’(t) separately. For the second component:

\(\int {{e^t}\;dt = {e^t} + {c_2}} \)

The 1^st component of r(0) is 1 so we can solve for \({c_2}\)

\(\begin{array}{l}{e^0} + {c_2} = 1\\1 + {c_2} = 1\\{c_2} = 0\\r(t) = {e^t}j\end{array}\)

Step 3: Integration for third component

Integrate each component of r’(t) separately. For the third component:

\(\begin{array}{l}\int {t{e^t}\;dt = uv - \int {vdu} } \\u = t\;\;\;\;\;\;\;\;\;\;\;\;du = dt\\dv = {e^t}dt\;\;\;\;\;\;v = {e^t}\end{array}\)

\(\begin{array}{l}\int {t{e^t}\;dt = uv - \int {vdu} } \\ = t{e^t} - \int {{e^t}} dt\\ = t{e^t} - {e^t} + {c_3}\end{array}\)

The 1^st component of r(0) is 1 so we can solve for \({c_3}\)

\(\begin{array}{l}0{e^0} - {e^0} + {c_3} = 1\\ - 1 + {c_3} = 1\\{c_3} = 2\\r(t) = (t{e^t} - {e^t} + 2)k\end{array}\)

So the final answer : \(\left( {\frac{1}{2}{t^2} + 1} \right)i + {e^t}j + (t{e^t} - {e^t} + 2)k\)