To solve this problem, we'll use related rates, which involves finding the rate of change of one variable with respect to another variable. We'll start by calculating the rate at which the oil is leaking from the oilcan.
Given:
- Diameter of the oilcan = 4 inches
- Height of the oilcan = 6 inches
- Rate of oil leakage = 2 cubic inches/hr
To find the rate at which the oil is leaking, we'll need to calculate the volume of the oilcan and differentiate it with respect to time.
Step 1: Calculate the volume of the oilcan
The oilcan is a cylinder, so its volume formula is V_cylinder = πr^2h, where r is the radius and h is the height.
Given:
Diameter of the oilcan = 4 inches
Radius (r) = diameter / 2 = 4 / 2 = 2 inches
Height (h) = 6 inches
Volume (V_cylinder) = π(2^2)(6) = 24π cubic inches
Step 2: Differentiate the volume equation with respect to time
dV_cylinder/dt = d(24Ï€)/dt
Since the oil is leaking out of the oilcan, the volume is decreasing. Therefore, the derivative will be negative.
Rate of oil leakage (dV_cylinder/dt) = -2 cubic inches/hr
Now, we know the rate at which the oil is leaking from the oilcan. Next, we'll calculate the rate at which the oil level is rising in the conical cup when the oil in the cup is 3 inches deep.
a. At what rate is the depth of the oil in the conical cup rising when the oil in the cup is 3 inches deep?
To find the rate of change of the depth of oil in the conical cup (dh_cone/dt), we'll use the concept of similar triangles.
Given:
Diameter of the conical cup = 8 inches
Height of the conical cup = 8 inches
Step 3: Set up a proportion between the oilcan and the conical cup
Since the oil is flowing from the oilcan into the conical cup, they share a similar shape. We can set up a proportion of their dimensions.
Let h_cone represent the height of the oil in the conical cup at any given time.
Using similar triangles, we can establish the following relationship:
(h_cone - 3) / h_cone = 2 / 8
Simplifying the equation:
(h_cone - 3) / h_cone = 1 / 4
Cross-multiplying:
4(h_cone - 3) = h_cone
Expanding:
4h_cone - 12 = h_cone
Simplifying:
3h_cone = 12
h_cone = 4 inches
Step 4: Differentiate the proportion equation with respect to time
Differentiating the equation 3h_cone = 12 with respect to time (t), we get:
3(dh_cone/dt) = 0
Simplifying:
dh_cone/dt = 0
Therefore, when the oil in the conical cup is 3 inches deep, the rate at which the depth of the oil in the conical cup is rising (dh_cone/dt) is 0. This means the oil level remains constant at that point.
b. When the oilcan is empty, what is the depth of the oil in the conical cup?
To find the depth of the oil in the conical cup when the oilcan is empty, we need to determine the maximum depth the oil reaches in the conical cup.
Since the oilcan is empty, the total volume of oil leaked from the oilcan will fill the conical cup.
Given:
Volume of the conical cup (V_cone) = 1/3 * π * r^2 * h
Diameter of the conical cup = 8 inches
Radius (r) = diameter / 2 = 8 / 2 = 4 inches
Height (h) = 8 inches
Substitute the values into the volume equation:
V_cone = 1/3 * π * (4^2) * 8 = 128π/3 cubic inches
Therefore, when the oilcan is empty, the depth of the oil in the conical cup is the maximum depth, which is equal to the height of the conical cup i.e., 8 inches.