# Related rates: dh/dt given time

• virulent
In summary, the rate at which the height of the solution is increasing after 3 minutes in the process is 1/9 cm/min. This is determined by using the formula V = (1/3)πr^2h and differentiating it with respect to time, where the rate of change of volume (dV/dt) is known to be 1 cm^3/min. By substituting the values for r and h, and solving for dh/dt, we get dh/dt = (1/9) cm/min.
virulent

## Homework Statement

A solution is being poured into a conical filter in a chemistry experiment at a rate of 5cm^3/min. The filter is 15 cm high with a diameter of 10 cm at the top. The solution is dropping out of the filter at a rate of 1 cm^3/min. Determine the rate at which the height of the solution is increasing 3 minutes in this process.

$$h = 15 cm$$
$$r = \frac{10cm}{2} = 5 cm$$
$$\frac{dV}{dt} = \frac{5cm^3}{min} - \frac{4cm^3}{min} = \frac{1cm^3}{min}$$
$$t = 3 minutes$$
$$\frac{dh}{dt} = ?$$

## Homework Equations

$$V = \frac{1}{3}\pi r^2h$$
$$\frac{r}{h} = \frac{5cm}{15cm} => r = \frac{h}{3}$$

## The Attempt at a Solution

Substitute.

$$V = \frac{1}{3}\pi (\frac{h}{3})^2h$$
...
$$V = \frac{\pi}{27} h^3$$

Differentiate.

$$\frac{dV}{dt} = \frac{\pi}{27} 3h^2 \frac{dh}{dt}$$

At this point I feel I went wrong somewhere because I have not incorporated t so I am unsure of where to go from there. Countless searches brought me no results which is even more confusing.

I feel I have to get a formula for h given t but that is also a problem, I think?

welcome to pf!

hi virulent! welcome to pf!
virulent said:
$$\frac{dV}{dt} = \frac{\pi}{27} 3h^2 \frac{dh}{dt}$$

that looks fine

but after all that work, you've lost the plot

you've forgotten that dV/dt is known!

## 1. What is "dh/dt" in related rates?

In related rates, "dh/dt" represents the rate of change of a variable h with respect to time t. It is a derivative that measures how quickly the value of h is changing at a specific moment in time.

## 2. How is "dh/dt" calculated in related rates?

To calculate "dh/dt" in related rates, you need to find the derivative of the equation that relates the variables h and t. This could involve using the chain rule, product rule, or quotient rule, depending on the specific problem. Once you have the derivative, you can substitute in the given values to find the specific rate of change at a given time.

## 3. What is the significance of "dh/dt" in related rates?

"dh/dt" is significant in related rates because it allows us to understand how a particular variable is changing over time. By calculating this rate of change, we can make predictions and solve problems related to real-life scenarios, such as the changing height of an object or the rate of water flowing into a tank.

## 4. How do you use "dh/dt" to solve related rates problems?

To solve related rates problems using "dh/dt", you first need to identify the variables involved and the equation that relates them. Then, take the derivative of this equation with respect to time, and substitute in the given values. Finally, solve for the unknown rate of change, which will be represented by "dh/dt."

## 5. Can "dh/dt" be negative in related rates?

Yes, "dh/dt" can be negative in related rates. This would indicate a decreasing rate of change, meaning the variable h is decreasing over time. It is important to pay attention to the sign of "dh/dt" in related rates problems, as it can affect the interpretation of the solution.

Replies
2
Views
902
Replies
24
Views
1K
Replies
1
Views
961
Replies
4
Views
2K
Replies
3
Views
3K
Replies
9
Views
3K
Replies
30
Views
3K
Replies
3
Views
2K
Replies
10
Views
1K
Replies
4
Views
2K