Mathematics, 23.06.2019 21:10 paypay319

Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min. it forms a pile in the shape of a right circular cone whose base diameter and height are always the same. how fast is the height of the pile increasing when the pile is 13 ft high?

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Answer from: whosdarrin9396

This involves relationship between rates using Calculus.

dh/dt = 0.3 ft/min

We are given;

Volumetric rate; dv/dt = 40 ft³/min

height of pile; h = 13 ft

We are not given the diameter here but as we are dealing with a right circular cone, we will assume that the diameter is equal to the height.

Thus; diameter; d = 13 ft

radius; r = h/2 = d/2 = 13/2

radius; r= 6.5 ft

Formula for volume of a cone is;

V = ¹/₃πr²h

We want to find how fast the height is increasing and this is dh/dt.

Thus, we will need to express r in the volume formula in terms of h;

V = ¹/₃π(h/2)²h

V = ¹/₃π(h²/4)h

V = ¹/₁₂πh³

differentiating both sides with respect to time t gives;

dV/dt = 3(¹/₁₂πh²)dh/dt

dV/dt = ¹/₄πh²(dh/dt)

Plugging in the relevant values, we have;

40 = ¹/₄π × 13² × (dh/dt)

dh/dt = (40 × 4)/(π × 13²)

dh/dt = 0.3 ft/min

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Answer from: dbhuggybearow6jng

$\frac{dh}{dt}=\frac{160}{169\pi } ft/min$

Step-by-step explanation:

This is a classic related rates problem. Gotta love calculus!

Start out with the formula for the volume of a cone, which is

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

and with what we know, which is $\frac{dV}{dt}=40$

and the fact that the diameter = height (we will come back to that in a bit).

We need to find $\frac{dh}{dt}$ when h = 13

The thing we need to notice now is that there is no information given to us that involves the radius. It does, however, give us a height. We need to replace the r with something in terms of h. Let's work on that first.

We know that d = h. Because d = 2r, we can say that 2r = h, and solving for r gives us that $r=\frac{h}{2}$ .

Now we can rewrite the formula with that replacement:

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

Simplify that all the way down to

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

The derivative of that function with respect to time is

$\frac{dV}{dt}=\frac{1}{12}\pi(3h^2)\frac{dh}{dt}$

Filling in what we have gives us this:

$40=\frac{1}{12}\pi (3)(13)^2\frac{dh}{dt}$

Solve that for the rate of change of the height:

$\frac{dh}{dt}=\frac{160}{169\pi } \frac{ft}{min}$

or in decimal form:

$\frac{dh}{dt}=.95\pi \frac{ft}{min}$

Answer from: Quest

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

What are the zeros of the function f(x)=x^3-2x^2-3x !

Answer from: Quest

answer: i ! the answer is 9000 seconds, also d

hope i

-lavira : )

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