A cup has a radius of 2" at the bottom and 6" on the top. It is 10" high. 4 Minutes ago, water started pouring at 10 cubic " per minute. How fast was the water level rising 4 minutes ago? How fast is the water level rising now? What will the rate be when the glass is full?
Rates in a Cup with a Base
- Thread starter Herdcules
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There is an easy way to solve for dh/dt in related rates problems.
It is based on the observation that \(\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}\)
Thus, \(\displaystyle \frac{dh}{dt}=\frac{dV/dt}{A(t)}\)
Where \(\displaystyle A(t)\) is the cross-sectional area of the water surface at some time or height.
So, when h=0, the area of the surface is the area of the bottom of the cup.
\(\displaystyle A={\pi}(2)^{2}=4\pi\)
So, \(\displaystyle \frac{dh}{dt}=\frac{10}{4\pi}=\frac{5}{2\pi}\)
When the glass is full, the area of the water surface will be \(\displaystyle 36\pi\)
So, \(\displaystyle \frac{dh}{dt}=\frac{5}{18\pi}\)
Intuitively, dh/dt should be slowing as the water rises.
Four minutes later, there will be \(\displaystyle 40 \;\ in^{3}\) of water in the cup.
You can treat this like a truncated cone. Extend the sides to form an imaginary tip on down to form a cone.
By similar triangles, \(\displaystyle \frac{r}{h}=\frac{6}{15}=\frac{2}{5}\)
\(\displaystyle r=\frac{2h}{5}\)
\(\displaystyle V=\frac{\pi}{3}\left(\frac{2h}{5}\right)^{2}h-V_{0}\)
\(\displaystyle V_{0}\) is the volume of the imaginary portion after the sides are extended. This imaginary part is 5 inches high. So, it has volume \(\displaystyle \frac{20\pi}{3}\). Upon differentiating, it is 0.
\(\displaystyle \frac{dV}{dt}=\frac{4\pi h^{2}}{25}\frac{dh}{dt}\)
\(\displaystyle \frac{dh}{dt}=\frac{25}{4\pi h^{2}}\frac{dV}{dt}\)
Now, find the height when the volume is 40 in^3 and you have it.
It is based on the observation that \(\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}\)
Thus, \(\displaystyle \frac{dh}{dt}=\frac{dV/dt}{A(t)}\)
Where \(\displaystyle A(t)\) is the cross-sectional area of the water surface at some time or height.
So, when h=0, the area of the surface is the area of the bottom of the cup.
\(\displaystyle A={\pi}(2)^{2}=4\pi\)
So, \(\displaystyle \frac{dh}{dt}=\frac{10}{4\pi}=\frac{5}{2\pi}\)
When the glass is full, the area of the water surface will be \(\displaystyle 36\pi\)
So, \(\displaystyle \frac{dh}{dt}=\frac{5}{18\pi}\)
Intuitively, dh/dt should be slowing as the water rises.
Four minutes later, there will be \(\displaystyle 40 \;\ in^{3}\) of water in the cup.
You can treat this like a truncated cone. Extend the sides to form an imaginary tip on down to form a cone.
By similar triangles, \(\displaystyle \frac{r}{h}=\frac{6}{15}=\frac{2}{5}\)
\(\displaystyle r=\frac{2h}{5}\)
\(\displaystyle V=\frac{\pi}{3}\left(\frac{2h}{5}\right)^{2}h-V_{0}\)
\(\displaystyle V_{0}\) is the volume of the imaginary portion after the sides are extended. This imaginary part is 5 inches high. So, it has volume \(\displaystyle \frac{20\pi}{3}\). Upon differentiating, it is 0.
\(\displaystyle \frac{dV}{dt}=\frac{4\pi h^{2}}{25}\frac{dh}{dt}\)
\(\displaystyle \frac{dh}{dt}=\frac{25}{4\pi h^{2}}\frac{dV}{dt}\)
Now, find the height when the volume is 40 in^3 and you have it.
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D
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Read carefully - instead of staring at the screen - start working with paper and pencil - step by step......
By similar triangles,
