Answer:
[tex]\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}[/tex]
Explanation:
The formula for the surface of the circle is:
[tex]A(r) = \pi\cdot r^{2}[/tex]
The rate of change of the spill area is obtained by deriving the previous formula in terms of time:
[tex]\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}[/tex]
Finally, variables are replaced by known data:
[tex]\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )[/tex]
[tex]\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}[/tex]
The question is incomplete! The complete question along with answers and explanation is provided below.
Question:
(a) If A is the area of a circle with radius r and the circle expands as time passes, find dA/dt in terms of dr/dt.
(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 39 m?
GIven Information:
Rate of change of oil spill radius = dr/dt = 1 m/s
Radius of oil spill = r = 39 m
Required Information:
dA/dt in terms of dr/dt = ?
Rate of change of oil spill area = dA/dt
Answer:
a) dA/dt = 2πr*dr/dt
b) Rate of change of oil spill area = 78π m²/s
Explanation:
(a) Since the oil spill spreads in a circular pattern, it can be modeled as circle.
Recall the the area of a circle is given by
A = πr²
Where r is the radius
Taking the derivative of radius with respect to time yields
dA/dt = 2πr*dr/dt
(b) Since r and dr/dt is given we can calculate the dA/dt
dA/dt = 2π(39)*(1)
dA/dt = 78π m²/s
Therefore, the area of the oil spill is increasing at the rate of 78π m²/s
