Question

(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?

Answer 1

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

Answer 2

Answer:

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

Explanation:

The formula for the surface of the circle is:

$A(r) = \pi\cdot r^{2}$

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$

Finally, variables are replaced by known data:

$\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )$

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$