If a cylindrical tank holds 100,000 gallons of water, whichcan be drained from the bottom of the tank in an hour, thenTorricelli's Law gives the volume V of water remaining inthe tank after t minutes as:V(t)=100,000(1−t/60)2for: 0 is less than or equal to t which is less than or equal to 60Find the rate at which the water is flowing out of the tank(the instantaneous rate of change of V with respect tot) as a function of t. What are its units? Fortimes t=0, 10, 20, 30, 40, 50, and 60 min, find the flow rate andthea mount of water remaining in the tank. Summarize your finding sin a sentence or two. At what time is the flow rate the greatest?The least?

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If a cylindrical tank holds 100,000 gallons of water, whichcan be drained from the bottom of the tank in an hour, thenTorricelli&#039;s Law gives the volume V of water remaining inthe tank after t minutes as:V(t)=100,000(1−t/60)2for: 0 is less than or equal to t which is less than or equal to 60Find the rate at which the water is flowing out of the tank(the instantaneous rate of change of V with respect tot) as a function of t. What are its units? Fortimes t=0, 10, 20, 30, 40, 50, and 60 min, find the flow rate andthea mount of water remaining in the tank. Summarize your finding sin a sentence or two. At what time is the flow rate the greatest?The least?

Aubree Mcintyre · Accepted Answer

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Jeffrey Jordon · Accepted Answer

Step 1Find V&#039;(t)V(t)=100000(1−160t)2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;0≤t≤60V′(a)=limt→aV(t)−V(a)t−a=limt→a100000(1−160t)2−100000(1−160a)2t−a=100000limt→a(1−160t)2−(1−160a)2t−aStep 2=100000limt→a1−260t+1602t2−(1−260a+1602a2)t−a=100000limt→a1−260t+1602t2−1+260a−1602a2t−a=100000limt→a−260t+1602t2+260a−1602a2t−a=100000602limt→a−120t+t2+120a−a2t−a=100000602limt→a−(t−a)120+t2−a2t−aStep 3=100000602limt→a−(t−a)120+t2−a2t−a=100000602limt→a(t−a)[−120+(t+a)]t−a=100000602limt→a[−120+t+a]=100000602[−120+(a)+a]=2509(2a−120)=5009(a−60)Step 4This can be expressed as a function of t.V′(t)=5009(t−60)evaluate at the given timesThe flow rate is greatest at the beginning , the least towards the end. The pressure from the weight of the higher volume at the beginning helps the water to push itself out the drain opening.

xleb123 · Accepted Answer

Answer:The flow rate of water out of the tank decreases over time, starting at −100,000 gal/min and gradually approaching 0 gal/min at t=60 minutes. The amount of water remaining in the tank also decreases over time, with the most significant decrease occurring in the first half of the hour. At t=60 minutes, the tank is completely drained.Explanation:To find the derivative, we can apply the power rule and the chain rule. Let&#039;s calculate it step by step:V(t)=100,000(1−t60)2dVdt=ddt(100,000(1−t60)2)Using the chain rule, we have:dVdt=100,000·2·(1−t60)·(−160)Simplifying further:dVdt=−260·100,000·(1−t60)dVdt=−130·100,000·(1−t60)Therefore, the instantaneous rate of change of V with respect to t, dVdt, is given by:V′(t)=−130·100,000·(1−t60)The units of the flow rate will be gallons per minute (gal/min).Now, let&#039;s calculate the flow rate and the amount of water remaining in the tank at specific times: t=0,10,20,30,40,50, and 60 minutes.For t=0 minutes:V(0)=100,000(1−060)2=100,000 gallons (amount of water remaining)V′(0)=−130·100,000·(1−060)=−100,000 gal/min (flow rate)For t=10 minutes:V(10)=100,000(1−1060)2≈88,889 gallons (amount of water remaining)V′(10)=−130·100,000·(1−1060)≈−77,778 gal/min (flow rate)Proceeding in the same manner, we can calculate the remaining values:For t=20 minutes:V(20)≈55,556 gallons (amount of water remaining)V′(20)≈−55,556 gal/min (flow rate)For t=30 minutes:V(30)≈22,222 gallons (amount of water remaining)V′(30)≈−33,333 gal/min (flow rate)For t=40 minutes:V(40)≈5,556 gallons (amount of water remaining)V′(40)≈−11,111 gal/min (flow rate)For t=50 minutes:V(50)≈1,389 gallons (amount of water remaining)V′(50)≈−2,778 gal/min (flow rate)For t=60 minutes:V(60)=0 gallons (amount of water remaining)V′(60)=0 gal/min (flow rate)

fudzisako · Accepted Answer

The rate at which water is flowing out of the tank, the instantaneous rate of change of V with respect to t, is given by the derivative of V(t) with respect to t:dVdt=ddt(100,000(1−t60)2)Simplifying this expression gives:dVdt=100,000·2·(1−t60)·(−160)Therefore, the rate at which water is flowing out of the tank, in gallons per minute, as a function of t is:dVdt=−100,00030(1−t60)To find the flow rate and the amount of water remaining in the tank at times t=0, 10, 20, 30, 40, 50, and 60 minutes, we substitute these values into the expressions for dVdt and V(t).For t=0 minutes:dVdt=−100,00030(1−060)=−100,00030V(0) = 100,000For t=10 minutes:dVdt=−100,00030(1−1060)=−80,00030V(10) = 81,481.48For t=20 minutes:dVdt=−100,00030(1−2060)=−60,00030V(20) = 64,444.44For t=30 minutes:dVdt=−100,00030(1−3060)=−40,00030V(30) = 49,382.72For t=40 minutes:dVdt=−100,00030(1−4060)=−20,00030V(40) = 36,296.30For t=50 minutes:dVdt=−100,00030(1−5060)=−10,00030V(50) = 25,925.93For t=60 minutes:dVdt=−100,00030(1−6060)=0V(60) = 0From the above calculations, we can summarize that as time passes, the flow rate decreases and eventually becomes zero at t=60 minutes. The amount of water remaining in the tank also decreases with time until the tank is empty at t=60 minutes.The greatest flow rate occurs at the beginning (t=0 minutes) and decreases steadily until it reaches zero at t=60 minutes. The least flow rate is zero at t=60 minutes when the tank is completely drained.

If a cylindrical tank holds 100,000 gallons of water, whichcan be drained from the bottom of the tank in an hour, thenTorricelli's Law gives the volum

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