A balloon and a bicycle. A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground , a bicycle moving at a constant rate of 17 ft/sec passes under it . How fast is the distance between the bicycle and balloon increasing 3 sec later ?

Explanation

Let at any instant bicycle has moves a distance x from below the balloon and height of balloon at that instant be y. Then distance between bicycle and balloon
`r=sqrt(x^(2)+y^(2))`
`:. " " r^(2)=x^(2)+y^(2)`
Differentiating ,
`2r(dr)/(dt)=2x(dx)/(dt)
=2y(dy)/(dt)`
`:. " " sqrt(x^(2)+y^(2) ) (dr)/(dt)=x(dx)/(dt)+y(dy)/(dt)`
After 3 sec, `" " y=65+1xx3=68 ft`
`x=17 xx3 =51 ft`
`:. " " sqrt((68)^(2)+(51)^(2))(dr)/(dt)=51xx17 +68 xx1`
`:. " " (dr)/(dt)=11ft//s`