An airforce plane is ascending vertically at the rate of 100 km/h. If the radius of the earth is `rk m ,` how fast is the area of the earth, visible from the plane, increasing at 3 minutes after it started ascending? Given that the visible area `A` at height `h` is given by `A=2pir^2h/(r+h)` .

Explanation

Let say` h=`height



`frac{d h}{d t}=100 `


`A=frac{2 pi r^{2} h}{r+h} `


`frac{d A}{d t}=frac{d A}{d h} frac{d h}{d t} `


`=2 pi r^{2} frac{d}{d} h(frac{h}{r+h}) cdot 100 `


`=2 pi r^{2}[frac{(r+h)-h}{(f+h)^{2}}] cdot 100`


`=frac{200 pi r^{3}}{(r+h)^{2}} `


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