A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later?

Solution:

Given,

A balloon is rising vertically above a straight road at constant rate of 1 ft/sec

When the balloon reaches 65ft above the ground, a bicycle passes under it at a constant rate of 17 ft/sec

Thus, it forms a right angled triangle.

Let, L be the hypotenuse of the right angled triangle which refers to the distance between the balloon and bicycle.

Let, x be the vertical distance of the balloon from the ground.

Using Pythagoras theorem,

L² = x² + h² --------------- (1)

Given, dx/dt = 17 ft/sec

dh/dt = 1 ft/sec

We have to find dL/dt,

Differentiating (1),

2L dL/dt = 2x dx/dt + 2h dh/dt

L dL/dt = x dx/dt + h dh/dt

dL/dt = [x dx/dt + h dh/dt]/L ---------------- (2)

Now, 3 seconds after the bicycle passes under the balloon,

x = 3 dx/dt

x = 3 (17)

x = 51 feet

h = 3 dh/dt

h = 3 (1)

h = 65 + 3 = 68 

h = 68 feet

Put the values of x and h in (1) to find L,

L²  = (51)² + (68)²

L² = 7225

Taking square root,

L = √7225

L = 85 feet

Substitute the values of L, x and h in (2),

dL/dt = [51(17) + 68(1)]/85

dL/dt = (867+68)/85

dL/dt = 11 ft/sec

Therefore, the rate of change of distance is 11 ft/sec.

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A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later?

Summary:

A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. The increase in distance between the bicycle and balloon after 3 sec is 11 ft/sec.