Question

Find \(f\)

\(f''\left( \theta \right) = \sin \theta + \cos \theta ,\;f\left( 0 \right) = 3,\;f'\left( 0 \right) = 4\)

Short answer

Required function is \(f\left( \theta \right) = - \sin \theta - \cos \theta + 5\theta + 4\).

Step-by-step solution

Find a general antiderivative of \(f''\)

Here we have,

\(f''\left( \theta \right) = \sin \theta + \cos \theta \)

Now, the general antiderivative of \(f''\left( \theta \right) = \sin \theta + \cos \theta \) is,

\(f'\left( \theta \right) = - \cos \theta + \sin \theta + C\)

Find the general antiderivative of \(f'\)

Now, we have, \(f'\left( \theta \right) = - \cos \theta + \sin \theta \)

Now, the general antiderivative of \(f'\left( \theta \right) = - \cos \theta + \sin \theta + C\) is,

\(f\left( \theta \right) = - \sin \theta - \cos \theta + C\theta + D\)

Find the value of constant \(C\) and \(D\).

Now, we have\(f'\left( \theta \right) = - \cos \theta + \sin \theta + C\)and \(f\left( \theta \right) = - \sin \theta - \cos \theta + C\theta + D\)

Also, we have \(f\left( 0 \right) = 3\) and \(f'\left( 0 \right) = 4\)

To find value of constant \(C\) put \(f'\left( 0 \right) = 4\) in \(f'\left( \theta \right) = - \cos \theta + \sin \theta + C\)

\(\begin{aligned}{l} \Rightarrow 4 &= - \cos 0 + \sin 0 + C\\ \Rightarrow C &= 5\end{aligned}\)

To find value of constant \(D\) put \(f\left( 0 \right) = 3\) in \(f\left( \theta \right) = - \sin \theta - \cos \theta + C\theta + D\)

\(\begin{aligned}{l} \Rightarrow 3 &= - \sin 0 - cos0 + 5\left( 0 \right) + D\\ \Rightarrow 3 &= - 1 + D\\ \Rightarrow D &= 4\end{aligned}\)

So, required function is \(f\left( \theta \right) = - \sin \theta - \cos \theta + 5\theta + 4\).