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Snow is piling on a driveway so its depth is changing at a rate of
r(t)=10sqrt(1-cos(0.5 t)) centimeters per hour, where
t is the time in hours,
0 <= t <= 5. At time
t=0, the depth of the snow is 20 centimeters.
What is the snow's depth at time
t=5 hours?
Use a graphing calculator and round your answer to three decimal places.
centimeters

Snow is piling on a driveway so its depth is changing at a rate of $r(t)=10 \sqrt{1-\cos (0.5 t)}$ centimeters per hour, where $t$ is the time in hours, $0 \leq t \leq 5$ . At time $t=0$ , the depth of the snow is $20$ centimeters. $\newline$ What is the snow's depth at time $t=5$ hours? $\newline$ Use a graphing calculator and round your answer to three decimal places. $\newline$ centimeters

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Find the rate of change of one variable when rate of change of other variable is given

Full solution

Q. Snow is piling on a driveway so its depth is changing at a rate of

r(t)=10 \sqrt{1-\cos (0.5 t)}

centimeters per hour, where

t

is the time in hours,

0 \leq t \leq 5

. At time

t=0

, the depth of the snow is

20

centimeters.

\newline

What is the snow's depth at time

t=5

hours?

\newline

Use a graphing calculator and round your answer to three decimal places.

\newline

centimeters

Set up integral: To find the snow's depth at $t=5$ hours, we need to integrate the rate of change function $r(t)$ from $t=0$ to $t=5$ .
Evaluate integral: Set up the integral of $r(t)$ from $0$ to $5$ : $\int_{0}^{5} 10\sqrt{1-\cos(0.5t)} \, dt$ .
Add initial depth: Use a graphing calculator to evaluate the integral: $\int_{0}^{5} 10\sqrt{1-\cos(0.5t)} \, dt \approx 70.685$ .
Calculate total depth: Add the initial depth of the snow ( $20\,\text{cm}$ ) to the result of the integral to find the total depth at $t=5\,\text{hours}$ : $20\,\text{cm} + 70.685\,\text{cm}.$
Calculate total depth: Add the initial depth of the snow ( $20\,\text{cm}$ ) to the result of the integral to find the total depth at $t=5\,\text{hours}$ : $20\,\text{cm} + 70.685\,\text{cm}$ .Calculate the total depth: $20\,\text{cm} + 70.685\,\text{cm} = 90.685\,\text{cm}$ .

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