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

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grapz

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Jan 13, 2007
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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
 
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
Click to expand...
Code: 
                                |
  *                             |          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.

 
grapz said:
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?
Click to expand...
Click to expand...
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:
 
Galactus: Thank you for your help! :D

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.
 
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