Question

The volume of water in a tank that is draining from the bottom is given : V(t): 150(1-(t/5)^2. At what rate is the water draining from the tank after 2 mins?

Answer 1

do i just sub in t = 2 into the derivative?

Answer 2

yes

Answer 3

thanks

Answer 4

is the derivative -60 +12t?

Answer 5

no its, \[150 - (1-t/5)^{2}\]
=\[150(1-2/5)^{2}\]
=??

Answer 6

the answer is supposed to be linear

Answer 7

the question is asking for a rate,i.e discharge per given time. so what do you mean by linear?

Answer 8

this is calculus so to derive a quadratic function you should get a new function; a linear function

Answer 9

but you've already been given the rate, for this question ,you don't need to derivate

Answer 10

but if the question says the volume in a tank that is draining form the bottom, doesn't that mean the functionn will tell me the volume, not the rate at t=2. and so i need to find the derivative

Answer 11

Is this the equation for the volume at time t?\[V(t)=150(1-(\frac{t}{5})^2)\]

Answer 12

thats the equalsion and the question says: where V is in litres and t is in mins.

Answer 13

then the /rate/ at which the volume changes at t=2 mins will be given by the value of the derivative of this at t=2.

Answer 14

actually its 150(1-t/5)^2

Answer 15

so if the question says the volume in a tank that is draining form the bottom, doesn't that mean the functionn will tell me the volume, not the rate at t=2. and so i need to find the derivative right?

Answer 16

ok - then your derivative is correct

Answer 17

if that's what the question says the Vis the litres then you use the derivative

Answer 18

the derivative is
300(1-t/5)

Answer 19

I believe @itsme3366 has the correct expression for the derivative

Answer 20

ok, and the reason i can't just plug t=2 in the original equation i sbecause that will just tell me the volume that is draining from the bottom and not the rate of it at t=2 right?

Answer 21

@gitahimart - I think you have forgotten to multiply by the derivative of the terms inside the braces

Answer 22

yes @itsme3366

Answer 23

only reason I'm confused is because the questions says "the volume of water in a tank that is draining" is draining makes me think that the rate equation already....

Answer 24

the sentence has two parts - "the volume of water in a tank" and its equation is then given.
the second part of the sentence tells you that this tank "is draining..."

Answer 25

lol ok good. and if i expand it, is the correct answer "300t^2/5 - 300t/5 + 150"?

Answer 26

no

Answer 27

\[(1-\frac{t}{5})^2=1-2\times\frac{t}{5}+(\frac{t}{5})^2=1-\frac{2t}{5}+\frac{t^2}{25}\]now just multiply by 150

Answer 28

which is which
\[(1-t/5)^{2}\]
or
\[1- (t/5)^{2}\]

Answer 29

so its 6t^2 -60 + 150?

Answer 30

you missed a 't' in the second term.

Answer 31

yeah sory. Ok once i factor i get t = 5. So that means when t=5, the volume is 0 meaning the water completely drained from the tank right?

Answer 32

yes

Answer 33

why did you need to expand it in the first place?

Answer 34

ok so i solve for t not from its derivative but from the original function given right? I expanded it because had to show how i got t = 5

Answer 35

you are given:\[V(t)=150(1-\frac{t}{5})^2\]so, to work out when the volume will be zero, you just need to solve this:\[150(1-\frac{t}{5})^2=0\implies 1-\frac{t}{5}=0\implies t=5\]

Answer 36

lmao yeah guess so :P alright thank you very much !

Answer 37

yw :)