Question

A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water is poured into the cup at a rate of 2 cm^3/sec, how fast is the water level rising when the water is 5 cm deep?

Answer 1

@tom982

Answer 2

Hey, so what have you done so far? What answer did you get?

Answer 3

I got 2.22

Answer 4

The guy got 2.79

Answer 5

so its a frustrum with radii 3 cm and 1.5 cm and the height is 5 cm

Answer 6

Okay that's what I kinda had

Answer 7

Similar triangles

Answer 8

10/3=h/r but h could maybe be 5 now that I'm thinking about it

Answer 9

in one second it fills \[2cm^3\]
frustrum area is \[(\pi/3)\times h \times (r^2+R^2 + (R \times r))\]
where r and R are bases

Answer 10

pi/3*h*(1.5^2 + R^2 +1.5r)
this is going to get messy

Answer 11

Hmmmm

Answer 12

@kayders1997, I've done this two different ways and got the same answer - it's different to both of yours. I may well be wrong, but here's my method:\[\frac{dv}{dt}=\frac{dv}{dh}\frac{dh}{dt}\]As the triangles are similar, we can find an equation for the radius given the height: \(r=h(\frac{3}{10})\) just using the properties of similar triangles. Now we have this, we input it into our volume equation:\[v=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(h\frac{3}{10})^2h=\frac{1}{3}\pi h^2\frac{9}{100}h=\frac{3}{100}\pi h^3\]Differentiating this gives us:\[\frac{dv}{dh}=\frac{9}{100}\pi h^2\]We already know \(\frac{dv}{dt}=2\) from the question, so we have enough to solve this:\[\frac{dv}{dt}=\frac{dv}{dh}\frac{dh}{dt}\]\[2=\frac{9}{100}\pi h^2\frac{dh}{dt}\]Giving us\[\frac{dh}{dt}=\frac{200}{9\pi h^2}\]At our height \(h=5\), this gives \[\frac{dh}{dt}=2.52313252202 cms^{-1}\]
Now for the other way. At height \(h=5\), we know the radius is \(r=1.5\) so the area of that circle is \(\pi \times1.5^2=2.25 \pi\). The circle is \(2.25\pi cm^2\) and the cone is filling up at \(2cm^3s^{-1}\), so we can find the rate by calculating \(\frac{2}{2.25\pi} = 2.52313252202cms^{-1}\).

Same number as before :)