Respuesta :
The rate of the change of the volume of the pile, in cubic feet per hour, when the circumference of the base is 8x feet, for this case, is 40x²/π³ cubic ft/ hour.
How to calculate the instantaneous rate of change of a function?
Suppose that a function is defined as;
[tex]y = f(x)[/tex]
Then, suppose that we want to know the instantaneous rate of the change 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]
What is chain rule of differentiation?
If a variable z depends on the variable y, which itself depends on the variable x, then we have:
[tex]\dfrac{dz}{dx} = \dfrac{dz}{dy} \times \dfrac{dy}{dx}[/tex]
(assuming all these derivatives exist).
For this case, we're given that:
[tex]V = \dfrac{r^3}{3}[/tex] (in cubic ft)
Assuming the cross section being circular, expressing radius 'r' in terms of circumference, we get:
[tex]C = 2\pi r\\\\r = \dfrac{C}{2\pi}\\\\V = r^3/3\\\\V = \dfrac{C^3}{24\pi^2} \: \rm ft^3[/tex]
It is given that rate of change of circumference with respect to time is 5.
Thus, symbolically, we get:
[tex]\dfrac{dC}{dt} = 5[/tex]
Finding rate of change of volume with respect to time:
[tex]\dfrac{dV}{dt} = \dfrac{d\left (\dfrac{C^3}{24\pi^2}\right) }{dC} \times \dfrac{dC}{dt} = \dfrac{3C^2}{24\pi^3} \dfrac{dC}{dt} = \dfrac{5C^2}{8\pi ^3} \: \rm ft^3/hour[/tex]
(using chain rule of differenciation).
At C = 8x ft, we get:
[tex]\dfrac{dV}{dt} = \dfrac{5C^2}{8\pi ^3} \: \text{ft}^3\text{/hour}\\\\\dfrac{dV}{dt} |_{C = 8x} = \dfrac{5(8x)^2}{8 \pi^3} = \dfrac{40x^2}{\pi^3} \: \rm ft^3/hour[/tex]
Thus, the rate of the change of the volume of the pile, in cubic feet per hour, when the circumference of the base is 8x feet, for this case, is 40x²/π³ cubic ft/ hour.
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