Question 672·Hard·Algebra

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A water tank is being filled at a constant rate. After $8$ minutes, the tank contains $230$ liters of water, and after $26$ minutes, it contains $710$ liters.

Let $t$ be the time, in seconds, since the tank started filling, and let $p(t)$ be the amount of water, in liters, in the tank. Which equation models $p(t)$ ?

A linear model represents a constant rate, with a rate multiplying time and a separate starting amount. Before finding either number, convert each observation’s time to the unit used by the function. In Desmos, fit a linear rule to both readings and read the rate and starting amount. Don’t mistake the first observed amount for the amount at time zero.

Hints

Hint 1
The input $t$ is measured in seconds, but both readings use minutes. A unit conversion multiplies each number of minutes by $60$ . What second counts go with the two water amounts?
Hint 2
A linear function for constant filling has the form $p(t)=mt+b$ . Here $m$ is liters added per second, while $b$ is the amount at the start. How can the two readings determine both?
Hint 3
A regression fits unknown numbers to the readings. In Desmos, match the two modeled amounts with the two recorded amounts, using the converted times. The fitted numbers will give you the rate and starting amount.

Step-by-step

Approach 1: Fit the two readings in Desmos
Step 1
Convert both times to seconds
The input $t$ counts seconds, so multiply each time by $60$ seconds per minute: $8(60)=480,\qquad 26(60)=1560$ .Match the rate’s time unit to the function’s input.
Step 2
Turn the readings into conditions for a line
A linear model fits the constant filling rate, so write $p(t)=mt+b$ . Here $m$ is liters per second and $b$ is the amount when $t=0$ . The readings mean $p(480)=230,\qquad p(1560)=710$ .Neither recorded amount is automatically $b$ : both were measured after filling began.
Step 3
Find the rate and starting amount
Type $[480m+b,1560m+b]\sim[230,710]$ . The $\sim$ tells Desmos to fit $m$ and $b$ to both readings. Under PARAMETERS, it shows $m\approx0.444444$ and $b\approx16.666667$ . Type $m$ and $b$ on separate lines and use their fraction buttons to read $m=\frac49$ and $b=\frac{50}{3}$ .
Step 4
Write the model in the requested units
Put the fitted rate and starting amount into $p(t)=mt+b$ : $p(t)=\frac49t+\frac{50}{3}$ .The tank starts with $\frac{50}{3}$ liters and gains $\frac49$ liter each second. Choice D.

Approach 2: Find the rate from the changes
Step 1
Divide the water gained by the time elapsed
The slope is liters gained per second, so divide the change in water by the change in time. Type $\frac{710-230}{1560-480}$ into Desmos; it shows $0.444444\ldots$ . Use its fraction button to read $\frac49$ . The rate is $\frac49$ liter per second, not liters divided by the total time since filling began.
Step 2
Use one reading to find the start
At $480$ seconds, the model must give $230$ liters, so $230=\frac49(480)+b$ .Type $230\sim\frac49(480)+b$ to find the intercept $b$ , the amount at $t=0$ . Desmos shows $b\approx16.666667$ under PARAMETERS. Type $b$ on the next line and use its fraction button to read $\frac{50}{3}$ .
Step 3
Combine the two parts
The rate multiplies the number of seconds, while the intercept is added once: $p(t)=\frac49t+\frac{50}{3}$ .That models the tank’s water in liters after $t$ seconds. Choice D.

Question 672·Hard·Algebra

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