Question

3. A cylindrical container filled with snow cone syrup has a diameter of 4

inches and a height of 10 inches. The container is full of syrup and
leaking at a rate of 2 in ³/hr. The syrup is leaking into an empty conical
snow cone cup with a diameter of 2.5 inches and a height of 3.5 inches.
Part A: How fast is the depth of the syrup in the cylinder changing when
the level in the cylindrical container is 8 inches high?
Part B: At what time will the cylindrical syrup container be empty?
Part C: At what rate is the depth of the syrup in the conical cup rising
when the syrup in the cup is 1 inch deep? Indicate units of measure.

Answer 1

Step-by-step explanation:

The rate the level in the tank falling when the height of the syrup level is 3 feet is 8/π feet per minute. The result is obtained by using the rate of change concept.

How to count the rate of change?

A conical tank containing syrup is draining.

The rate of water volume decreasing, dV/dt = 8 ft³/min.

Height of conical tank = 6 ft.

Radius of conical tank = 2 ft.

Find the rate of syrup level falling (dh/dt) when h = 3 ft!

By the similar triangles, we get

r/h = 2/6

r/h = 1/3

r = 1/3 h

Find the volume of the syrup left at a certain height.

V = 1/3 πr²h

V = 1/3 π(1/3 h)²h

V = 1/3 π(1/9) h² h

V = 1/27 πh³

The rate of syrup level falling is

dV/dt = dV/dh . dh/dt

dV/dt = d/dt [1/27 πh³] . dh/dt

dV/dt = 1/27 π 3h² dh/dt

8 = 1/27 π 3(3)² dh/dt

8 = π dh/dt

dh/dt = 8/π ft/min

Hence, the rate of syrup level falling is 8/π feet per minute.

Hope this helps!

Answer 2

Part A: The rate at which the depth of the syrup in the cylindrical container is changing when the level is 8 inches high is 0 in/hr. Part B: The cylindrical syrup container will be empty after 20π hours. Part C: The rate at which the depth of the syrup in the conical cup is rising when the syrup is 1 inch deep is 25/3π in³/hr.

Step-by-step explanation:

Part A: To find the rate at which the depth of the syrup in the cylindrical container is changing, we can use the formula for the volume of a cylinder: V = πr^2h, where r is the radius and h is the height. First, we need to find the radius of the cylindrical container. The diameter is given as 4 inches, so the radius is half of that, which is 2 inches. The volume of the syrup in the cylindrical container is constant, as it is leaking at a rate of 2 in³/hr into the conical snow cone cup. So, the rate at which the depth of the syrup in the cylindrical container is changing when the level is 8 inches high is 0 in/hr.

Part B: To find the time at which the cylindrical syrup container will be empty, we can calculate the volume of syrup in the container and divide it by the rate at which it is leaking. The volume of a cylinder is given by V = πr^2h, and the radius and height of the cylinder are given as 2 inches and 10 inches respectively. Plugging these values into the formula, we get V = π(2^2)(10) = 40π in³. Since the syrup is leaking at a rate of 2 in³/hr, it will take (40π/2) = 20π hours for the syrup to be completely empty.

Part C: To find the rate at which the depth of the syrup in the conical cup is rising when the syrup in the cup is 1 inch deep, we can use the formula for the volume of a cone: V = (1/3)πr^2h, where r is the radius and h is the height. The radius of the conical cup is given as 2.5 inches and the height is given as 3.5 inches. We need to find the rate at which the volume of the syrup in the conical cup is changing, so we take the derivative with respect to time. Using the chain rule, we have dV/dt = (1/3)π(2.5^2)(dh/dt). We know that dh/dt = 2 in/hr, as the syrup is leaking from the cylindrical container into the conical cup at a rate of 2 in³/hr. Plugging in the values, we get dV/dt = (1/3)π(2.5^2)(2) = 25/3π in³/hr.