Question

A spherical tank contains 81.637 gallons of water at timet=0 minutes. For the next 6 minutes, water flows out ofthe tank at the rate of 9sin square root(t+1) gallons per minute. How many gallonsof water are in the tank at the end of the 6 minutes?
why is it 36.606?

Answer 1

Okay after 6 minutes:
There is 9*Sin(sqrt(6+1)) amount of water leaked.

Answer 2

So you have original:
81.637 - amount leaked = 81.637 - 9*Sin(sqrt(7))

Answer 3

that does not equal 36.606

Answer 4

It doesn't? Let me see

Answer 5

its a related rates question

Answer 6

Ohh

Answer 7

this is so weird.. I've never encountered one like this.

Answer 8

yeah ap calculus sucks

Answer 9

I am pretty good at RR, but I don't remember seeing anything like this.
V = (4/3)*pi*r^3
dV/dt = 4*pi*r^2(dr/dt)

Answer 10

Water is flowing out so:
dv/dt = -9sin[sqrt(t+1)]

Answer 11

that gives me nothing. Sorry i'm lost. Let's see if others can help. @amistre64 , or @satellite73 or @KingGeorge . Can any of you guys take a look?

Answer 12

Since you're given a rate, you should see that as a derivative. So \[{d \over dt}=9 \sin(\sqrt{t+1}) \]Just to make sure, this is for a calculus class correct?

Answer 13

yea, i think. RR is calc

Answer 14

So integrate that from 0 to 6?

Answer 15

And that should have a negative in front.

But yes, I would integrate that from 0 to 6 first.

Answer 16

And then you add what you get to 81.637

Answer 17

It's a rather nasty integration problem, given that wolfram seems to be doing two u-substitutions followed by an integration by parts.

Answer 18

thx i'll try that

Answer 19

Although it seems to me as if you could get by with only one u-sub then an integration by parts if you let \(u=\sqrt{t+1}\)

Answer 20

In conclusion, you should be doing \[V= 81 + \int\limits_0^6 -9\sin(\sqrt{t+1}) \quad dt\]Where \(V\) is the amount of water left.