Question

Find f. f ''(θ) = sin θ + cos θ, f(0) = 3, f '(0) = 4

Answer 1

`\bf f''(\theta)=sin(\theta)+cos(\theta)\qquad thus \\\\\\ \displaystyle \int [sin(\theta)+cos(\theta)]d\theta\implies -cos(\theta)+sin(\theta)+C =\underline{f'(\theta)} \\\\\\ \textit{now, we know \underline{f'(0)}=4}\implies =-cos(0)+sin(0)+C=\underline{4} \\\\\\ -1+0+C=4\implies C=5 \\\\\\ thus\qquad sin(\theta)-cos(\theta)+5=f'(\theta)\\\\ -----------------------------\\\\`


`\bf \displaystyle \int [sin(\theta)-cos(\theta)+5]d\theta\implies -cos(\theta)-sin(\theta)+5\theta+C=\underline{f(\theta)} \\\\\\ \textit{now, we know that \underline{f(0)=3}}\implies -cos(0)-sin(0)+5(0)+C=\underline{3} \\\\\\ -1-0+0+C=3\implies C=4 \\\\\\ thus\qquad -cos(\theta)-sin(\theta)+5\theta+4=f(\theta)`

Answer 2

`f''(\theta)=\sin\theta+\cos\theta`

`f'(\theta)=\displaystyle\int f''(\theta)\,\mathrm d\theta=\int(\sin\theta+\cos\theta)\,\mathrm d\theta`

`f'(\theta)=-\cos\theta+\sin\theta+C_1`

Given that
`f'(0)=4`, you have


`4=-\cos0+\sin0+C_1\implies C_1=5`


`f(\theta)=\displaystyle\int f'(\theta)\,\mathrm d\theta=\int(-\cos\theta+\sin\theta+5)\,\mathrm d\theta`

`f(\theta)=-\sin\theta-\cos\theta+5\theta+C_2`

and given that
`f(0)=3`, you get


`3=-\sin0-\cos0+5(0)+C_2\implies C_2=4`

and so


`f(\theta)=-\sin\theta-\cos\theta+5\theta+4`