A pool is being filled. The following function gives the pool's water level (in centimeters) after $t$ seconds: $\newline$ $W(t)=4 \sqrt{t}+\ln (t+1)$ $\newline$ What is the instantaneous rate of change of the pool's water level after $100$ seconds? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ . $21$ centimeters $\newline$ (B) $0$ . $21$ centimeters per second $\newline$ (C) $44$ . $6$ centimeters $\newline$ (D) $44$ . $6$ centimeters per second

Full solution

Q. A pool is being filled. The following function gives the pool's water level (in centimeters) after

t

seconds:

\newline

W(t)=4 \sqrt{t}+\ln (t+1)

\newline

What is the instantaneous rate of change of the pool's water level after

100

seconds?

\newline

Choose

1

answer:

\newline

(A)

0

21

centimeters

\newline

(B)

0

21

centimeters per second

\newline

(C)

44

6

centimeters

\newline

(D)

44

6

centimeters per second

Differentiate function $W(t)$ : To find the instantaneous rate of change, we need to differentiate the function $W(t)$ with respect to $t$ .
$W(t) = 4\sqrt{t} + \ln(t+1)$
Let's differentiate it.
$\frac{dW}{dt} = \frac{d}{dt} [4\sqrt{t}] + \frac{d}{dt} [\ln(t+1)]$
$\frac{dW}{dt} = 4 \cdot (\frac{1}{2}) \cdot t^{-\frac{1}{2}} + \frac{1}{(t+1)}$
$\frac{dW}{dt} = \frac{2}{t^{\frac{1}{2}}} + \frac{1}{(t+1)}$
Substitute $t = 100$ : Now we substitute $t = 100$ into the derivative to find the rate of change at $t = 100$ seconds. $\newline$ $\frac{dW}{dt} = \frac{2}{100^{(1/2)}} + \frac{1}{100+1}$ $\newline$ $\frac{dW}{dt} = \frac{2}{10} + \frac{1}{101}$ $\newline$ $\frac{dW}{dt} = 0.2 + 0.00990099$
Calculate rate of change: Let's add these up to get the instantaneous rate of change. $\newline$ $\frac{dW}{dt} = 0.2 + 0.00990099$ $\newline$ $\frac{dW}{dt} \approx 0.209901$
Round to two decimal places: We round this to two decimal places as per the options given. $\frac{dW}{dt} \approx 0.21$