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Find the function with the given derivative whose graph passes through the point P.r'(theta)=9+sec^2(theta), P(pi/4,2)

User Atorscho
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1 Answer

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The given derivative is:


r^(\prime)(\theta)=9+\sec ^2\theta

Integrate w. r. t. theta to get:


\begin{gathered} r(\theta)=\int (9+\sec ^2\theta)d\theta \\ =9\int d\theta+\int \sec ^2\theta d\theta \end{gathered}

Use the formulae given by:


\int d\theta=\theta+c,\int \sec ^2\theta d\theta=\tan \theta+c

To get the solution as follows:


r(\theta)=9\theta+\tan \theta+c

The given point is P(pi/4,2) so it follows:


\begin{gathered} 2=9((\pi)/(4))+\tan (\pi)/(4)+c \\ 2-(9\pi)/(4)-1=c \\ c=1-(9\pi)/(4) \end{gathered}

Substitute the value to get the required answer as follows:


r(\theta)=9\theta+\tan \theta+1-(9\pi)/(4)

The above function is the required answer.

User Rodrigo Lopetegui
by
8.4k points

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