Question

A cylindrical tank holds 20,000 gallons of water, which can be drained from the bottom of the tank in 20 minutes. The function gives the volume V of water (in gallons) in the tank at time t in minutes, where time t = 0 corresponds to the instant the tank starts to drain.

At time t = 3 minutes, at what rate is the volume of water in the tank changing?
(Points : 1)
–2200 gallons/minute
1700 gallons/minute
2200 gallons/minute
–1800 gallons/minute
1800 gallons/minute

Answer 1

May I plz have some help?

Answer 2

u will need a calculator, I jus dont know how to srart it...

Answer 3

do you have the function of V respect to t?

Answer 4

yes one sec

Answer 5

V(t)=20,000 (1-(t/20) )squared

Answer 6

no

Answer 7

yes

Answer 8

\[ V(t) = 20000(1-\frac{t}{20})^2\] at t =3, V =14450

Answer 9

at t = 3?

Answer 10

so the volume of water drains out is 20000-14450=5550
5550gallons in 3minutes, that means 5550/3 = 1850/minute which is not one of your options. I don't know why

Answer 11

@amistre64 @ash2326

Answer 12

@mathstudent55

Answer 13

Let wait for those mathematicians, hehehe, sorry for my uselessness

Answer 14

could u help me with another problem plz?

Answer 15

Question 4.4. The displacement s (in meters) of a particle moving in a straight line is a function of time t (in seconds). The table below gives the particle's displacement for various times:

AP Calculus AB Semester 1
The average velocity of the particle between time t = 1 and time t = 1.8 seconds is:


(Points : 1)
11.3536
14.192
12.7534
–11.3536
–14.192

Answer 16

lets make sure the first one is good

Answer 17

ok

Answer 18

A cylindrical tank holds 20,000 gallons of water,

which can be drained from the bottom of the tank in 20 minutes.

The function gives the volume V of water (in gallons) in the tank at time t in minutes, where time t = 0 corresponds to the instant the tank starts to drain.

At time t = 3 minutes, at what rate is the volume of water in the tank changing?
(Points : 1)

oh, they give you the function .... what is the function?

Answer 19

V(t)=20,000 (1-(t/20) )squared

Answer 20

\[V = 20000(1-\frac{t}{20})^2\]
take an implicit derivative with respect to time

Answer 21

how do I do this?

Answer 22

\[V = k(1-\frac{t}{n})^2\]
\[V' = 2k(-\frac 1n)(1-\frac{t}{n})t'\]
we know t = 3, and t' = dt/dt = 1

\[V' = 2k(-\frac 1n)(1-\frac{3}{n})\]

Answer 23

ok, that makes sense, so is this the derivative equation I use?

Answer 24

without all the big numbers yes

Answer 25

a rate of change should give you the idea of taking a derivative

Answer 26

l am a bit stuck

Answer 27

http://www.wolframalpha.com/input/?i=20000%28-2%2F20%29%281-3%2F20%29

Answer 28

?

Answer 29

Could u help me with the second one plz?

Answer 30

Question 4.4. The displacement s (in meters) of a particle moving in a straight line is a function of time t (in seconds). The table below gives the particle's displacement for various times:

AP Calculus AB Semester 1
The average velocity of the particle between time t = 1 and time t = 1.8 seconds is:


(Points : 1)
11.3536
14.192
12.7534
–11.3536
–14.192

Answer 31

an average is the slope of the line between the desired end points

Answer 32

ok

Answer 33

given 2 points of reference; the slope between them was well defined back in algebra 2

Answer 34

The average velocity of the particle between time t = 1 and time t = 1.8 seconds is: 12.7?

Answer 35

whats the value at t=1? and at t= 1.8?

Answer 36

13.44, and 2.0864

Answer 37

well, in order to go from high to low, we have to slow down ... our slope has to be negative for starters

Answer 38

what is the difference from 13.44 to 2.0864 ?

Answer 39

11.3536

Answer 40

wow, that was easy!!!

Answer 41

yeah, we could have estimated prolly at 11.5 divide that by the .8 seconds it took to get there

Answer 42

yup

Answer 43

could u assist me with a few more plz?

Answer 44

slope is about -11.5/.8