Respuesta :
The rate at which the height of water tank changing is [tex]0.4244 \; \text {ft./hr}[/tex]
Step-by-step explanation:
Given information:
[tex]\text{\bold{The diameter of cone}}=10\;\rm{ ft.}[/tex]
[tex]\rm{The\; height\; of\; the\; cone}= 15\;\rm{ ft.}[/tex]
[tex]\text{The rate at which the water is leaking }= 12 \; \rm ft^3/hr.[/tex]
[tex]\text{Volume of water tank}=27 . \pi \; \rm ft^3[/tex]
Now,
If ,we take a right circular cone with radius r and height h , its volume
[tex]\text{volume (V)} = (1/3) \times \pi \times \rm{r^3 \times h[/tex]
Now, the radius of cone when the volume of water tank is given
[tex]\frac{1}{3}\times \pi \rm\times{r^2 \times h = 27 \times \pi[/tex]
The ratio of height to radius of the cone :
[tex]\text {h/r} =3[/tex]
[tex]\text r = \text h/3[/tex]
Now calculate :
[tex]dV/dt= dV/dh \times dh/dt[/tex]
Which gives:
[tex]12 = \frac{d(\text h^3/27 \pi)}{dh} \times \frac{d\text h}{dt} \\\\\frac{d\text h}{dt} = \frac{12}{\pi \text h^2/9}[/tex]
As ,
[tex]27=\frac{1}{27 \times h}[/tex]
[tex]\text h^3=27^2\\\text h= 9\\[/tex]
Now,
[tex]dh/dt=12/(\pi.\text h^2/9)\\dh/dt=4/3\pi\\dh/dt=0.4244\\[/tex]
Hence, The rate at which the height of water tank changing is [tex]0.4244 \; \text {ft./hr}[/tex]
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