Question

Gravel is being dumped from a conveyor belt at a rate of 25 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 ft high?

Answer 1

Answer:

$\frac{25}{49\pi}\approx 0.1624$ feet per minute.

Step-by-step explanation:

Let x represent height of the cone.

We have been given that gravel is being dumped from a conveyor belt at a rate of 25 ft^3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal.

Since radius is half the diameter, so radius of cone would be $\frac{x}{2}$.

Now we will use volume of cone formula.

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi(\frac{x}{2})^2 (x)$

$V=\frac{1}{3}\pi(\frac{x^2}{4}) (x)$

$V=\frac{1}{12}\pi x^3$

Now, we will take derivative with respect to time.

$\frac{dV}{dt}=\frac{1}{12}\pi \cdot 3x^2\frac{dx}{dt}$

$\frac{dV}{dt}=\frac{1}{4}\pi x^2\frac{dx}{dt}$

Upon substituting our given values, we will get:

$25=\frac{1}{4}\pi (14)^2\frac{dx}{dt}$

$25=\frac{1}{4}\pi (196)\frac{dx}{dt}$

$25=49\pi\frac{dx}{dt}$

$\frac{25}{49\pi}=\frac{dx}{dt}$

$\frac{dx}{dt}=\frac{25}{49\pi}$

$\frac{dx}{dt}=0.16240$

Therefore, the height of the pile is increasing at a rate of $\frac{25}{49\pi}\approx 0.1624$ feet per minute.

Answer 2

The rate at which the height of the pile is increasing when the pile is 14 ft high is;

dh/dt = 0.1624

Let h be the height of the cone.

We are told that the diameter and height are always equal.

Thus; d = h

Radius;r = d/2 = h/2

Now, formula for volume of a cone is;

V = ⅓πr²h

Putting h/2 for r to get;

V = ⅓π(h/2)²h

V = (1/12)πh³

Now, let's differentiate with respect to t since we are given rate at which volume is increasing and we want to find rate at which height of 14 ft is increasing.

Thus;

dV/dt = (3/12)πh²(dh/dt)

We are given; dV/dt = 25 ft³/min and h = 14 ft

Thus;

25 = ¼π(14²)(dh/dt)

dh/dt = 100/196π

dh/dt = 0.1624

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