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A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank
t minutes after the drain is opened:

V(t)=3000(1-0.05 t)^(2)
What is the instantaneous rate of change of the volume after 10 minutes?
Choose 1 answer:
(A) -150 liters per minute
(B) -150 minutes per liter
(C) 750 liters per minute
(D) 750 minutes per liter

A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank $t$ minutes after the drain is opened: $\newline$ $V(t)=3000(1-0.05 t)^{2}$ $\newline$ What is the instantaneous rate of change of the volume after $10$ minutes? $\newline$ Choose $1$ answer: $\newline$ (A) $-150$ liters per minute $\newline$ (B) $-150$ minutes per liter $\newline$ (C) $750$ liters per minute $\newline$ (D) $750$ minutes per liter

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Write linear functions: word problems

Full solution

Q. A water tank is drained. The following function gives the volume, in liters, of the water remaining in the tank

t

minutes after the drain is opened:

\newline

V(t)=3000(1-0.05 t)^{2}

\newline

What is the instantaneous rate of change of the volume after

10

minutes?

\newline

Choose

1

answer:

\newline

(A)

-150

liters per minute

\newline

(B)

-150

minutes per liter

\newline

(C)

750

liters per minute

\newline

(D)

750

minutes per liter

Differentiate $V(t)$ : To find the instantaneous rate of change, we need to differentiate the volume function $V(t)$ with respect to time $t$ .
Apply chain rule: Differentiate $V(t) = 3000(1 - 0.05t)^{2}$ using the chain rule. $\frac{dV}{dt} = 3000 \times 2(1 - 0.05t)^{2-1} \times -0.05$
Simplify the derivative: Simplify the derivative. $\frac{dV}{dt} = -300 \times (1 - 0.05t)$

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