suppose you are drinking root beer from a conical cup. the cup has a diameter of 8 cm and a depth of 10 cm. as you suck the straw, root beer leaves the cup at the rate of 7 cm^3/second. at what rate is the level of the liquid in the cup changng when the liquid is 6 cm deep?
rate of change of volume = surface area * rate of change of depth
(look at thin slice of area pi r^2 and depth delta h)
what is r?
4/10 = r/6 similar triangles
r = 12/5
surface area = pi(144/25)
so
-7 cm^3/s = pi(144/25) cm^2 * dh/dt
so
dh/dt = -(25*7)/(144 pi) cm/s
To find the rate at which the level of the liquid in the cup is changing, you need to use the concept of related rates. You can use the formula for the volume of a cone to relate the rate at which the liquid is leaving the cup to the changing depth of the liquid.
1. Let's start by finding the formula for the volume of a cone: V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height (or depth).
2. The diameter of the cup is given as 8 cm, so the radius (r) is half of that: r = 8 cm / 2 = 4 cm.
3. The rate at which the root beer leaves the cup is given as 7 cm^3/second.
4. We are asked to find the rate at which the level of the liquid is changing when the depth is 6 cm. Let's call this rate dh/dt (change in height with respect to time).
5. Since the depth is changing over time, dh/dt is the rate we need to find.
6. To relate the rate at which the root beer leaves the cup to dh/dt, we differentiate both sides of the volume formula with respect to time (t):
dV/dt = (1/3)(d/dt)πr^2h
7. The left side of the equation represents the rate at which the volume is changing, which is the same as the rate at which the root beer leaves the cup: dV/dt = 7 cm^3/second.
8. The right side of the equation can be expressed in terms of dh/dt:
7 = (1/3)πr^2(dh/dt)
Now, we can substitute the values for r and solve for dh/dt.
9. Plug in the values:
7 = (1/3)π(4 cm)^2(dh/dt)
10. Solve for dh/dt:
7 = (1/3)π(16 cm^2)(dh/dt)
7 = (16/3)π(dh/dt)
Divide both sides by (16/3)π:
(16/3)π(dh/dt) = 7
Then, divide by (16/3)π to solve for dh/dt:
dh/dt = 7 / [(16/3)π]
Simplify:
dh/dt = (7 * 3) / (16 * π)
dh/dt = 21 / (16π)
So, the rate at which the level of the liquid in the cup is changing when the liquid is 6 cm deep is 21 / (16π) cm/second (approximately 0.416 cm/second).
To find the rate at which the level of the liquid is changing when the liquid is at a depth of 6 cm, we will use the concept of related rates.
Let's denote the depth of the liquid in the cup as 'h' (in cm), the radius of the conical cup as 'r' (in cm), and the rate at which the liquid is leaving the cup as 'dV/dt' (in cm^3/second). We know that the diameter of the cup is 8 cm, so the radius is half the diameter, which is 4 cm.
Now, we can start by outlining the given information:
- The diameter of the cup, d = 8 cm, so the radius, r = d/2 = 8/2 = 4 cm.
- The rate at which the liquid is leaving the cup, dV/dt = 7 cm^3/second.
- We need to find the rate at which the level of the liquid is changing when the liquid is at a depth of 6 cm, dH/dt (in cm/second).
To solve this problem, we will use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
We need to differentiate this formula with respect to time, t, to obtain an expression that relates the rate of change of the volume, dV/dt, to the rate of change of the depth, dH/dt.
Differentiating the volume formula, we get:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dH/dt)
Since we know dV/dt = 7 cm^3/second and r = 4 cm, we can rearrange the formula to solve for dH/dt.
7 = (1/3) * π * (2 * 4 * dr/dt * h + 4^2 * dH/dt)
To find dH/dt when h = 6 cm, we need to solve for dr/dt, which is related to the rate of change of the radius.
Since the cup is conical, the radius changes similarly to the depth. Therefore, at any given time, the ratio of the radius to the depth remains constant:
r/h = 4/10 = 2/5
Now, we can find dr/dt:
dr/dt = (2/5) * dH/dt
Replacing dr/dt in the volume formula equation:
7 = (1/3) * π * (2 * 4 * (2/5) * dH/dt * h + 4^2 * dH/dt)
Simplifying the equation:
7 = (8/15) * π * (8dH/dt * h + 16dH/dt)
7 = (64/15) * π * dH/dt * (h + 2)
Now, we can solve for dH/dt by rearranging the equation:
dH/dt = 7 * (15/64) * (1/π) * (h + 2)
Substituting the value of h = 6 cm:
dH/dt = 7 * (15/64) * (1/π) * (6 + 2)
Performing the calculations:
dH/dt = 2.578 cm/second (rounded to three decimal places)
Therefore, when the liquid is 6 cm deep, the level of the liquid in the cup is changing at a rate of approximately 2.578 cm/second.