Question

Water runs into a conical tank at the rate of
`9ft^(3)/min.` The tank stands point down and has a height of 10 ft and a base of 5 ft. How fast is the water level rising when the water is 6 ft deep?

Answer 1

Hey there!

We can create some variables to denote some of the things we are working with.

v = volume

h = height

r = radius

We know that [ dv/dt = (π/3)r²h ]. We also know that [ r/h = 5/10 = 1/2 ] which resembles the parts of a triangle.

Solution:

v = π/3*h/2²

h = πh³/(4)(3)

~Differentiate both sides

dv/dt = (3πh³/4*3)(dh/dt)

~Simplify

dh/dt = (4dv/dt)/πh²

~Use chain rule

(4)(9)/π6²

1/π

Thus, the speed of the water level rising is [ 1/π ft/min ] when the water is 6ft deep.

Best of Luck!

Answer 2

`(1)/(\pi) \frac{\text{ft}}{\text{min}}` is the rate at which the water is rising when the water is 6 ft deep.

Explanation:

See the attached diagram that I drew to represent this problem.

To solve related rates problems, let's use the recommended steps:

1. Draw a diagram.
2. Label all quantities and their rates of change.
3. Relate all quantities in the same equation.
4. Differentiate (implicitly) with respect to time.
5. Use the resulting equation to answer the question in context.

Step 1:

I already drew the diagram; see attached image.

Step 2:

I labeled the quantities we are given in the problem. h = 10 ft, and r = 5 ft. We are also told that the change in volume is 9 ft³/min; dV/dt = 9 ft³/min.

We want to find dh/dt when h = 6.

• 
  `(dh)/(dt)\ \vert \ _h_=_6 =\ ?`

Step 3:

We know that we are dealing with a cone in this problem, and we are given the volume of the cone. Therefore, we can use the formula for the volume of a cone in order to relate all of the quantities in the same equation.

• 
  `V=(1)/(3) \pi r^2 h`

Since we only want the two variables, V and h, we can solve for r in terms of h and substitute this value for r in the formula.

This is because when we perform implicit differentiation, we do not have the change in r (dr/dt) but we do have dh/dt, which is what we are trying to solve for.

We know that r = 5 ft, and h = 10 ft. Therefore, we can say that
`(r)/(h)=(5)/(10) \rightarrow (r)/(h) = (1)/(2)`. Multiply h to both sides to solve for the variable r:
`r=(h)/(2)`.

Substitute this into the volume of a cone equation:

• 
  `V=(1)/(3) \pi((h)/(2))^2 h`

Simplify this equation.

• 
  `V=(\pi)/(3)\cdot (h^2)/(4) \cdot h`
• 
  `V=(\pi)/(12)h^3`

Step 4:

Perform implicit differentiation on the volume equation.

• 
  `(dV)/(dt) =(\pi)/(12)3h^2 \cdot (dh)/(dt)`

Step 5:

Substitute known values and solve for dh/dt to find the change in height, or the rise in water level, when the water is 6 ft deep (h = 6).

We know that:

• 
  `(dV)/(dt) =9`
• 
  `h=6`

Plug these values into the implicitly differentiated volume equation.

• 
  `9=(\pi)/(12)3(6)^2\cdot (dh)/(dt)`
• 
  `9=(3\pi)/(12)\cdot 36 \cdot (dh)/(dt)`
• 
  `9=9\pi(dh)/(dt)`
• 
  `(dh)/(dt) =(9)/(9\pi)= (1)/(\pi)`

Answering the question in context:

Since dh/dt = 1/π when h = 6 ft, we can say that the water is rising at a rate of
`(1)/(\pi) \frac{\text{ft}}{\text{min}}` when the water is 6 ft high.