Respuesta :
Answer:[tex]43.54 ft^3/min[/tex]
Step-by-step explanation:
Given
Thickness of oil slick=0.07 foot
radius of slick=110 ft
[tex]\frac{\mathrm{d} r}{\mathrm{d} t}=0.9 ft/s[/tex]
Let V be the volume of oil slick
so, [tex]V=0.07\pi \cdot r^2[/tex]
rate of oil flowing is
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \cdot r\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=0.07\times 2\pi \times 110\times 0.9=43.54 ft^3/min[/tex]
Rate of something is always with respect to other thing. The rate at which the oil is flowing from the site of the accident is [tex]\approx 43.54\: ft^3/min[/tex]
How to calculate the instantaneous rate of growth of a function?
Suppose that a function is defined as;
y = f(x)
Then, suppose that we want to know the instantaneous rate of the growth of the function with respect to the change in x, then its instantaneous rate is given as:
[tex]\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}[/tex]
The thickness of the slick is 0.07 foot(constant), and the radius of the spread is increasing at 0.09 foot/minute(constant growth).
Let at some time, the radius be of 'r' units, then
Volume of the oil at that time = [tex]\pi r^2 h = \pi r^2 \times 0.07 =0.07\pi r^2 \: \rm ft^3[/tex]
It is known that
[tex]\dfrac{dr}{dt} = 0.9[/tex] (as rate of increment of radius with respect to time is constant value 0.9 foot/minute)
Thus,
[tex]\dfrac{dV}{dt} = \dfrac{d(0.07\pi r^2)}{dt} = 0.07\pi \dfrac{d(r^2)}{dt} = 0.14\pi r\dfrac{dr}{dt}\\\\\dfrac{dV}{dt} = 0.14 \times \pi \times r \times 0.9 = 0.126\pi r \: \rm ft^3/min[/tex] (instantaneous rate of change of volume when radius is r units)
At r = 110, the rate of volume increment is:
[tex]\dfrac{dV}{dt} = 0.126\pi r \: \rm ft^3/min\\\\\dfrac{dV}{dt}|_{r=110} = 0.126 \pi (110) \approx 43.54\: ft^3/min[/tex]
Thus, the rate at which the oil is flowing from the site of the accident is [tex]\approx 43.54\: ft^3/min[/tex]
Learn more about instantaneous rate here:
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