sand is falling into a conical pile so that the radius of the base of the pile is always equal to one half its altitude. of the sand is falling at the rate of 10 cubic feet per minute, how fast is the altitude of the pile increasing when the pile is 5 feet deep?

given: r = h/2

V = (1/3)π r^2 h
= (1/3)π (h/2)^2 (h)
= (1/12) π h^3

dV/dt = (1/4)π h^2 dh/dt
for the given data ...

10 = (1/4)π(25)dh/dt
dh/dt = 10(4)/((25π) = 1.6/π feet/min

check my arithmetic

Just to appreciate the good work. I wanted to hammer it in a congruent way.

Well, well, well, it seems like we have a conical pile on our hands! And not just any conical pile, but one that likes to keep things proportional – with its radius and altitude playing nicely together. How delightful!

Now, let me sprinkle some humor into this mathematical mix. If the sand is falling at a rate of 10 cubic feet per minute, we can imagine it as tiny, brave sand soldiers diving into the pile. They are volunteering for this adventure, one cubic foot at a time. Salute!

Alright, where were we? Ah, yes! We need to find how fast the altitude is increasing. Let's call the altitude "h". Since the radius is always half the altitude, we can say the radius "r" is equal to "h/2". We're all good so far, right?

To find out how fast the altitude is increasing (dh/dt), we'll differentiate both sides of the equation with respect to time. You with me? Here we go:

V = (1/3)πr²h

Differentiating both sides:

dV/dt = (1/3)π(2rh)(dh/dt)

But hey, we know dV/dt is 10 cubic feet per minute, so let's put that in there:

10 = (1/3)π(2rh)(dh/dt)

Now, at this point, we need some extra info. We need to find the values of r and h when the pile is 5 feet deep. That way, we can solve this equation and unveil the answer!

So, my friend, since r = h/2, we can substitute 5 for h and get r = 5/2. Now, let's plug these values into our equation:

10 = (1/3)π(2(5/2)(5))(dh/dt)

Simplifying this expression:

10 = (1/3)π(25)(dh/dt)

Now, we just need to solve for (dh/dt):

dh/dt = (10 * 3) / [(1/3)π(25)]

And voila! That's how fast the altitude of the pile is increasing when it's 5 feet deep. Keep in mind that this answer might be a bit "cone-fusing" to calculate exactly, but with a little bit of math magic, I'm sure you'll get it!

Hope this made you smile, my friend, and remember - never limit yourself to mathematical conundrums; keep spreading laughter like sand in the wind!

To solve this problem, we need to relate the variables involved. Let's denote the altitude as "h" and the radius as "r". We are given that the radius is always equal to half the altitude, so we have the equation:

r = (1/2)h

Next, we need to find a relationship between the volume of the cone and its dimensions. The volume of a cone is given by the formula:

V = (1/3)πr^2h

Substituting the value of "r" from the first equation, we get:

V = (1/3)π((1/2)h)^2h

Simplifying, we have:

V = (1/3)π(1/4)h^3
V = (1/12)πh^3

Taking the derivative of both sides with respect to time, we have:

dV/dt = (1/12)π(3h^2)(dh/dt)

The rate at which sand is falling into the pile is given as 10 cubic feet per minute, so we have dV/dt = 10. Substituting this into the equation, we get:

10 = (1/12)π(3h^2)(dh/dt)

Simplifying, we have:

dh/dt = (12/πh^2)

Now we can find the rate at which the altitude is increasing when the pile is 5 feet deep by substituting h = 5 into the equation:

dh/dt = (12/π(5)^2)
dh/dt = (12/π25)
dh/dt = 0.152 ft/min

Therefore, the altitude of the pile is increasing at a rate of 0.152 feet per minute when the pile is 5 feet deep.

To find the rate at which the altitude of the pile is increasing, we can use the concept of related rates. We are given that the sand is falling at a rate of 10 cubic feet per minute. Let's assign some variables to the quantities involved:

Let h be the altitude of the pile (in feet).
Let r be the radius of the base of the pile (in feet).

We know that the radius is always equal to one half the altitude: r = 1/2 * h.

The volume of a cone is given by V = (1/3) * π * r^2 * h. We can differentiate this equation with respect to time (t) to find the rate at which the volume of the pile is changing with time:

dV/dt = d/dt [(1/3) * π * r^2 * h]
= (1/3) * π * (2rh * dh/dt + r^2 * dh/dt).

Since we are given that the sand is falling at a rate of 10 cubic feet per minute (dV/dt = 10), we can substitute this value:

10 = (1/3) * π * (2rh * dh/dt + r^2 * dh/dt).

Now, we need to find dh/dt, the rate at which the altitude of the pile is changing with respect to time. We can rearrange the equation and substitute the known values at the given moment when the pile is 5 feet deep (h = 5) to find the answer:

10 = (1/3) * π * (2 * (1/2 * 5) * dh/dt + (1/2 * 5)^2 * dh/dt)
10 = (1/3) * π * (5 * dh/dt + (25/4) * dh/dt)
10 = (1/3) * π * (20/4 + 25/4) * dh/dt
10 = (1/3) * π * (45/4) * dh/dt

Now, isolate dh/dt:

dh/dt = (10 * 3)/(π * (45/4))
dh/dt = 120/(π * 45/4)
dh/dt = 35.56 ft/min (rounded to two decimal places)

Therefore, the altitude of the pile is increasing at a rate of approximately 35.56 feet per minute when the pile is 5 feet deep.