
Water is flowing into a vertical cylindrical tank of radius $2ft.$ at the rate of $8$ cubic ft. per minute. The water level is rising at the speed of (ft . per minute)
A. $\dfrac{\pi}{2}$
B. $\dfrac{2}{\pi}$
C. $2$
D. None of these
Answer
162.9k+ views
Hint: A cylindrical tank volume is determined by multiplying its height by the area of its circular base. The rate of change of one variable with another is measured by Derivatives. This means we can determine the rate of change in a function by finding its derivative at the given point.
Formula Used: The volume of the cylinder $=\pi r^2h$
Complete step by step solution: It is given that the vertical cylindrical tank has a radius of 2ft. and the rate of change is 8 cubic ft.
$r = 2 ft$
$\dfrac{\text{d}V}{\text{d}t}=8 ft.^3/min$
The volume of the cylinder $V=\pi r^2h$
$V = \pi (2)^2 h$
$V = 4 \pi h$
Taking the derivative of both sides with respect to time (t):
$\dfrac{\text{d}V}{\text{d}t}=4\pi(\dfrac{\text{d}h}{\text{d}t})$
Substituting values we get;
$\Rightarrow8=4\pi(\dfrac{\text{d}h}{\text{d}t})\\
\Rightarrow\dfrac{8}{4\pi}=\dfrac{\text{d}h}{\text{d}t}\\
\Rightarrow\dfrac{\text{d}h}{\text{d}t}=\dfrac{2}{\pi}$
Hence, the water level is rising at the speed of$ \dfrac{2}{\pi} ft/min$.
So, option B is correct.
Note: The term "rate of change" refers to the rate at which one quantity changes with another. $Rate~of~change=\dfrac{change~in~y}{change~in~x}$ if x is the independent variable and y is the dependent variable. The rate is expressed in feet per minute. A derivative's units are always a ratio of the dependent quantity (in this case, feet) to the independent quantity (e.g. min).
Formula Used: The volume of the cylinder $=\pi r^2h$
Complete step by step solution: It is given that the vertical cylindrical tank has a radius of 2ft. and the rate of change is 8 cubic ft.
$r = 2 ft$
$\dfrac{\text{d}V}{\text{d}t}=8 ft.^3/min$
The volume of the cylinder $V=\pi r^2h$
$V = \pi (2)^2 h$
$V = 4 \pi h$
Taking the derivative of both sides with respect to time (t):
$\dfrac{\text{d}V}{\text{d}t}=4\pi(\dfrac{\text{d}h}{\text{d}t})$
Substituting values we get;
$\Rightarrow8=4\pi(\dfrac{\text{d}h}{\text{d}t})\\
\Rightarrow\dfrac{8}{4\pi}=\dfrac{\text{d}h}{\text{d}t}\\
\Rightarrow\dfrac{\text{d}h}{\text{d}t}=\dfrac{2}{\pi}$
Hence, the water level is rising at the speed of$ \dfrac{2}{\pi} ft/min$.
So, option B is correct.
Note: The term "rate of change" refers to the rate at which one quantity changes with another. $Rate~of~change=\dfrac{change~in~y}{change~in~x}$ if x is the independent variable and y is the dependent variable. The rate is expressed in feet per minute. A derivative's units are always a ratio of the dependent quantity (in this case, feet) to the independent quantity (e.g. min).
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