Answer:
[tex]\frac{1}{\pi} \frac{\text{ft}}{\text{min}}[/tex] is the rate at which the water is rising when the water is 6 ft deep.
Step-by-step explanation:
See the attached diagram that I drew to represent this problem.
To solve related rates problems, let's use the recommended steps:
- Draw a diagram.
- Label all quantities and their rates of change.
- Relate all quantities in the same equation.
- Differentiate (implicitly) with respect to time.
- 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.
- [tex]\frac{dh}{dt}\ \vert \ _h_=_6 =\ ?[/tex]
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.
- [tex]V=\frac{1}{3} \pi r^2 h[/tex]
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 [tex]\frac{r}{h}=\frac{5}{10} \rightarrow \frac{r}{h} = \frac{1}{2}[/tex]. Multiply h to both sides to solve for the variable r: [tex]r=\frac{h}{2}[/tex].
Substitute this into the volume of a cone equation:
- [tex]V=\frac{1}{3} \pi(\frac{h}{2})^2 h[/tex]
Simplify this equation.
- [tex]V=\frac{\pi}{3}\cdot \frac{h^2}{4} \cdot h[/tex]
- [tex]V=\frac{\pi}{12}h^3[/tex]
Step 4:
Perform implicit differentiation on the volume equation.
- [tex]\frac{dV}{dt} =\frac{\pi}{12}3h^2 \cdot \frac{dh}{dt}[/tex]
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:
- [tex]\frac{dV}{dt} =9[/tex]
- [tex]h=6[/tex]
Plug these values into the implicitly differentiated volume equation.
- [tex]9=\frac{\pi}{12}3(6)^2\cdot \frac{dh}{dt}[/tex]
- [tex]9=\frac{3\pi}{12}\cdot 36 \cdot \frac{dh}{dt}[/tex]
- [tex]9=9\pi\frac{dh}{dt}[/tex]
- [tex]\frac{dh}{dt} =\frac{9}{9\pi}= \frac{1}{\pi}[/tex]
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 [tex]\frac{1}{\pi} \frac{\text{ft}}{\text{min}}[/tex] when the water is 6 ft high.
