Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 4.9t0.8 - 2.6t + 18, with t ≥ 0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? What is the rate of mass change at (c) t = 1.9 s and (d) t = 4.9 s?

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Asked Nov 16, 2019
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Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 4.9t0.8 - 2.6t + 18, with t ≥ 0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? What is the rate of mass change at (c) t = 1.9 s and (d) t = 4.9 s?

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Expert Answer

Step 1

Write the given function of time for the mass of the water, and differentiate with respect to time.

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m 4.9t.8 2.6t +18 (1) dm d(4.9f08 -2.6t +18) dt dt dm = 3.92f02 -2.6 dt (2)

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Step 2

(a)

 

The water mass will be greatest when the rate of mass change will be equal to zero. Thus, time for the greatest water mass can be calculated as,

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dm =0 dt 3.92t02 2.6 0 t0.2 0.66 2 t=0.66 t 7.985 s

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Step 3

(b)

 

Substitute the value of time for the greatest water mass in the given ...

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m 4.9f0.8-2.6t 18 =49(7.985)-2.6(7.985)+18 max 23.0 kg

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