Question

A cylindrical tank with radius 3 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing?

Answer 1

\[
A = \pi r^2h
\]Can you differentiate?

Answer 2

uhm i mean i can plug the numbers in and get the answer but i don't know which parts to differentiate?

Answer 3

Think about it like this: \[
A(t) = \pi \times [r(t)]^2 \times [h(t)]
\]

Answer 4

and then plug in the variables?

Answer 5

You want to use the chain rule to differentiate.

Answer 6

how do you tell when to use the chain rule?

Answer 7

so would my answer be 1728pi?

Answer 8

Because we are differentiating functions of functions.

Answer 9

nevermind i know thats the wrong answer

Answer 10

i dont know that to do

Answer 11

We want to use: \[
\frac{dA}{dt} = \frac{\partial A}{\partial r}\frac{dr}{dt} + \frac{\partial A}{\partial h}\frac{dh}{dt}
\]

Answer 12

what is that weird symbol?

Answer 13

Basically, \[
A(t) = f(r(t),h(t))
\]

Answer 14

oh

Answer 15

Hmmm, well \(\partial \) is used for partial derivatives. Perhaps you have not learned about them yet?

Answer 16

i havent learned that yet

Answer 17

In that case, we'll have to use a bit more complicated method.

Answer 18

Okay, so this is what we can do... \[
\frac{d}{dt}\pi r^2 h = \left(\frac{d}{dt} \pi r^2 \right)h +\pi r^2 \left(\frac{d}{dt}h\right)
\]

Answer 19

Then we need to simplify: \[
\frac{d}{dt} \pi r^2 = \pi 2r \frac{d}{dt}r
\]

Answer 20

Tell me, does this make sense at all to you so far?

Answer 21

The chain rule tells us that: \[
\frac{d}{dt} \pi r^2 = \frac{d}{dr} \pi r^2 \frac{d}{dt}r
\]And we know: \[
\frac{d}{dr}\pi r^2 = 2\pi r
\]Which means \[
\frac{d}{dr} \pi r^2 \frac{d}{dt}r = 2\pi r \frac{d}{dt}r
\]

Answer 22

okay i understand that so far like it makes sense... but then which equation do we place the numbers into

Answer 23

would it be the 2pir d/dt r

Answer 24

Hold on. So up above we have: \[
\frac{d}{dt}\pi r^2 h = \left(\frac{d}{dt} \pi r^2 \right)h +\pi r^2 \left(\frac{d}{dt}h\right)
\]
And this equals: \[
= \left(2\pi r \frac{d}{dt}r \right)h +\pi r^2 \left(\frac{d}{dt}h\right)

\]

Answer 25

okay

Answer 26

Hmmm, hold on... I think I made things more complicated than necessary...

It seems that \(\frac{dr}{dt}=0\) from the start....

Answer 27

So actually, we just have to do: \[
A = \pi r^2h
\]
And we will get:
\[
\frac{dA}{dt} = \pi r^2 \frac{dh}{dt}
\]

Answer 28

Since \(r\) is constant with respect to time, all my other stuff wasn't necessary. This problem is actually very simple. They already tell us \(r=3\text{m}\) and \(dA/dt = 4\text{m}^3/\text{min}\)

Answer 29

You just solve for \(dh/dt\).

Answer 30

so then would my answer be 4-9pi?

Answer 31

this is so frustrating i really dont know what to do

Answer 32

No, you don't want to use subtraction to solve for \(dh/dt\).

Answer 33

Since it is being multiplied, you want to divide out the other factor.

Answer 34

so its then 4/9pi

Answer 35

thank you so much