Respuesta :
Answer:
dV/dt = 9 cubic inches per second
Explanation:
Let the height of the cylinder is h
Diameter of cylinder = height of the cylinder = h
Radius of cylinder, r = h/2
dh/dt = 3 inches /s
Volume of cylinder is given by
[tex]V = \pi r^{2}h[/tex]
put r = h/2 so,
[tex]V = \pi \frac{h^{3}}{4}[/tex]
Differentiate both sides with respect to t.
[tex]\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}[/tex]
Substitute the values, h = 2 inches, dh/dt = 3 inches / s
[tex]\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3[/tex]
dV/dt = 9 cubic inches per second
Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.
The speed at which the volume "V" is increasing when the height is 2 inches is 9πin³/s
Given the data in the question;
Height is increasing at;[tex]\frac{dh}{dt} = 3in/s[/tex]
Height; [tex]h = 2 in[/tex]
We know that the Volume of a Cylinder can be calculated using the formula:
[tex]V = \pi r^2h[/tex]
Where [tex]V[/tex] is the volume, [tex]\pi[/tex] is pie, [tex]r[/tex] is radius and [tex]h[/tex] is the height.
Given that the height is always equal to the diameter of the base of the cylinder.
Radius [tex]r[/tex] as a function of diameter is [tex]\frac{d}{2}[/tex], since d always equals h
Hence, Radius [tex]r = \frac{h}{2}[/tex]
So, [tex]V = \pi r^2h = \pi (\frac{h}{2})} ^2h = (\frac{\pi }{4})h^3[/tex]
Now, lets differentiate both side with respect to time "t"
[tex]V = (\frac{\pi }{4})h^3[/tex]
[tex]\frac{dv}{dt} = (\frac{\pi }{4})3h^2\frac{dh}{dt}[/tex]
We substitute in our given values
[tex]\frac{dv}{dt} = (\frac{\pi }{4})3(2in)^2(3in/s)\\\\ \frac{dv}{dt} = (\frac{\pi }{4})36in^3/s\\\\\frac{dv}{dt} = 9\pi in^3/s[/tex]
Therefore, the speed at which the volume, V , is increasing when the height is 2 inches is 9πin³/s
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