Water is being withdrawn from a conical reservoir, 3 metres radius and 10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?
help me! Thank you so much
Oh, boy, sounds like you're taking a deep dive into some math problems! Don't worry, I'm here to clown around and help you out. Let's tackle these questions one by one:
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. This can be calculated using related rates. Now, let the depth of water in the cone be represented by 'h', and the rate at which water is being withdrawn be 'dV/dt'.
We know that the volume of a cone is given by V = (1/3)πr^2h, where r is the radius of the cone. So, the volume of the water in the cone at any time 't' is given by V = (1/3)π(3^2)h = 3πh.
We are given that dV/dt = -4 cubic meters per minute, which means that the rate of change of volume with respect to time is negative because water is being withdrawn. So, dV/dt = -4.
To find dh/dt (the rate of change of the depth of water with respect to time), we can differentiate the volume equation with respect to time 't'. This gives us dV/dt = 3π(dh/dt).
Plugging in known values, we have -4 = 3π(dh/dt).
Now, we need to find the value of dh/dt when h = 6 meters. Substitute h = 6 into the equation and solve for dh/dt:
-4 = 3π(dh/dt)
dh/dt = -4/(3π)
dh/dt ≈ -0.424 meters per minute.
So, the surface of the water is falling at a rate of approximately 0.424 meters per minute when the depth of water is 6 meters.
b) To find how fast the radius of the surface is diminishing, we need to find the rate of change of the radius with respect to time. We'll call this dr/dt.
We know that the volume equation for a cone is V = (1/3)πr^2h.
Differentiating this equation with respect to time 't', we get dV/dt = (2/3)πrh(dr/dt) + (1/3)πr^2(dh/dt).
We also know that dV/dt = -4 cubic meters per minute, and we've already calculated dh/dt to be approximately -0.424 meters per minute.
Plugging in these known values and h = 6, we can solve for dr/dt:
-4 = (2/3)π(3)(dr/dt) + (1/3)π(3^2)(-0.424)
-4 = 2π(dr/dt) - 9π(0.424)
-4 = 2π(dr/dt) - 3.192π
Now, solve for dr/dt:
2π(dr/dt) = -4 + 3.192π
dr/dt = (-4 + 3.192π) / (2π)
I don't have a calculator handy, but you can plug in the value of π and evaluate the expression. Remember to round to an appropriate number of decimal places!
Hope this helps, and remember, math is just a bunch of numbers clowning around!
To solve the problem, we can apply related rates, where we relate the change in one variable with the change in another variable.
a) Let's find the rate at which the surface of the water is falling when the depth of water is 6 meters.
Given:
- Radius of the conical reservoir (r) = 3 meters
- Depth of water (h) = 6 meters
- Rate of water withdrawal (V) = 4 cubic meters per minute
We need to find the rate at which the surface of the water (S) is falling, which can be represented by dS/dt.
We can relate the radius (r), depth (h), and surface area (S) of the water using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since the rate of water withdrawal is given as 4 cubic meters per minute (dV/dt = -4), and we need to find the rate at which the surface of the water is falling (dS/dt), we can substitute the given values into the equation:
-4 = (1/3) * π * (2 * 3 * dr/dt * 6 + 3^2 * dh/dt)
Simplifying the equation and solving for dh/dt, we get:
-12π = 12π * dr/dt + 9 * dh/dt
dh/dt = (-12π - 12π * dr/dt) / 9
Now, to find dh/dt, we need to determine dr/dt.
b) Let's find the rate at which the radius of the surface is diminishing when the depth of water is 6 meters.
To do this, we can use the formula for the relationship between the radius (r) and depth (h) of the cone:
h/r = 10/3
Taking the derivative of both sides with respect to time (t), we get:
dh/dt * (1/r) - h * (dr/dt)/r^2 = 0
Simplifying the equation and solving for dr/dt, we have:
dr/dt = -(h/r^2) * (dh/dt)
Substituting the given values (h = 6 meters and r = 3 meters) into the equation, we get:
dr/dt = -(6/3^2) * (dh/dt)
dr/dt = -2 * (dh/dt)
Now, we can substitute the value of dr/dt into the previously obtained equation for dh/dt:
dh/dt = (-12π - 12π * (-2 * dh/dt) / 9
We can now solve for dh/dt by rearranging the equation:
dh/dt + 12π * (2/9) * dh/dt = -12π * (1/9)
dh/dt * (1 + 8π/9) = -12π/9
dh/dt = -12π/9 / (1 + 8π/9)
dh/dt = -12π/9 / (9 + 8π)/9
dh/dt = -12π / (9 + 8π)
So, the rate at which the surface of the water is falling when the depth of water is 6 meters is approximately -12π / (9 + 8π) meters per minute.
To solve these questions, we can use related rates, which involves differentiating both sides of an equation with respect to time. Let's start with question a) and then move on to question b).
a) To find how fast the surface of the water is falling, we need to determine the rate of change of the depth of water with respect to time (dH/dt), where H is the depth of water.
Let's derive an equation that relates the depth of water and the radius of the water's surface. Considering the shape of the conical reservoir, we can apply similar triangles to establish the following relationship:
r / H = R / H'
Here, r is the radius of the water's surface, R is the radius of the conical reservoir, and H' is the decreasing depth of water. We are given that R = 3m, and we want to find dH'/dt when H' = 6m.
Differentiating both sides of the equation with respect to time (t):
(dR / dt) / H = (R / H') * (dH' / dt)
We need to find dH' / dt, so let's solve for it:
(dH' / dt) = (dR / dt) * (H / R)
Given that dR/dt = 0 (because the radius of the conical reservoir remains constant), we can rewrite the equation as:
(dH' / dt) = 0 * (H / 3)
(dH' / dt) = 0
Therefore, the surface of the water isn't changing at this moment, and the rate of change is 0.
b) Now, let's find how fast the radius of the water's surface is diminishing. We need to determine the rate of change of the radius with respect to time, denoted by dr/dt.
Again, we can use the same similar triangles relationship:
r / H = R / H'
Differentiating both sides with respect to time:
(dR / dt) / H = (R / H') * (dr / dt) + (r / H') * (dH' / dt)
We are interested in finding dr/dt. Rearranging the equation, we have:
(dr / dt) = [(dR / dt) / H] - [(R / H') * (dH' / dt)] / (r / H')
Substituting the given values:
dR/dt = 0 (as mentioned earlier), H = 6m, and R = 3m:
(dr / dt) = [0 / 6] - [(3 / 6) * (0)] / (r / 6)
(dr / dt) = 0
Therefore, the radius of the water's surface is not changing at this instant, and the rate of change is 0.
To summarize:
a) The surface of the water is not changing, so it is falling at a rate of 0 m/min when the depth of water is 6m.
b) The radius of the water's surface is not changing, so it is diminishing at a rate of 0 m/min at this instant.
Let me know if you need further clarification or assistance!