Sand falling from a hooper at 10Pi ft^3/s forms a conial sand pile whose radius is always equal to its height.

Question

How fast is the radius increasing when radius is 5ft?

Answer 1

This is obviously a related rates problem. Your first step would be to declare your givens.

Since sand is falling at a rate of 10pi ft^3 per second, is it given that

dV/dt = 10pi

What we want to find is how fast the radius is increasing; that is
dr/dt = ?

Our goal is to relate V and r.

Since it is given that the sand is forming a conical sand pile, we relate V and r with the volume of a cone. The volume of a cone is

V = (1/3) pi (r^2) h

However, it is given that the radius is always equal to its height; that is, r = h. Therefore, we can REPLACE the h with r, to give us

V = (1/3) pi (r^2) (r) , which we can simplify to

V = (1/3) pi (r^3)

Now, we differentiate implicitly. In related rates problems, we ALWAYS differentiate with respect to t, or time.

Differentiating the left hand side is simply dV/dt. Differentiating the right hand side means we can ignore the constants and just differentiate the r^3.

dV/dt = (1/3)(pi)(3r^2)(dr/dt). Simplifying this, we get

dV/dt = pi(r^2)(dr/dt). Also, we know what dV/dt is equal to (it's equal to 10pi), so we can plug this value in.

10pi = pi(r^2) (dr/dt)

This step is extremely important. In *ANY* related rates problem, you have your "when" statement. Our "when" statement goes as follows:

WHEN THE RADIUS IS 5 ft ....

It is at this point that you plug in r = 5 for your existing equation. It is EXTREMELY important you don't plug the value in too early; you have to do it AFTER the when statement.

So we have:
10pi = pi (r^2) (dr/dt)

Plugging in r = 5, we get

10pi = pi (5^2) (dr/dt)
10pi = pi (25) (dr/dt)

Dividing both sides by 25pi, we essentially are able to isolate (dr/dt), to get

(10pi) (25pi) = dr/dt

Simplifying the left hand side,

10/25 = dr/dt, and
2/5 = dr/dt

dr/dt is what we wanted to solve for, and we got an answer of 2/5. Your concluding statement would go as follows;

The radius of the conical sand pile is increasing at a rate of (2/5) ft / second.

It is equally valid to say that, since 2/5 works out to a nice non-repeating decimal number (2/5 = 4/10 = 0.4), you can also say the following:

The radius of the conical sand pile is increase at a rate of 0.4 feet per second.

Answer 2

t = time, r = radius and h = height = r

Volume of falling sand at time t = 10(pi) (t)
Volume of conical pile at time t
= 1/3 (pi) r^2 (h)
= 1/3 (pi) r^2 (r) = 1/3 (pi) r^3

1/3 (pi) r^3 = 10(pi) (t)
r^3 = 30t
Differentiating both sides with respect to t,
3r^2 (dr/dt) = 30

Plugging r = 5,
3 (25) dr/dt = 30
dr/dt = 30/75 = 2/5 = 0.4

Radius is increasing at the rate of 0.4 ft/s

Answer 3

i'm not extremely confident yet i imagine rocks existed beforehand them. that's some thing to do with the rock cycle and the lava below the earth. finally the rocks split into sand- by some ability. So after rocks sand existed. I must have payed extra interest in technological knowledge.