An air force plane is ascending vertically at the rate of \(100\text{ km/hr}\). If the radius of the Earth is \(R \text{ km}\), how fast is the area of the Earth visible from the plane increasing at 3 minutes after it started ascending?

Take visible area \(A=\dfrac { 2\pi R^ 2H }{ R+H } \), where \(H\) is the height of the plane (in \(\text{km}\)) above the Earth.

P : \(\frac { 200\pi r^ 3 }{ (R+5)^ 2 } \) \(\frac { km^ 2 }{ h } \)

Q : \(\frac { 200\pi r^ 3 }{ (R+5) } \) \(\frac { km^ 2 }{ h } \)

R : \(\frac { 400\pi r^ 3 }{ (R+5) } \) \(\frac { km^ 2 }{ h } \)

S : \(\frac { 400\pi r^ 3 }{ (R+5)^ 2 } \) \(\frac { km^ 2 }{ h } \)

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