A kite 100 feet above the ground is being blown away from the person holding its string in a direction parallel to the ground at a rate of 10 feet per second. At what rate must the string be let out when the length of string already let out is 200 feet? (Hint: start by drawing a picture. There should be a triangle involved.) You need not simplify your answer.

  5√3 ft/s

Step-by-step explanation:

Let h represent the horizontal distance of the kite from the person. Let s represent the string length. Then the Pythagorean theorem tells us ...

  s^2 = 100^2 + h^2

  2s·ds/dt = 0 + 2h·dh/dt

  ds/dt = (h/s)·dh/dt

So, we need to know the horizontal distance when s=200.

  200^2 = 100^2 + h^2

  40000 -10000 = h^2

  h = 100√3

Substituting known values into the equation for ds/dt, we have ...

  ds/dt = (100√3)/(200)(10 ft/s) = 5√3 ft/s

The string must be let out at 5√3 ft/s when it is already 200 ft long.

circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation: