Answers: 2
, you have
, you get
![\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)\\\\ -----------------------------\\\\ ](/tpl/images/0234/1401/c5ebf.png)
![\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)](/tpl/images/0234/1401/e887f.png)
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Find f. f ''(θ) = sin θ + cos θ, f(0) = 3, f '(0) = 4...
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