Question

Air is being pumped into a spherical balloon at the rate of 7 cm3/sec. What
is the rate of change of the radius at the in stant the volume equals 36π cm3
? The volume of
the sphere of radius r is (4 over PI )r power of 3

Answer 1

Answer:

Step-by-step explanation:

$V=\frac{4}{3} \pi r^3\\\frac{dV}{dt} =\frac{4}{3} ~\pi \times ~3r^2\frac{dr}{dt} \\\frac{dV}{dt} =4\pi r^2\frac{dr}{dt} \\7=4\pi r^2\frac{dr}{dt} \\\frac{dr}{dt} =\frac{7}{4\pi r^2} \\when~V=36 \pi cm^3\\36\pi =\frac{4}{3} \pi r^3\\r^3=\frac{36\pi \times 3}{4\pi } =27=3^3\\r=3~cm\\\frac{dr}{dt} =\frac{7}{4\pi \tmes 3^2} \\\frac{dr}{dt} =\frac{7}{36 \pi } cm/sec$