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Answer:
[tex]v = \frac{1}{3}bh[/tex]
since base is pi r^2
[tex]v = \frac{1}{3} \pi \: r {}^{2} h[/tex]
it's given that h=4r
[tex]v = \frac{1}{3} \pi \: r^{2} (4r) = \frac{4}{3} \pi \: {r}^{3} [/tex]
now find derivative
[tex] \frac{dv}{dt} = 4\pi \: r {}^{2} [/tex] × dr/dt
r=8 , dv/dt = 3
dv/dt = 4pi (8)^2 ×3 = 768pi
Answer:
768 pi in^3/min
Step-by-step explanation:
Volume of cone=1/3 pi×r^2×h
Differentiating this gives:
dV/dt=1/3×pi×2r dr/dt×h+1/3×pi×r^2×dh/dt
We are given the following:
dr/dt = 3 in./min.
h=4r when r = 8 inches
If h=4r then dh/dt=4dr/dt=4(3 in/min)=12 in/min
If h=4r and r=8 in, then h=4(8)=32 in for that particular time.
Plug in:
dV/dt=1/3×pi×2r dr/dt×h+1/3×pi×r^2×dh/dt
dV/dt=1/3×pi×2(8)(3)×32+1/3×pi×(8)^2×12
dV/dt=pi×2(8)(32)+pi×(8)^2(4)
dV/dt=pi(256×2)+pi(64×4)
dV/dt=pi(512)+pi(256)
dV/dt=pi(768)
dV/dt=768pi
dV/dt=768/pi in^3/min