# A paper cup has the shape of a cone with

Aufopferaq 2021-11-22
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/s, how fast is the water level rising when the water is 5 cm deep?

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

content_user

Step 1
If the water level is x and the rsdius of the cone at this height is $$r_{x}$$, then we can use the properties of similar triangles to write
$$\frac{r_{x}}{3}=\frac{x}{10}$$
Multiply both sides by 3
$$r_{x}=0.3x$$
The volume of the water in the cone is given by
$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(0.3x)^{2}x$$
Differentiate both sides with respect to t
$$\frac{dV}{dt}=0.03\pi\frac{d(x^{3})}{dt}$$
Since the water is being poured at $$2cm^{3}/s, dV/dt=2$$
Use chain rule in the right-hand side
$$2=0.03\pi(3x^{3-1})\cdot\frac{dx}{dt}$$
Solve for $$dx/dt$$
$$\frac{dx}{dt}=\frac{2}{0.09\pi x^{2}}=\frac{200}{9\pi x^{2}}$$
When the water is 5 cm deep, we have x=5
$$\frac{dx}{dt}|_{x=5}=\frac{200}{9\pi\cdot5^{2}}\approx0.283 cm/s$$
Result
weter level is rising at the rate of $$\approx0.283 cm/s$$

###### Have a similar question?
user_27qwe

Step 1
V=volume of water
h=heidht of water
$$V=\frac{1}{3}\pi r^{2}h$$
Step 2
Find r in terms of h.
$$\frac{r}{h}=\frac{3}{10}$$
$$r=\frac{3h}{10}$$
Step 3
Substitute for r.
$$V=\frac{1}{3}\pi(\frac{3h}{10})^{2}h$$
$$V=\frac{3}{100}\pi h^{3}$$
Step 4
Differentiate
$$\frac{dV}{dt}=\frac{3}{100}\pi\cdot3h^{2}\frac{dh}{dt}$$
Step 5
Substitute given information
$$2=\frac{3}{100}\pi\cdot3(5)^{2}\cdot\frac{dh}{dt}$$
Step 6
Solve for $$\frac{dh}{dt}$$
$$2=\frac{9\pi}{4}\cdot\frac{dh}{dt}$$
$$\frac{dh}{dt}=\frac{8}{9\pi}\approx.283$$
Result
$$\frac{8}{9\pi}\approx.283 cm/sec$$