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if a baloon is rising at 1 ft/sec and exactly when it reaches 65 ft above ground, a bicycle moving at 17ft/sec passes under it, how fast is the distance between the bicycle and the balloon INCREASING 3 seconds later?!

2006-12-22 05:42:52 · 2 answers · asked by Anonymous in Education & Reference Homework Help

2 answers

You write the equation for the elevation of the balloon y=v1*t+65. Write the equation for the horizontal position of the bicycle x = v2*t (at t=0, y=65 and x=0) The distance between the balloon and the bicycle is d(t) = √[x^2 + y^2]. Substituting for x and y, get

d(t) = √[(v1*t+65)^2 + (v2*t)^2] =
√[v1^2*t^2+2*65*v1*t+65^2+v2^2*t*2]

Find the derivative of d(t) and evaluate that at t=17sec

2006-12-22 19:31:06 · answer #1 · answered by gp4rts 7 · 0 0

you have to use a derivative~but the equations i would use are from physics not calculus (though i would use calculus to solve the equations, if that makes sense). what class is this for?

2006-12-22 14:47:26 · answer #2 · answered by dally1025 3 · 1 0

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