A leaky cylindrical oilcan has a diameter of 4 inches and a height of 6 inches. The can is full of oil and is leaking at the rate of 2 cubic inches/hr. The oil leaks into an empty conical cup with a diameter of 8 inches and a height of 8 inches.


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?
b. When the oilcan is empty, what is the depth of the oil in the conical cup?

Please show step-by-step work in your answers. Thanks in advance! :)

let the height of oil in the cone be h in, let the radius of the surface of oil in the cone be r in. Let the volume in the cone be V in^3

by ratios: r/h = 4/8
r = h/2

given d(V)/d(time) = 2 cu inches/hour
find dh/dt when h = 3 inches

V = (1/3)pi(r^2)h subbing in r=h/2
V = (1/12)pi(h^3)

dV/dt = (1/4)pi(h^2)dh/dt

sub in dV/dt = 2 and h = 3 to get
dh/dt = 8/(9pi) inches/hour

for b)

find the volume of the cylinder.
sub that into V=(1/12)pi(h^3) and solve for h

Let me know if you got h = 6.6

For part (b), I don't understand why h = 6.6 ft because that would mean that the volume of the oilcan had a volume of 75.27 in^3. In actuality, the oilcan had a volume of 25.13 in^3, so the height should be 4.6 ft. **

Ryan is wrong. Reiny is correct. h = 6.6

A drum can in the shape of a rectangular shape with a square base is required to have a volume of 100 cubic feet. Draw a diagram of the drum indicating x, the side of the base and h, the height.

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(24π)/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.