The number of gallons of wat.cr in a tank t minutes after the tank has started to drain is /Q(t) = 200(30 - t) 2• How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min? report this ad
3.0
BLACKSTAR 1
We have to find the derivative of this to find rate of change.
2
Factor it out to use only power rule.
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Q'(t)= -12000+400t
Used power rule, this is derivative/rate of change of the tank flowing.
-12000+400(10)=-8000 gallons/min
Plug in t as 10, and get -8000 gallons/min. This is the answer to the first question.
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Plug it in and solve.
7
-10000 gal/min
Average is units gal/min, and -1000 is obtained from equation above.
RE S ULT
-8000 gal/min, -10000 gal/min
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john
0
It's (30-t)^2 instead of (30-t^2)
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2.0
T Y N A N A L E X… 1
Find the first derivative of Q(t) and plug t=10 to get the water's draining speed at 10 seconds. To find the average velocity over the first 10 minutes, take the change in water divided by the change in time over that period.
RE S ULT
Instantaneous = -8000 gallons/minute, Average = -10000 gallons/minute
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aznboy
0
When you used the power rule to take the derivative of Q(t), how does it become negative 400?
