Water is falling on a surface, wetting a circular area that is expanding at a rate of 9 mm2
/s. How fast is
the radius of the wetted area expanding when the radius is 120 mm? (Round your answer to two
decimal places.)
To find out how fast the radius of the wetted area is expanding, we need to use the formula for the area of a circle:
A = π * r^2
Given that the area is expanding at a rate of 9 mm^2/s, we can express this as:
dA/dt = 9 mm^2/s
We want to find out how fast the radius (dr/dt) is changing when the radius is 120 mm.
To solve this problem, we need to differentiate both sides of the area formula with respect to time (t).
dA/dt = d/dt (π * r^2)
Since the derivative of a constant (π) is zero, and differentiating r^2 with respect to time gives us 2r(dr/dt), we can write:
dA/dt = 2πr(dr/dt)
Now we can substitute the given values: dA/dt = 9 mm^2/s, and r = 120 mm:
9 = 2π(120)(dr/dt)
To find dr/dt, we can rearrange the equation:
dr/dt = 9 / (2π * 120)
Now we can calculate the value of dr/dt:
dr/dt = 9 / (2 * 3.14 * 120)
≈ 0.0119 mm/s
Therefore, the radius of the wetted area is expanding at a rate of approximately 0.0119 mm/s when the radius is 120 mm.