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?

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content_user
Answered 2021-11-25 Author has 1944 answers

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 \(\approx0.283 cm/s\)

 
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user_27qwe
Answered 2021-11-29 Author has 1922 answers

Step 1
V=volume of water
r=radius 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
\(\frac{8}{9\pi}\approx.283 cm/sec\)

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