Respuesta :
Answer:
The volume is decreasing at a rate of 48 cubic meters per year.
Step-by-step explanation:
We are given the following information in the question:
A pyramid has height h and a square base with side x.
[tex]\text{Volume of square pyramid} = \displaystyle\frac{1}{3}x^2 h[/tex]
The height of the square pyramid remains constant.
[tex]\displaystyle\frac{dh}{dt} = 0[/tex]
The side of the base is decreasing by 0.004 meter/yr
[tex]\displaystyle\frac{dx}{dt} = -0.004\text{ meter per year}[/tex]
We have to find the rate at which the volume is decreasing when the height is 120 meters and the width is 150 meters.
[tex]V = \displaystyle\frac{1}{3}x^2 h\\\\\frac{dV}{dt} = \frac{1}{3}\bigg(2x\frac{dx}{dt}h + x^2 \frac{dh}{dt}\bigg)\\\\\text{Putting h = 120 meters and x = 150 meters}\\\\\frac{dV}{dt} = \frac{1}{3}\bigg(2(150)(-0.004)(120) + (150)^2(0)\bigg) = -48\text{ cubic meter per year}[/tex]
The volume is decreasing at a rate of 48 cubic meters per year.
The volume is decreasing at a rate of 48 m³/yr
Volume of a square pyramid
The volume of the square base pyramid is V = 1/3x²h where
- x = side of base and
- h = height of pyramid
Rate at which volume is decreasing
To find the rate at which the volume of the pyramid is decreasing we differentiate the volume with respect to x and h so,
dV/dt = d(1/3x²h)/dt
= 1/3hdx²/dt + 1/3x²dh/dt
= (2xh/3)dx/dt + 1/3x²dh/dt
Given that the height is constant, dh/dt = 0
So, dV/dt = (2xh/3)dx/dt + 1/3x²dh/dt
dV/dt = (2xh/3)dx/dt + 1/3x²(0)
dV/dt = (2xh/3)dx/dt + 0
dV/dt = (2xh/3)dx/dt
Since we need to find the rate at which the volume is decreasing when the side of the base is decreasing at a rate of 0.004 meter/yr,
- dx/dt = -0.004 m/yr and
- the height, h = 120 meters and
- its width, x = 150 meters.
Substituting the values of the variables into the equation, we have
dV/dt = (2xh/3)dx/dt
dV/dt = (2 × 150 m × 120 m/3) × (-0.004 m/yr)
dV/dt = (2 × 150 m × 40 m) × (-0.004 m/yr)
dV/dt = 12000 m² × (-0.004 m/yr)
dV/dt = -48 m³/yr
So, the volume is decreasing at a rate of 48 m³/yr
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