Question

A certain cylindrical tank holds 20,000 gallons of water which can be drained from the bottom of the tank in 20 minutes the volume tea of water remaining in the tank after T minutes is given by the function (v)= 20,000(1-(t/20)² where B is in gallons zero is less than or equal to T which less than or equal to 20 is in a minutes and T equals zero represent the instant the tank starts draining. The average rate of change in volume of water in the tank from time t=0 to t=20 is (v(20)-v(0))/20-0=-1000 gallons/minute. At what time t is the instantaneous rate of the water draining from the tank at 1000 gallons/minute​

Answer 1

Answer:The volume of the water in the tank after t minutes is given by

\displaystyle \sf{ v(t) = 20000 {\bigg(1 - \frac{t}{20} \bigg)}^{2} }v(t)=20000(1−

20

t

)

2

Differentiating both sides with respect to t we get

\displaystyle \sf{ \frac{dv}{dt} = 20000 \times 2 \times \bigg( - \frac{1}{20} \bigg)\times {\bigg(1 - \frac{t}{20} \bigg)} }

dt

dv

=20000×2×(−

20

1

)×(1−

20

t

)

\implies \: \displaystyle \sf{ \frac{dv}{dt} = - 2000 \times {\bigg(1 - \frac{t}{20} \bigg)} }⟹

dt

dv

=−2000×(1−

20

t

)

\implies \: \displaystyle \sf{ \frac{dv}{dt} = 100t - 2000 } \: \: \: \: ....(1)⟹

dt

dv

=100t−2000....(1)

Which the equation of instantaneous rate of the water draining from the tank

Now

\displaystyle \sf{ v(20) = 20000 {\bigg(1 - \frac{20}{20} \bigg)}^{2} = 0 }v(20)=20000(1−

20

20

)

2

=0

\displaystyle \sf{ v(0) = 20000 {\bigg(1 - \frac{0}{20} \bigg)}^{2} = 20000 }v(0)=20000(1−

20

0

)

2

=20000

So

\displaystyle \sf{ \frac{v(20) - v(0)}{20 - 0} }

20−0

v(20)−v(0)

\displaystyle \sf{ = \frac{0 - 20000 }{20} }=

20

0−20000

\displaystyle \sf{ = \frac{ - 20000 }{20} }=

20

−20000

\displaystyle \sf{ = - 1000 }=−1000

Therefore

\displaystyle \sf{ \frac{v(20) - v(0)}{20 - 0} = - 1000 }

20−0

v(20)−v(0)

=−1000

Now let at time t = T the instantaneous rate of the water draining from the tank at 1000 gallons/minute

In other words we have to determine time t such that

\displaystyle \sf{ At \: \: \: t = T \: \: \: we \: \: have \: \: \frac{dv}{dt} = 1000}Att=Twehave

dt

dv

=1000

From Equation (1) we get

\sf{1000 = 100 T - 2000}1000=100T−2000

\implies \: \sf{100 T = 3000}⟹100T=3000

\implies \: \sf{T = 30}⟹T=30

Hence the required time is 30 min

Step-by-step explanation:

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Answer 2

Answer:

30 mins.

Step-by-step explanation:

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The volume of the water in the tank after t minutes is given by

$\text v(t) = 20000(1- \frac {t}{20} )^2$

Differentiating both sides w.r.t. t we get

$\text { \frac {dv}{dt} = 20000 * 2 * (1- \frac {t}{20} ) * \frac {-1}{20} }$

=> $\text{ \frac {dv}{dt} = -2000 * (1- \frac {t}{20} ) }$

=> $\text { \frac {dv}{dt} = 100t - 2000 }$ ------------(1)

Which is the equation of instantaneous rate of the water draining from the tank.

Now,

====> $\text { v(20) = 20000(1- \frac {20}{20} )^2 }$ = 0

====> $\text { v(0) = 20000(1- \frac {0}{20} )^2 }$ = 20000

So,

==> $\frac {v(20) - v(0)}{20-0}$

==> $\frac {0-20000}{20}$

==> $\frac {-20000}{20}$

==> -1000

Now let at time t = T the instantaneous rate of the water draining from the tank at 1000 gallons/minute

.

In other words we have to determine time t such that

,

At t = T we have $\text \frac {dv}{dt} = 1000$ -------(2)

From (1) and (2)

1000 = 100t - 2000

3000 = 100t

t = 30mins.

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