Related Rates Help

1. Jan 3, 2005

ashleyk

A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

Last edited: Jan 3, 2005
2. Jan 3, 2005

quasar987

Do you know what the answer is?

Personally I would just say: "There is no telling what the rate of change of the baloon and of the car will be 1 second later"... maybe a big wind will blow the balloon sideway and maybe in 0.5 seconds later the car will crash into a tree.

3. Jan 3, 2005

here is my suggestion, have to wait for some one else to check to see if its right.

i formed a right angled triangle
the vertical side has length 200+15t, the horizontal side has length 66t and let the hypotenuse be x.
we want to find dx/dt
using pythagorus we get x^2=291t^2+600t+40000
from this calculate dx/dt
pop in t=1 second to get dx/dt=2.9ft/s

4. Jan 11, 2005

gatechguy

I did sort of the same thing, creating a right triangle with sides x, y, and hyp of d. the taking d^2=x^2+y^2, then taking the derivative, getting d(dd/dt)=x(dx/dt)+y(dy/dt), then subbing in the given numbers, and finding d after one second, I get (224.9ft)(dd/dt)=(66ft)(66ft/sec)+(215ft)(15ft/sec), and when solved, I get (dd/dt)=33.7ft/sec at t=1