Question

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

Answer 1

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:

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

Substitute the value to get the required answer as follows:

$r(\theta)=9\theta+\tan \theta+1-\frac{9\pi}{4}$

The above function is the required answer.