Question

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

Answer 1

Answer:

DL/dt  =  11 feet/sec

Step-by-step explanation:

We have a right triangle between    balloon   ( h  =  65 feet)  from the point straight down the ballon,  and the bicycle position. Hypothenuse is L the distance between balloon and bicycle

Let  x be distance between (vertical from the balloon to ground) then

according to Pythagoras Theorem

L²  =  x² +  h²       (1)

and we know

Dx/dt  =  17   ft/sec

Dh/dt =  1  ft/sec

We need to find   DL/dt

From equation (1)

L²   =  x²   +  h²

Differentiating on both sides of the equation

2L DL/dt   =  2xDx/dt  +  2yDy/dt

L*DL/dt   =  x*Dx/dt + y*Dy/dt

DL/dt    =   [  x*Dx/dt + y*Dy/dt ] /L    (2)

Now  3 sec after the bicycle passed under the balloon

x  =  3 * Dx/dt     =  3 * 17   = 51 feet

and  h  =  65 + 1* 3    =  68 feet

then   L  = √ √(68)²  + (51)²         L  =  √ 4624  +  2601    L  = 85 feet

Now plugging in these values in equation  (2)

DL/dt  =  [ 51*17  +  68* 1]  / 85

DL/dt  =  11 feet/sec