Question

find the rate of change of volume of a cone if dr/dt is 3 in./min. and h=4r when r = 8 inches

Answer 1

Answer:

$v = \frac{1}{3}bh$

since base is pi r^2

$v = \frac{1}{3} \pi \: r {}^{2} h$

it's given that h=4r

$v = \frac{1}{3} \pi \: r^{2} (4r) = \frac{4}{3} \pi \: {r}^{3}$

now find derivative

$\frac{dv}{dt} = 4\pi \: r {}^{2}$ × dr/dt

r=8 , dv/dt = 3

dv/dt = 4pi (8)^2 ×3 = 768pi

Answer 2

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