hard Rate of Change question (1 Viewer)

[c0okies]

Water is poured into a cone vessel at a constant rate of 24cm^3 per second. The depth of water is h cm at any time t seconds. What is the rate of increase of the area of the surface S of the liquid when the depth is 16cm...


Thanks ahead for any help.

 

[c0okies]

ok i finished it lol; if you dont kno how to do it PM me.

 

[Riviet]

I'm stuck as well. XD

 

[c0okies]

hm; i was stuck.. i spent 30 mins on it and i almost went crazy...... but then i had a break and came back to it

but anyways.. ill try not to waste forum space =/

 

[c0okies]

btw the answer is 3cm^2/s; oh probably because i forgot to mention the ratio of r and h is r=h

sorry!

 

[Riviet]

btw the answer is 3cm^2/s

? Where did I go wrong?

 

[deadnature]

Correct me if im worng but isnt the volume of a cone 1/3 * pi * r^2 * h

 

[c0okies]

u want me to post the solution?

 

[insert-username]

Water is poured into a cone vessel at a constant rate of 24cm^3 per second. The depth of water is h cm at any time t seconds. What is the rate of increase of the area of the surface S of the liquid when the depth is 16cm...

V = (1/3)πr²h

but h = r, so
V = (1/3)πr³

dV/dr = πr²
so dr/dV = 1/πr²

dV/dt = 24

So dr/dt = dV/dt . dr/dV (chain rule)
= 24/πr²

S = πr²
so dS/dr = 2πr

dS/dt = dr/dt . dS/dr (chain rule)

= 24/πr² . 2πr

= 48/r cm²/s

but r = h = 16

so dS/dt = 48/16 cm²/s

= 3 cm²/s


There we go. It can be done without knowing that h = r (you substitute h = 16 into the first equation for V, and that eliminates it completely), but then you end up with the rate of change of area with respect to time being 4.5 cm²/s.


I_F