Answer:
Water is being pored in the cone with the rate of 401.92 cm³ per second.
Step-by-step explanation:
There is a cone having height h = 10 cm and radius r = 10 cm
That means h = r
Rate of change of water level [tex]\frac{dh}{dt}=2[/tex] cm per second
Since volume of a cone is represented by the formula,
[tex]V=\frac{1}{3}\pi r^{2}h[/tex]
Now we have to find the value of [tex]\frac{dV}{dt}[/tex].
[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\pi r^{2}h)[/tex]
Since r = h
[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\pi h^{3} )[/tex]
[tex]\frac{dV}{dt}=\frac{3}{3}\pi h^{2}.\frac{dh}{dt}[/tex]
For h = 8 cm and [tex]\frac{dh}{dt}=2[/tex] cm per second.
[tex]\frac{dV}{dt}=\frac{3}{3}\pi (8)^{2}\times (2)[/tex]
[tex]\frac{dV}{dt}=128\pi[/tex] = 401.92 cubic cm per second
Therefore, water is being pored with the rate of 401.92 cm³ per sec.
At the rate of [tex]128\pi\ cm^{3}/sec[/tex] water being poured into the cup when the water level is [tex]8[/tex] cm
The above figure shows the variables
Radius of conical cup([tex]R[/tex])[tex]=10[/tex] cm
Height of the conical cup () cm
Height of water level in cup at time , cm
Radius of water at time [tex]t[/tex] [tex]=r[/tex] cm
Volume of water at time [tex]t[/tex][tex]=V^{3}[/tex]
[tex]\dfrac{\partial h}{\partial t}=2m/sec[/tex]
Triangle [tex]ABC[/tex] is similar to triangle [tex]PBQ[/tex]
[tex]\frac{r}{h} = \frac{r}{h}[/tex]
[tex]\frac{r}{h} = \frac{10}{10}[/tex]
[tex]r=h\\r=8[/tex]
find [tex]\dfrac{dV}{dt}[/tex] which is rate of water being poured in cup.
The volume of a [tex]PBQ[/tex] cone is,
[tex]\begin{aligned}Volume&= Area\;of\;circle \times h\\&=\pi r^2 h\end{aligned}[/tex]
Differentiate with respect to [tex]t[/tex] we get:
[tex]\dfrac{dV}{dt} = \pi r^{2}\frac{dh}{dt}[/tex]
[tex]\dfrac{dV}{dt} = \pi (8)^{2}\times 2[/tex]
Therefore , [tex]\dfrac{dV}{dt}=128\pi\ cm^{3}/sec[/tex]
Learn more about Volume related concepts here:
https://brainly.com/question/13338592?referrer=searchResults
