A weather balloon is rising vertically at 60 meters per minute. An observer is standing on the ground 6 meters from the point at which the balloon was released. Determine (in meters per minute) the rate at which the distance between the feet of the observer and the balloon is changing when the balloon is 8 meters high.

Question

anonymous · Answer

speed is 60m/s. We'll use it later.
Let the balloon's height be x m from the ground. Assume the eye at the ground level. Let it be E. Assume at any given moment the balloon's position is P. And the point from where it was released be R. Now EPR is a right angled triangle. Find its hypotenuse using Pythagoras theorem and find let it be y (x is the height of the balloon).Now find x in terms of y and differentiate it. That gives velocity of the balloon. When balloon is 8m high y is 10m. Put this in the equation which you have just diffentiated. That directly gives the velocity of the balloon at x=8

anonymous · Answer

Similarly you can diffentiate the x and y equation other way round. And finally you'll get the answer.

anonymous · Answer

dy/dt= dy/dx X dx/dt where t is the time.

anonymous · Answer

remember that you have to only find dy/dx . dx/dt is given.

anonymous · Answer

ok im a little lost... how do i know what y is

anonymous · Answer

you'll get 
y=$$sqrt{36+x ^{2}}$$
find dy/dx at x=8. dx/dt is already given

anonymous · Answer

dx/dt is the vertical velocity of the balloon going upwards