Answer the following problems.

Question

1.A child is flying a kite at a height of 40 ft, that is, moving horizontally at a rate of 

3ft /sec. If the string is taut, at what rate is the string being paid out when the length of the string released is 5 ft? 

2.A spherical snowball is being made so that its volume is increasing at the rate of 8ft3/min and the rate at which the radius is increasing when the snowball is 4t in diameter. 

3.A stone is dropped into a still pond. Concentric circular ripples spread out, and the radius of the disturbed region increases at the rate of 16cm/sec. At what rate does the area of the disturbed region increase when its radius is 4 cm? 

Show Complete Solution:

Defined variables-                                                                           

Numerical facts known about the variables and their

           derivatives with respect to t-                                                

The unknown (what you wish to find)-                                           

The equation to relate the variable that depend on t-                       

Final answer (includes the solution on differentiation)-                   

Interpretation-          

Answer 1

2)

Given: Rate of volume increase with respect to time

V' = dV/dt = 8 ft³/s

I.e., the snowball is increasing at a constant rate of 8 cubic feet per second

The volume of a sphere can be described by the following equation:

V = (4/3) * πr³

Take the time derivative; make sure to apply the chain rule on the radius, such that you get:

V' = 4πr²r'

Isolate r' to get the time derivative of r:

r' = V'/(4πr²)

Now, we can plug in our known value for V' at the desired radius. When d = 4 ft, r = 2 ft:

r' = (8 ft³/s)/(4π * (2 ft)²)

r' = 1/(2π) ft

When the snowball is 4 feet in diameter, its radius is growing at a rate of 1/(2π) feet per second.

3)

Given: Rate that radius of ripple increases with respect to time:

r' = dr/dt = 16 cm/s

I.e., the radius of the disturbed area is increasing at a constant rate of 8 cubic feet per second

The area of a circle can be described by the following equation:

A = πr²

Take the time derivative; make sure to apply the chain rule on the radius, such that you get:

A' = 2πrr'

Now, we can plug in our known value for A' at the desired radius. When r = 4 cm:

A' = 2π * (4 cm) * (16 cm/s)

A' = 128π cm²/s

When the radius of the area of the pond disturbance is 4 centimeters, the area is growing at a rate of 128π square centimeters per second.