Aballoon is rising vertically above a level, straight road at a constant rate of 4 ft/sec. just when the balloon is 80 ft above the ground, a bicycle moving at a constant rate of 13 ft/sec passes under it. how fast is the distance s(t) between the bicycle and balloon increasing 6 seconds later?

The distance s(t) is increasing at 11 ft/sec

Step-by-step explanation:

Since the balloon is moving vertically and the bicycle is moving horizontally, the distance between them is the hypotenuse of the triangle that is shown in the diagram attached at the bottom of this answer. By Pythagoras:

s(t)²=a(t)²+b(t)²

being a(t) the speed of the balloon at any given time t and b(t) the speed of the bicycle at any given time t.

To know how fast this distance is changing we need to find the derivative s'(t):

Evaluating it at t = 6 sec we get:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

step-by-step explanation: