The circular area of vision on the ground by a surveillance drone fitted with camera increases at the rate of $44 𝑚^2$ per metre as it gains vertical height. The radius of the circle on the ground increases at the rate of 6 m/s. Find the rate at which the height of the drone increases with time, at the instant when the radius of the circular area on the ground is 14 m. (𝑈𝑠𝑒 𝜋 =22/7)
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Detailed Solution
Step 1: Define Variables and Given Rates
Let $A$ be the area of the circle, $r$ be the radius, and $h$ be the height of the drone. We are given: $$ \frac{dA}{dh} = 44 \, \text{m}^2/\text{m} $$ $$ \frac{dr}{dt} = 6 \, \text{m/s} $$ We need to find $\frac{dh}{dt}$ when $r = 14$ m.
Step 2: Relate Area to Radius
The area of the circle is given by $A = \pi r^2$. Differentiating with respect to $h$ using the chain rule: $$ \frac{dA}{dh} = \frac{dA}{dr} \cdot \frac{dr}{dh} $$ $$ 44 = (2\pi r) \cdot \frac{dr}{dh} $$
Step 3: Calculate dr/dh
Substitute $\pi = 22/7$ and $r = 14$: $$ 44 = 2 \cdot \frac{22}{7} \cdot 14 \cdot \frac{dr}{dh} $$ $$ 44 = 88 \cdot \frac{dr}{dh} \implies \frac{dr}{dh} = \frac{44}{88} = 0.5 $$
Step 4: Relate Rates with Time
Using the chain rule: $$ \frac{dr}{dt} = \frac{dr}{dh} \cdot \frac{dh}{dt} $$ Substitute the known values: $$ 6 = 0.5 \cdot \frac{dh}{dt} $$ $$ \frac{dh}{dt} = \frac{6}{0.5} = 12 \, \text{m/s} $$
Final Answer: 12 m/s
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