related rates: Water is leaking from a conical tank....

[KNS]

Okay, this is the question;

Water is leaking out a conical tank (vertex of the cone pointing down) at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2m, find the rate at which water is being pumped into the tank.

V=(1/3)(pie)r^2h pie is 3.142

if the height is 2 m then the radius is 2/3 according to like triangles (with the cone at height 6 and diameter 4)

I need to get the equation V=(1/3)(pie)r^2h down to one variable (either r or h)
I think it is h=(20)/((1/3)(pie)r^2)

then I need the derivative of the equation 1/3)(pie)r^2h =(1/3)(pie)r^2*(20)/((1/3)(pie)r^2) ??

 

[morson]

Let's say the water is being pumped into a tank at k cubed centimetres per minute. For each
10 000 cubic centimetres that leaves the tank in a minute, k cubic centimetres enter. So
(10, 000 - k) cubic centimetres per minute is the rate that water is leaving, making the rate
that water volume is entering (k - 10, 000). In math symbols, this is:

\(\displaystyle \L\ \frac{dV}{dt}\ = k - 10000\)

Let's assume the tank has height h (the water level) at any time t (in minutes since the water
begins pouring in and out). Let's also say that the radius of the surface of the water at time
t is r. So, as you pointed out, we reach the following relationship:

\(\displaystyle \L\ V = \frac{1}{3}\ \pi\ r^2 h\)

Differentiating both sides with respect to time, t, we get:

\(\displaystyle \L\ \frac{dV}{dt}\ = \frac{1}{3}\ \pi\ (2rh \frac{dr}{dt}\ + r^2 \frac{dh}{dt}\)\)

We're given \(\displaystyle \frac{dh}{dt}\ = 20\), \(\displaystyle h = 200\), and we deduced above that \(\displaystyle \L\ \frac{dV}{dt}\ = k - 10000\).

So, making these substitutions, we obtain:

\(\displaystyle \L\ k = \frac{1}{3}\ \pi\ (400r \frac{dr}{dt}\ + 20r^2) + 10 000\)

Now, you need to find a relationship relating r and h. By similar triangles, \(\displaystyle \frac{r}{h}\ = \frac{1}{3}\\)

Hence: \(\displaystyle r = \frac{h}{3}\\)

Differentiating with respect to time:

\(\displaystyle \frac{dr}{dt}\ = \frac{1}{3}\ \frac{dh}{dt}\ = \frac{20}{3}\\)

Also, when h = 200, \(\displaystyle r = \frac{200}{3}\\)

So substitute the values obtained above and find k.

 

[KNS]

thank you morson.

 

[galactus]

Re: related rates problem

Okay, this is the question;

Water is leaking out a conical tank (vertex of the cone pointing down) at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2m, find the rate at which water is being pumped into the tank.

V=(1/3)(pie)r^2h pie is 3.142

if the height is 2 m then the radius is 2/3 according to like triangles (with the cone at height 6 and diameter 4)

I need to get the equation V=(1/3)(pie)r^2h down to one variable (either r or h)
I think it is h=(20)/((1/3)(pie)r^2)

then I need the derivative of the equation 1/3)(pie)r^2h =(1/3)(pie)r^2*(20)/((1/3)(pie)r^2) ??

Not once, but 7 times.
Come on, :roll: , since when do pastries and desserts represent the ratio of a circles diameter to its circumference?.
You're in calculus and you don't know it's spelled 'pi' (the 13th letter of the Greek alphabet)?.