Solve the initial value problem.
-$$\frac { d r } { d \theta } = - \frac { \pi } { 2 } \cos \frac { \pi } { 2 } \theta , r ( 0 ) = - 10$$
A) $r = \sin \frac { \pi } { 2 } \theta - 10$
B) $r = - \sin \frac { \pi } { 2 } \theta - 10$
C) $r = - \frac { \pi } { 2 } \sin \frac { \pi } { 2 } \theta - 10$
D) $r = \cos \frac { \pi } { 2 } \theta - 11$

Question

Quizplus · Accepted Answer

We can solve the initial value problem by integrating both sides with respect to $	heta$:
$$r(	heta)=-\frac{\pi}{2}\int\cos\frac{\pi}{2}	heta\ d	heta+C$$
Using the substitution $u=\frac{\pi}{2}	heta$ and $du=\frac{\pi}{2} d	heta$, we get
$$r(	heta)=-\int\cos u\ du+C=-\sin u+C=-\sin\frac{\pi}{2}	heta+C$$
Using the initial condition $r(0)=-10$, we get $C=-10$. Therefore,
$$r(	heta)=-\sin\frac{\pi}{2}	heta-10$$
which matches with choice B.

Solve the Initial Value Problem drdθ=−π2cos⁡π2θ,r(0)=−10\frac { d r } { d \theta } = - \frac { \pi } { 2 } \cos \frac { \pi } { 2 } \theta , r ( 0 ) = - 10dθdr​=−2π​cos2π​θ,r(0)=−10

Solve the Initial Value Problem $\frac { d r } { d \theta } = - \frac { \pi } { 2 } \cos \frac { \pi } { 2 } \theta , r ( 0 ) = - 10$