Water is being withdrawn from a conical reservoir, 3 metres radius and 10
metres deep at 4 cubic metres per minute. a) How fast is the surface of the
water falling when the depth of water is 6 metres?
b) How fast is the radius of this surface diminishing at this instant?
To solve this problem, we can use related rates. Let's break it down step by step:
a) To find how fast the surface of the water is falling, we need to find the rate of change of the depth of water with respect to time.
Given:
Radius of the conical reservoir (r) = 3 meters
Rate of withdrawal (V) = 4 cubic meters per minute
We want to find:
Rate of change of the surface area of the water (dA/dt) when the depth is 6 meters.
First, let's find the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Differentiating both sides with respect to time (t):
dV/dt = (1/3) * π * r^2 * dh/dt
Since dV/dt is given as -4 cubic meters per minute (negative because the volume is decreasing), and r = 3 meters, we can substitute these values in the equation above:
-4 = (1/3) * π * (3^2) * dh/dt
Now, we can find dh/dt (rate of change of the depth):
dh/dt = -4 * 3^2 / [(1/3) * π * 6^2]
= -4 * 9 / (2π)
= -36 / π
Therefore, the rate at which the surface of the water is falling when the depth is 6 meters is approximately -11.46 meters per minute.
b) To find how fast the radius is diminishing at this instant, we need to find the rate of change of the radius with respect to time.
We already know the rate of change of the depth (dh/dt) from part a. Now, we need to find dr/dt (rate of change of the radius).
We can use the formula for the similar triangles in the conical reservoir:
r / h = R / H
where R is the radius of the surface of the water and H is the depth of the conical reservoir.
Substituting the known values R = 3 m (given as the radius of the conical reservoir) and H = 10 m (given depth), we get:
3 / 10 = 3 / h
Cross-multiplying:
3h = 30
h = 10
So, at the instant when the depth of the water is 6 meters, the height (h) is 6 meters.
Using the similar triangles formula, we can find the radius (r) at this instant:
r / 6 = 3 / 10
Cross-multiplying:
10r = 18
r = 18 / 10
r = 1.8 meters
Now, we can differentiate both sides of the similar triangles equation with respect to time (t):
dr/dt / 6 = -3 / 10
Solving for dr/dt:
dr/dt = -18 / 60
= -0.3 meters per minute
Therefore, at the instant when the depth is 6 meters, the radius of the surface of the water is diminishing at a rate of approximately 0.3 meters per minute.
To solve these problems, we can use the concepts of related rates. Let's break down the problems:
a) How fast is the surface of the water falling when the depth of water is 6 meters?
To find the rate at which the surface of the water is falling, we need to find the rate of change of the height of the water in terms of time. We are given that the rate of water withdrawal is 4 cubic meters per minute, but we need to express it in terms of height.
The volume of a cone can be found using the formula V = (1/3) * pi * r^2 * h, where V is the volume, r is the radius, and h is the height of the cone.
Given that the radius of the cone is 3 meters and the rate of change of the water level is dh/dt (the rate at which the height of the water is changing), we can differentiate the volume formula with respect to time:
dV/dt = (1/3) * pi * (2r * dr/dt) * h + (1/3) * pi * r^2 * dh/dt.
Since we want to find dh/dt, we need to isolate it:
dh/dt = -((2r * dr/dt) * h) / (r^2) - (dh/dt).
We know that the rate of water withdrawal (dV/dt) is -4 cubic meters per minute (since the water level is decreasing). Plugging in the values we know, we get:
-4 = -((2 * 3 * dr/dt) * h) / (3^2) - dh/dt.
Since we are interested in dh/dt when the depth of water is 6 meters (h = 6), we can plug in these values and solve for dh/dt:
-4 = -((2 * 3 * dr/dt) * 6) / (3^2) - dh/dt.
Simplifying this equation, we get:
-12 = -12 * (dr/dt) - dh/dt.
Rearranging the equation, we finally obtain:
dh/dt = -(12 + 12 * (dr/dt)).
So, when you know the rate at which the radius is changing (dr/dt), you can substitute that value into the equation and find the rate at which the surface of the water is falling (dh/dt).
b) How fast is the radius of this surface diminishing at this instant?
To find the rate at which the radius of the surface is diminishing, we need to find the rate of change of the radius with respect to time (dr/dt) when the depth of water is 6 meters (h = 6).
Using the formula for the volume of a cone (V = (1/3) * pi * r^2 * h), we can differentiate with respect to time to get:
dV/dt = (1/3) * pi * (2r * dr/dt) * h + (1/3) * pi * r^2 * dh/dt.
Since we are interested in dr/dt, we need to isolate it:
(1/3) * pi * (2r * dr/dt) * h = - (1/3) * pi * r^2 * dh/dt.
Multiplying both sides by 3/(pi * h), we get:
2r * dr/dt = - r^2 * (dh/dt) / h.
Simplifying this equation, we get:
dr/dt = - (r * (dh/dt) / (2h)).
At the instant when the water depth is 6 meters (h = 6), we can substitute the known values and solve for dr/dt:
dr/dt = - (3 * (dh/dt) / (12)).
So, when you know the rate at which the height of the water is changing (dh/dt), you can substitute that value into the equation and find the rate at which the radius is diminishing (dr/dt).
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