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Sagot :
To determine how fast the height of the gravel pile is increasing when the pile is 20 feet high, we'll use the given parameters. Let's break down the problem step-by-step:
### Step 1: Identify the Given Information
1. Gravel is being dumped at a rate of [tex]\( 20 \)[/tex] cubic feet per minute. This is the rate of change of the volume ([tex]\( \frac{dV}{dt} \)[/tex]).
2. The height of the pile is [tex]\( h = 20 \)[/tex] feet when we are evaluating.
3. The base diameter and height of the cone are always equal. Therefore, the diameter [tex]\( d = h \)[/tex], and the radius [tex]\( r = \frac{d}{2} = \frac{h}{2} \)[/tex].
### Step 2: Write the Volume Formula
The volume [tex]\( V \)[/tex] of a right circular cone is given by:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Since [tex]\( r = \frac{h}{2} \)[/tex]:
[tex]\[ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \left(\frac{h^2}{4}\right) h = \frac{1}{12} \pi h^3 \][/tex]
### Step 3: Differentiate the Volume with Respect to Time
To find how fast the height [tex]\( h \)[/tex] is increasing with respect to time, we need to differentiate [tex]\( V \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{1}{12} \pi h^3 \right) \][/tex]
Using the chain rule:
[tex]\[ \frac{dV}{dt} = \frac{1}{12} \pi \cdot 3h^2 \cdot \frac{dh}{dt} \][/tex]
[tex]\[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \][/tex]
### Step 4: Solve for [tex]\( \frac{dh}{dt} \)[/tex]
Given [tex]\( \frac{dV}{dt} = 20 \)[/tex] cubic feet per minute, and [tex]\( h = 20 \)[/tex] feet, we can substitute these values into the equation:
[tex]\[ 20 = \frac{1}{4} \pi (20)^2 \frac{dh}{dt} \][/tex]
Simplify and solve for [tex]\( \frac{dh}{dt} \)[/tex]:
[tex]\[ 20 = \frac{1}{4} \pi \cdot 400 \cdot \frac{dh}{dt} \][/tex]
[tex]\[ 20 = 100 \pi \cdot \frac{dh}{dt} \][/tex]
[tex]\[ \frac{dh}{dt} = \frac{20}{100 \pi} \][/tex]
[tex]\[ \frac{dh}{dt} = \frac{1}{5 \pi} \][/tex]
Convert this to a decimal:
[tex]\[ \frac{dh}{dt} \approx 0.06366197723675814 \][/tex]
Therefore, the height of the pile is increasing at a rate of approximately [tex]\( 0.06366 \)[/tex] feet per minute when the pile is 20 feet high.
### Step 1: Identify the Given Information
1. Gravel is being dumped at a rate of [tex]\( 20 \)[/tex] cubic feet per minute. This is the rate of change of the volume ([tex]\( \frac{dV}{dt} \)[/tex]).
2. The height of the pile is [tex]\( h = 20 \)[/tex] feet when we are evaluating.
3. The base diameter and height of the cone are always equal. Therefore, the diameter [tex]\( d = h \)[/tex], and the radius [tex]\( r = \frac{d}{2} = \frac{h}{2} \)[/tex].
### Step 2: Write the Volume Formula
The volume [tex]\( V \)[/tex] of a right circular cone is given by:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Since [tex]\( r = \frac{h}{2} \)[/tex]:
[tex]\[ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \left(\frac{h^2}{4}\right) h = \frac{1}{12} \pi h^3 \][/tex]
### Step 3: Differentiate the Volume with Respect to Time
To find how fast the height [tex]\( h \)[/tex] is increasing with respect to time, we need to differentiate [tex]\( V \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{1}{12} \pi h^3 \right) \][/tex]
Using the chain rule:
[tex]\[ \frac{dV}{dt} = \frac{1}{12} \pi \cdot 3h^2 \cdot \frac{dh}{dt} \][/tex]
[tex]\[ \frac{dV}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt} \][/tex]
### Step 4: Solve for [tex]\( \frac{dh}{dt} \)[/tex]
Given [tex]\( \frac{dV}{dt} = 20 \)[/tex] cubic feet per minute, and [tex]\( h = 20 \)[/tex] feet, we can substitute these values into the equation:
[tex]\[ 20 = \frac{1}{4} \pi (20)^2 \frac{dh}{dt} \][/tex]
Simplify and solve for [tex]\( \frac{dh}{dt} \)[/tex]:
[tex]\[ 20 = \frac{1}{4} \pi \cdot 400 \cdot \frac{dh}{dt} \][/tex]
[tex]\[ 20 = 100 \pi \cdot \frac{dh}{dt} \][/tex]
[tex]\[ \frac{dh}{dt} = \frac{20}{100 \pi} \][/tex]
[tex]\[ \frac{dh}{dt} = \frac{1}{5 \pi} \][/tex]
Convert this to a decimal:
[tex]\[ \frac{dh}{dt} \approx 0.06366197723675814 \][/tex]
Therefore, the height of the pile is increasing at a rate of approximately [tex]\( 0.06366 \)[/tex] feet per minute when the pile is 20 feet high.
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