Question

for r(t) = e^(-3t)i + sin(5t)j + cos(6t)k, determine r(t) and ∫r(t)dt = i + j + k + c with c a constant vector.

Explanation:

Step1: Integrate the \(i\) - component

Integrate \(e^{-3t}\) with respect to \(t\). Using the formula \(\int e^{at}dt=\frac{1}{a}e^{at}+C_1\) for \(a = - 3\), we have \(\int e^{-3t}dt=-\frac{1}{3}e^{-3t}+C_1\).

Step2: Integrate the \(j\) - component

Integrate \(\sin(5t)\) with respect to \(t\). Using the formula \(\int\sin(at)dt=-\frac{1}{a}\cos(at)+C_2\) for \(a = 5\), we get \(\int\sin(5t)dt=-\frac{1}{5}\cos(5t)+C_2\).

Step3: Integrate the \(k\) - component

Integrate \(\cos(6t)\) with respect to \(t\). Using the formula \(\int\cos(at)dt=\frac{1}{a}\sin(at)+C_3\) for \(a = 6\), we obtain \(\int\cos(6t)dt=\frac{1}{6}\sin(6t)+C_3\).

Answer:

\(-\frac{1}{3}e^{-3t}\mathbf{i}-\frac{1}{5}\cos(5t)\mathbf{j}+\frac{1}{6}\sin(6t)\mathbf{k}+\mathbf{C}\)