grain is being poured through a chute at rate of 8 ft^3/min and forms a conical pile whose radius of the bottom is always 1/2 the height of pile. How fast is circumference increasing when the pile is 10 feet high?
given:
r = (1/2)h
2r = h
Let V be the volume
V = (1/3)π r^2 h
V = (1/3)π(r^2)(2r) = (2/3)π r^3
dV/dt = 2πr^2 dr/dt
when h = 10, r = 5
8 = 2π(25)dr/dt
dr/dt = 4/(25π)
Let circumference be C
C = 2πr
dC/dt = 2π dr/dt
so at that instance,
dC/dt = 2π(4/(25π) = 8/25 ft/min
To find the rate at which the circumference is increasing, we need to use the concept of related rates.
Let's denote the height of the pile as h and the radius of the bottom as r. According to the problem, the radius is always half the height, so we have r = h/2.
The volume of a cone can be given by the formula V = (1/3)πr^2h. Since the grain is being poured at a rate of 8 ft^3/min, we have dV/dt = 8 ft^3/min.
We are interested in finding dh/dt, the rate at which the height is changing when the pile is 10 feet high.
Differentiating the volume equation with respect to time (t), we get:
dV/dt = (1/3)[2πrh(dr/dt) + πr^2(dh/dt)]
Substituting the given values, we have:
8 = (1/3)[2π(1/2h)(dh/dt) + π(h/2)^2(dh/dt)]
Simplifying:
8 = (1/3)[π(h/2)(dh/dt) + π(h^2/4)(dh/dt)]
Multiplying by 3 to remove the fraction:
24 = π(h/2)(dh/dt) + π(h^2/4)(dh/dt)
Combining like terms:
24 = π(h/2 + h^2/4)(dh/dt)
We know that h = 10 (when the pile is 10 feet high). Substituting this value, we have:
24 = π(10/2 + 10^2/4)(dh/dt)
Simplifying:
24 = π(5 + 100/4)(dh/dt)
24 = π(5 + 25)(dh/dt)
24 = π(30)(dh/dt)
To find dh/dt, rearrange the equation:
dh/dt = 24 / (30π)
dh/dt = 0.08 ft/min (rounded to two decimal places)
Therefore, the height of the pile is increasing at a rate of 0.08 ft/min when it is 10 feet high.
To determine how fast the circumference is increasing when the pile is 10 feet high, we can use related rates. Let's break down the problem step-by-step.
First, let's set up some variables:
- Let h be the height of the pile (in feet).
- Let r be the radius of the bottom of the pile (in feet).
- Let V be the volume of the pile (in cubic feet).
We are given that the grain is being poured at a rate of 8 ft^3/min. Hence, dV/dt = 8 ft^3/min.
We are also given that the radius of the bottom of the pile is always 1/2 the height of the pile. So, r = 1/2 * h.
To find an equation involving V, h, and r, we can use the formula for the volume of a cone: V = (1/3) * pi * r^2 * h.
Substituting r = 1/2 * h into the equation for V, we get:
V = (1/3) * pi * (1/2 * h)^2 * h
V = (1/12) * pi * h^3
Differentiating both sides of the equation with respect to t (time), we can find dV/dt in terms of dh/dt (rate of change of height with respect to time):
dV/dt = (1/12) * pi * 3 * h^2 * dh/dt
8 = (1/4) * pi * h^2 * dh/dt
32 = pi * h^2 * dh/dt
Now, we need to find the rate of change of the circumference (∂C/∂t) with respect to time when the pile is 10 feet high. Since we want ∂C/∂t, we need an equation relating the circumference and height of the cone.
The circumference (C) of a cone can be found using the formula: C = 2 * pi * r.
Substituting r = 1/2 * h, we have:
C = 2 * pi * (1/2 * h)
C = pi * h
Differentiating both sides of the equation with respect to t (time), we can find dC/dt in terms of dh/dt:
dC/dt = pi * dh/dt
Now we have a relationship between dC/dt and dh/dt. We can find ∂C/∂t at h = 10 by plugging the value of t into the equation for dC/dt.
Substituting h = 10, we get:
dC/dt = pi * dh/dt
dC/dt = pi * (32 / (pi * 10^2))
Simplifying the equation, we get:
dC/dt = 32 / (10^2)
dC/dt = 32 / 100
dC/dt = 0.32 ft/min
Therefore, when the pile is 10 feet high, the circumference is increasing at a rate of 0.32 ft/min.