A constant drain rate signals a linear model: starting amount minus liters lost per minute times minutes. If the later amount is a fraction of the amount at an earlier time, use the earlier amount as the fraction’s base, not automatically the starting amount. Write that comparison as an equation, then use a Desmos regression to find the rate. Because water drains, the rate must be subtracted.
Hints
- Hint 1
An amount that falls by the same number of liters each minute has a linear model. Let be the positive number of liters drained per minute. How would you use and the starting amount to write ?
- Hint 2
A fraction of an amount needs a base, the amount you take the fraction of. Here, “as much water as it held after 3 minutes” makes the base. How does that connect and ?
- Hint 3
Replace both function values using your model. That leaves one unknown rate. In Desmos, a regression uses in place of to find that rate.
Step-by-step
Model the drain and solve for its rate
Step 1Write the draining model
The reservoir starts with liters and loses the same number of liters each minute. Let be that positive number of liters per minute. A linear model subtracts the total drained, , from the starting amount:
- Step 2
Identify which amount is being compared
The words “after 3 minutes” identify the amount that the fraction is taken from. Compare with , not with the starting amount :
- Step 3
Put both times into the model
The model gives and . Substitute those into the comparison, keeping the entire minute- amount inside the fraction:
- Step 4
Find the drain rate in Desmos
Type in Desmos. The tells Desmos to find the value of that makes both sides equal. Under PARAMETERS, it shows . That is the drain rate in liters per minute, not the water remaining.
- Step 5
Use the exact rate in the function
Type on the next Desmos line and tap its fraction button to read . Subtract this exact rate times from the starting liters:
The reservoir starts with liters and loses liters per minute. Choice C.