Answer :
Answer:
[tex]\frac{2}{5}\pi \:m/min[/tex]
Step-by-step explanation:
Volume of a Cylinder,V=[tex]\pi r^2 h[/tex]
Height = 2m
Radius=2m
When h=1m, [tex]\frac{dh}{dt} =-10cm/min=-0.1m/min[/tex]
The radius of the cylinder does not change, so we treat it as a constant.
[tex]\frac{dV}{dt}=\pi r^2 \frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt}=\pi* 2^2 (-0.1)[/tex]
[tex]\frac{dV}{dt}=-0.4\pi\\\frac{dV}{dt}=-\frac{2}{5}\pi[/tex]
The water is leaking out at a rate of [tex]\frac{2}{5}\pi \:m/min[/tex]
The rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is 10 cm/min when the height is 1 m is -1.256cm³/min
The formula for calculating the volume of the cylinder is expressed as:
[tex]V_c=\pi r^2 h[/tex]
Taking the derivative of the function with respect to the height
[tex]\frac{dV_c}{dh} = \pi r^2 \frac{dh}{dt}[/tex]
Given the following parameters
[tex]h = 2m\\r = 2m\\\frac{dh}{dt} = 10cm/min = -0.1m/min[/tex]
Since the radius is constant, hence:
[tex]\frac{dV_c}{dh} = (3.14) (2)^2 (-0.1)\\\frac{dV_c}{dh}=-1.256cm^3/min[/tex]
Hence the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is 10 cm/min when the height is 1 m is -1.256cm³/min
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