Problem and Solution
We are given a sector $OAB$ of a circle with center $O$ and radius $r$ cm. The angle $$ \angle AOB = \frac{\pi}{6} $$ radians, and the lengths $$ OA = OB = r $$ cm.
The area of the sector is increasing such that the angle $$ \angle AOB $$ remains constant, and the lengths $$ OA \text{ and } OB $$ are both increasing at a constant rate of $$ 0.2 \text{ cm/s} $$
We are tasked with finding the exact rate of change, in $$ \text{cm}^2/\text{s} $$ of the area of the sector when the length of arc $$ AB $$ is $$ \frac{5\pi}{2} $$
Solution
Step 1: Write the formula for the area of the sector
The area $A$ of a sector of a circle is given by
$$ A = \frac{1}{2}r^2\theta $$
where $r$ is the radius and $\theta$ is the central angle in radians. Here,
$$ \theta = \frac{\pi}{6} $$
which remains constant.
Therefore,
$$ A = \frac{1}{2}r^2 \cdot \frac{\pi}{6} = \frac{\pi}{12}r^2 $$
Step 2: Differentiate the area with respect to time $t$
To find the rate of change of the area, differentiate with respect to $t$:
$$ \frac{dA}{dt} = \frac{d}{dt} \left( \frac{\pi}{12}r^2 \right) = \frac{\pi}{6}r\frac{dr}{dt} $$
We are given
$$ \frac{dr}{dt} = 0.2 \quad \text{cm/s} $$
Therefore,
$$ \frac{dA}{dt} = \frac{\pi}{6}r \cdot 0.2 = \frac{\pi}{30}r $$
Step 3: Find the radius $r$ when the arc length is $ \frac{5\pi}{2} $
The formula for arc length is
$$ \text{Arc length} = r\theta $$
Substitute
$$ \theta = \frac{\pi}{6} $$
so
$$ \text{Arc length} = r \cdot \frac{\pi}{6} $$
We are given that the arc length is
$$ \frac{5\pi}{2} $$
Thus,
$$ r \cdot \frac{\pi}{6} = \frac{5\pi}{2} $$
Solving for $r$:
$$ r = \frac{5\pi}{2} \cdot \frac{6}{\pi} = 15 \text{ cm} $$
Step 4: Substitute $r = 15$ into the rate of change of the area
Now substitute
$$ r = 15 $$
into
$$ \frac{dA}{dt} = \frac{\pi}{30}r $$
$$ \frac{dA}{dt} = \frac{\pi}{30} \cdot 15 = \frac{\pi}{2} \text{ cm}^2/\text{s} $$
Final Answer
Thus, the exact rate of change of the area is
$$ \boxed{ \frac{\pi}{2} \text{ cm}^2/\text{s} } $$
Post a Comment