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An airforce plane is ascending vertically atthe 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 3minutes after it started ascending? Given that the visible area `A`at height `h`is given by `A=2pir^2h/(r+h)`. |
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Answer» Let say h=height and A= area `(dh)/dt=100` `A=(2pir^2h)/(r+h` `(dA)/(dt)=(dA)/(dh)(dh)/(dt)` `=2pir^2d/dh(h/(r+h))*100` `=2pir^2[((r+h)-h)/(f+h)^2]*100` `=(200pir^3)/(r+h)^2` `h=100/60*3=5km` `(dA)/(dt)=(200pir^3)/(r+5)^2``km^2`/h. |
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