Related Rates Q's: level falling in can, rising in pool

[grapz]

Use Chain Rule differienation or other differienation methods to solve the following.

1) A conical soda glass is 20 cm high with diameter at the top of 8cm. If a soda is being consumed through a thin straw at a rate of 4 pie mL/s, how fast is the soda level falling when the level is 10 cm?

this is what i did:

Dd/ Dt = Dd/Dv Dv/Dt
Dv / Dt = 4 Pie mL/s

But i am not sure how to get Dd/Dv

answer is 1 cm/s

2) A circular wading pool with sides sloping at a slope of 1/10 upward from its centre point is being filled by a hose at a rate of 1/4 L/s. How fast is the depth at the centre increasing when the depth is 15 cm?

I don't understand what this is asking.

answer = 1/90 pie cm/s

 

[soroban]

Hello, grapz!

I'll try to explain #2 . . .

2) A circular wading pool with sides sloping at a slope of 1/10 upward
from its centre point is being filled by a hose at a rate of 1/4 L/s.
How fast is the depth at the centre increasing when the depth is 15 cm?

Answer: \(\displaystyle \frac{1}{90\pi}\) cm/s

                                |
  *                             |          R                  *
        *-----------------------+-----------------------*
              *                 |                 *     :
                    *          H|           *           : rise = 1
                          *     |     *                 :
                                * - - - - - - - - - - - +
                                        run = 10

The bottom of the pool has a slope of \(\displaystyle \frac{1}{10}\)
.That is: \(\displaystyle rise\,=\,1,\;run\,=\,10\)

The volume of water is a cone: \(\displaystyle \:V \:=\:\frac{1}{3}\pi r^2h\;\) [1]

Our cone has radius \(\displaystyle R\) and height \(\displaystyle H\).
From the slope of the sides, we see that: \(\displaystyle \,R\,=\,10H\)

Substitute into [1]: \(\displaystyle \:V \;=\;\frac{1}{3}\pi(10H)^2H \;=\;\frac{100\pi}{3}H^3\)

Differentiate with respect to time: \(\displaystyle \:\frac{dV}{dt} \;=\;100\pi H^2\left(\frac{dH}{dt}\right)\;\) [2]


We are told that: \(\displaystyle \:\frac{dV}{dt}\,=\,\frac{1}{4}\) L/s \(\displaystyle \,=\,250\) cm³/s .and \(\displaystyle H \,=\,15\) cm

Substitute into [2]: \(\displaystyle \:250 \:=\:100\pi(15^2)\left(\frac{dH}{dt}\right)\;\;\Rightarrow\;\;\frac{dH}{dt}\:=\:\frac{250}{100\pi(15^2)}\)

Therefore: \(\displaystyle \:\frac{dH}{dt}\:=\:\frac{1}{90\pi}\) cm/s.

 

[galactus]

Use Chain Rule differienation or other differienation methods to solve the following.

1) A conical soda glass is 20 cm high with diameter at the top of 8cm. If a soda is being consumed through a thin straw at a rate of 4 pie mL/s, how fast is the soda level falling when the level is 10 cm?

Use similar triangles.

Since the diameter of the cone is 8, it's radius is 4.

\(\displaystyle \L\\\frac{r}{h}=\frac{4}{20}\)

\(\displaystyle \L\\r=\frac{h}{5}\)...[1]

Volume of cone: \(\displaystyle \L\\\frac{1}{3}{\pi}r^{2}h\)...[2]

Now, sub [1] into [2]. The formula will be entirely in terms of h.

Differentiate with respect to time and solve for dh/dt.

By the way, don't ever write 'pie' again. :roll:

 

[grapz]

Galactus: Thank you for your help!

Soroban: Thank you for your reply, but I have one question: I do not get this part:

dV/dT = 100pi H^2 (dH/dt)

How did you know to multiply the 100piH^2 by dH/dT ?

Thank you.

 

[grapz]

is it because of the chain rule?

H^3 = 3 H ^2 ( dH/ dT)?

 

[skeeter]

is it because of the chain rule?

H^3 = 3 H ^2 ( dH/ dT)?

I'll be %$#@*& ... another calculus student figures it out!