Question

9.5 integrating vector - valued functions
calculus
for problems 1 - 6, find the vector - valued function ( f(t) ) that satisfies the given initial conditions.

1. ( f(0)=langle 2,4

angle, f^{prime}(t)=langle 2e^{t},3e^{3t}
angle )

1. ( f(0)=langle\frac{1}{2}, - 1

angle, f^{prime}(t)=langle te^{-t^{2}},-e^{-t}
angle )

Explanation:

Problem 1

Step1: Integrate the x - component of \(f'(t)\)

To find the x - component of \(f(t)\), we integrate \(f'_x(t)=2e^{t}\) with respect to \(t\). The integral of \(e^{t}\) is \(e^{t}\), so \(\int 2e^{t}dt = 2e^{t}+C_1\).

Step2: Integrate the y - component of \(f'(t)\)

To find the y - component of \(f(t)\), we integrate \(f'_y(t) = 3e^{3t}\) with respect to \(t\). Using the substitution \(u = 3t\), \(du=3dt\), then \(\int 3e^{3t}dt=\int e^{u}du=e^{u}+C_2=e^{3t}+C_2\). So the vector - valued function before applying initial conditions is \(f(t)=\langle2e^{t}+C_1,e^{3t}+C_2
angle\).

Step3: Apply the initial condition \(f(0)=\langle2,4

angle\)
For the x - component: When \(t = 0\), \(2e^{0}+C_1=2\). Since \(e^{0}=1\), we have \(2\times1 + C_1=2\), so \(2 + C_1=2\), which gives \(C_1 = 0\).
For the y - component: When \(t = 0\), \(e^{3\times0}+C_2=4\). Since \(e^{0}=1\), we have \(1 + C_2=4\), so \(C_2=3\).

Step1: Integrate the x - component of \(f'(t)\)

To find the x - component of \(f(t)\), we integrate \(f'_x(t)=te^{-t^{2}}\) with respect to \(t\). Use the substitution \(u=-t^{2}\), then \(du=-2tdt\), and \(-\frac{1}{2}du = tdt\). So \(\int te^{-t^{2}}dt=-\frac{1}{2}\int e^{u}du=-\frac{1}{2}e^{u}+C_1=-\frac{1}{2}e^{-t^{2}}+C_1\).

Step2: Integrate the y - component of \(f'(t)\)

To find the y - component of \(f(t)\), we integrate \(f'_y(t)=-e^{-t}\) with respect to \(t\). The integral of \(-e^{-t}\) is \(e^{-t}+C_2\) (because \(\int - e^{-t}dt=\int e^{-t}(-dt)=e^{-t}+C_2\)). So the vector - valued function before applying initial conditions is \(f(t)=\langle-\frac{1}{2}e^{-t^{2}}+C_1,e^{-t}+C_2
angle\).

Step3: Apply the initial condition \(f(0)=\langle\frac{1}{2}, - 1

angle\)
For the x - component: When \(t = 0\), \(-\frac{1}{2}e^{-0^{2}}+C_1=\frac{1}{2}\). Since \(e^{0}=1\), we have \(-\frac{1}{2}\times1 + C_1=\frac{1}{2}\), so \(-\frac{1}{2}+C_1=\frac{1}{2}\), which gives \(C_1 = 1\).
For the y - component: When \(t = 0\), \(e^{-0}+C_2=-1\). Since \(e^{0}=1\), we have \(1 + C_2=-1\), so \(C_2=-2\).

Answer:

\(f(t)=\langle2e^{t},e^{3t}+3
angle\)

Problem 2