related rate...end part

[legacyofpiracy]

Alright so I already found out most of the problem, I just got confused with the final question. The problem was a cone with a height of 10 and radius 5 where the water within it was evaporating so that the depth was changing at a constant rate of -3/10cm/hr. I found out the volume of the water in the container when h=5 (height of water) to be 65.4498, and the rate of change of this volume still when h=5 to be -11.78cm.

But i was also asked to find the rate of change of the volume of water in the container was directly proportional to the exposed surface area of the water. Also what is the constant of proportionality?

Any thoughts? Thanks

 

[tkhunny]

Alright so I already found out most of the problem, I just got confused with the final question. The problem was a cone with a height of 10 and radius 5 where the water within it was evaporating so that the depth was changing at a constant rate of -3/10cm/hr. I found out the volume of the water in the container when h=5 (height of water) to be 65.4498, and the rate of change of this volume still when h=5 to be -11.78cm.

But i was also asked to find the rate of change of the volume of water in the container was directly proportional to the exposed surface area of the water. Also what is the constant of proportionality?

Any thoughts? Thanks

Is it sitting on the base or on the point?

If you are referring to a previous post, please reply on that thread. If not, please provide a complete problem statement.

 

[legacyofpiracy]

A container has the shape of an open right circular cone with a height of 10 and a radius of 5cm at the opening. Water in the container is dripping so the depth h is changing at a constant rate of -3/10 cm/hr

it is sitting on its point

show the rate of change of the volume of water in the container is directly proportional to the exposed surface of the water.

I have already found the rate of change of the volume to be -11.78cm/min

sorry about that

 

[soroban]

Hello, legacyofpiracy!

A container has the shape of an inverted right circular cone with a height of 10 and a radius of 5cm.
Water in the container is leaking so the depth \(\displaystyle h\) is changing at a constant rate of -3/10 cm/hr

Show the rate of change of the volume of water in the container
is directly proportional to the exposed surface of the water.

            :  5  :
    - *-----+-----*
    :  *    | r  *
    :   *   +---*
   10    *  |h *
    :     * | *
    :      *|*
    -       *

From the similar right triangles, we have: \(\displaystyle \;\frac{h}{r}\,=\,\frac{10}{5}\;\;\Rightarrow\;\;h\,=\,2r\) .[1]

Differentiate [1] with respect to time: \(\displaystyle \;\frac{dh}{dt}\,=\,2\cdot\frac{dr}{dt}\)
. . Since \(\displaystyle \frac{dh}{dt}\,=\,-0.3,\;\frac{dr}{dt}\,=\,-0.15\) .[2]


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

. . Substitute [1]: \(\displaystyle \;V\:=\:\frac{\pi}{3}r^2(2r)\:=\:\frac{2\pi}{3}r^3\)

. . Differentate with respect to time: \(\displaystyle \;\frac{dV}{dt}\:=\:2\pi r^2\left(\frac{dr}{dt}\right)\)

. . Substitute [2]: \(\displaystyle \;\frac{dV}{dt}\:=\:2\pi r^2(-0.15)\;\;\Rightarrow\;\;\L\frac{dV}{dt}\:=\:-0.3\pi r^2\) .(a)


The area of the surface of the water is: \(\displaystyle \L\;A\;=\;\pi r^2\) .(b)


The ratio of (a) to (b) is:\(\displaystyle \L\;\frac{\frac{dV}{dt}}{A}\;=\;\frac{-0.3\pi r^2}{\pi r^2}\;=\;-0.3\)
. . The ratio is a <u>constant</u>.

Therefore, the rate of change of the volume \(\displaystyle \frac{dV}{dt}\) is directly proportional to the area \(\displaystyle A\).

 

[legacyofpiracy]

Wow soroban you have helped me so much in just this past week alone I cannot even say how much I appreciate it. Your method of doing things is just very neat and clear and I always understand it. Thank you for all of your help.