A water tank holds 375 gallons. A drain is opened to let water out of the tank. The drain is closed 60 minutes later. The graph below shows the rate at which water drains from the tank as a function of time.

How much water remains in the tank after the drain is closed?
$\square$ gallons

Question

A water tank holds 375 gallons. A drain is opened to let water out of the tank. The drain is closed 60 minutes later. The graph below shows the rate at which water drains from the tank as a function of time.

How much water remains in the tank after the drain is closed?
$\square$ gallons

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The rate of water drainage is given by the function on the graph, which starts at 10 gal/min at $t = 0$ and decreases linearly to 2 gal/min at $t = 60$ minutes. ‖ ‖ step 2 ⋮ To find the total amount of water drained, we need to integrate the rate function over the interval from $t = 0$ to $t = 60$. ‖ ‖ step 3 ⋮ The rate function can be approximated as a linear function: $r(t) = 10 - \frac{8}{60}t = 10 - \frac{2}{15}t$. ‖ ‖ step 4 ⋮ Integrate $r(t)$ from $0$ to $60$: $$ \int_{0}^{60} \left(10 - \frac{2}{15}t\right) dt $$ ‖ ‖ step 5 ⋮ Compute the integral: $$ \int_{0}^{60} 10 \, dt - \int_{0}^{60} \frac{2}{15}t \, dt = \left[10t\right]_{0}^{60} - \left[\frac{2}{15} \cdot \frac{t^2}{2}\right]_{0}^{60} $$ ‖ ‖ step 6 ⋮ Evaluate the definite integrals: $$ 10 \cdot 60 - \frac{2}{15} \cdot \frac{60^2}{2} = 600 - \frac{2}{15} \cdot 1800 = 600 - 240 = 360 \text{ gallons} $$ ‖ ‖ step 7 ⋮ Subtract the total drained water from the initial amount: $$ 375 - 360 = 15 \text{ gallons} $$ ‖ ∻Answer∻ ⚹ 15 gallons ⚹ ∻Key Concept∻ ⚹Integration of a linear rate function⚹ ∻Explanation∻ ⚹To find the remaining water in the tank, we integrated the rate of drainage over the given time interval and subtracted the total drained water from the initial amount.⚹◊What is the rate of change of the water drainage at 30 minutes⍭ Generate me a similar question◊

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