Respuesta :
Answer:
The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.
Step-by-step explanation:
The volume of a cone is given by the following formula:
[tex]V = \frac{\pi r^{2}h}{3}[/tex]
We have that the diameter and the height are equal, so [tex]r = \frac{h}{2}[/tex]
So
[tex]V = \frac{\pi h^{3}}{12}[/tex]
Let's derivate this equation, using implicit derivatives.
[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]
In this problem, we have to:
Find [tex]\frac{dh}{dt}[/tex], when [tex]\frac{dV}{dt} = 20, h = 14[/tex]. So
[tex]\frac{dV}{dt} = \frac{\pi h^{2}}{4}\frac{dh}{dt}[/tex]
[tex]20 = \frac{196\pi}{4}\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{20}{49\pi}[/tex]
The height of the pile is increasing [tex]\frac{20}{49\pi}[/tex] a minute when the pile is 14ft high.
This involves relationship between rates using Calculus.
dh/dt = 0.13 ft/min
- We are given;
Volumetric rate; dv/dt = 20 ft³/min
height of pile; h = 14 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 = 14 ft
radius; r = h/2 = d/2 = 14/2
radius; r= 7 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;
20 = ¹/₄π × 14² × (dh/dt)
dh/dt = (20 × 4)/(π × 14²)
dh/dt = 0.13 ft/min
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