Respuesta :
Answer:
[tex]125\pi \frac{\text{ft}^3}{\text{min}}[/tex].
Step-by-step explanation:
Let x represent height of the cone.
We have been given that Sand pouring from a chute forms a conical pile whose height is always equal to the diameter.
We know that radius is half the diameter, so radius of cone would be [tex]\frac{x}{2}[/tex].
We will use volume of cone formula to solve our given problem.
[tex]V=\frac{1}{3}\pi r^2h[/tex]
Upon substituting the value of height and radius in terms of x, we will get:
[tex]V=\frac{1}{3}\pi (\frac{x}{2})^2(x)[/tex]
[tex]V=\frac{1}{3}\pi\frac{x^2}{4}(x)[/tex]
[tex]V=\frac{1}{12}\pi x^3[/tex]
Now, we will take the derivative of volume with respect to time as:
[tex]\frac{dV}{dt}=\frac{1}{12}\pi\cdot 3x^2\cdot \frac{dx}{dt}[/tex]
[tex]\frac{dV}{dt}=\frac{1}{4}\pi\cdot x^2\cdot \frac{dx}{dt}[/tex]
Upon substituting [tex]x=10[/tex] and [tex]\frac{dx}{dt}=5[/tex], we will get:
[tex]\frac{dV}{dt}=\frac{1}{4}\pi\cdot (10)^2\cdot 5[/tex]
[tex]\frac{dV}{dt}=\frac{1}{4}\pi\cdot 100\cdot 5[/tex]
[tex]\frac{dV}{dt}=\pi\cdot 25\cdot 5[/tex]
[tex]\frac{dV}{dt}=125\pi[/tex]
Therefore, the sand is pouring from the chute at a rate of [tex]125\pi \frac{\text{ft}^3}{\text{min}}[/tex].
