Question

A balloon rising 15sec/ft is 200 feet high when a car traveling 66ft/sec passes directly beneath it. How fast is the distance between the car and the balloon increasing 1 second later?

Answer 1

is this calculus?

Answer 2

yep

Answer 3

aw sorry i hate that part of calc =))))

Answer 4

Is it 15sec/ft or 15ft/s?

Answer 5

ft/sec

Answer 6

Use the chain rule:
Distance, D=sqrt((h^2+x^2))
differentiate with respect to t:
dD/dt=dD/dh*dh/dt + dD/dx dx/dt
Both dx/dt and dh/dt are known.
dD/dh=2h/sqrt(h^2+x^2)
and
dD/dx=2x/sqrt(h^2+x^2)

Answer 7

can you show me how to use the rule for this problem?

Answer 8

First draw a tree diagram of what function is dependent on what.
We know that the distance D is dependent on h and x.
In turn, x is dependent on t, and h as well.
So if we need to differentiate D with respect to t, we musnt't forget the relationship of h and x with t, so
dD/dt=dD/dx * dx/dt +... according to the chain rule
Since D is also dependent on h, we do the same thing, but add the effects of each, so
dD/dt=dD/dh*dh/dt + dD/dx dx/dt
dh/dt=15 fps
dx/dt=66mph=? fps
and dD/dx and dD/dh have been given above, so you can take it from here?

Answer 9

solve for dr/dt. You know x and dx/dt, y and dy/dt. you can find r (at the given time, 1 sec after the car passes beneath the balloon)