Solve the initial value problem.
-$\frac { \mathrm { dr } } { \mathrm { d } \theta } = \csc ^ { 2 } 15 \theta \cot 15 \theta , \quad \mathrm { r } \left( \frac { \pi } { 4 } \right) = 3$
A) $\mathrm { r } = \frac { 1 } { 6 } \csc ^ { 3 } 15 \theta \cot ^ { 2 } 15 \theta + 3$
B) $r = \frac { 1 } { 30 } \cot ^ { 2 } \theta + \frac { 89 } { 30 }$
C) $r = - \frac { 1 } { 30 } \tan ^ { 2 } 15 \theta + \frac { 91 } { 30 }$
D) $\mathrm { r } = - \frac { 1 } { 30 } \cot ^ { 2 } 15 \theta + \frac { 91 } { 30 }$

Question

Quizplus · Accepted Answer

We can separate the variables and integrate both sides to get:
$$\int r \, dr = \int \csc^2 15 	heta \cot 15 	heta \, d	heta$$
Integrating the right-hand side requires a substitution of $u = 15 	heta$, which gives us:
$$\int \csc^2 15 	heta \cot 15 	heta \, d	heta = \int \frac{1}{\sin^2 u} \cdot \frac{\cos u}{15} \, du = -\frac{1}{15} \cot u + C = -\frac{1}{15} \cot 15 	heta + C$$
Plugging in the initial condition $r\left(\frac{\pi}{4}ight)=3$, we get:
$$3 = -\frac{1}{15} \cot\frac{\pi}{4} + C = -\frac{1}{15} + C$$
Solving for $C$, we have $C = \frac{91}{15}$. Therefore, our solution is:
$$r = -\frac{1}{15} \cot 15 	heta + \frac{91}{15} = -\frac{1}{30} \cot 15 	heta + \frac{91}{30}$$
Since we want the value of $r$ when $	heta = \frac{\pi}{4}$, we can substitute this value into our solution to get:
$$r \left( \frac{\pi}{4} ight) = -\frac{1}{30} \cot \frac{3\pi}{4} + \frac{91}{30} = -\frac{1}{30} \cdot 0 + \frac{91}{30} = \frac{91}{30}$$
Therefore, the best choice is D.

Solve the Initial Value Problem A) r=16csc⁡315θcot⁡215θ+3\mathrm { r } = \frac { 1 } { 6 } \csc ^ { 3 } 15 \theta \cot ^ { 2 } 15 \theta + 3r=61​csc315θcot215θ+3

Solve the Initial Value Problem A) $\mathrm { r } = \frac { 1 } { 6 } \csc ^ { 3 } 15 \theta \cot ^ { 2 } 15 \theta + 3$