<h2>9.</h2><h3>Given</h3>
<h3>Find</h3>
- linear approximation to the volume when the radius increases 0.4 cm
<h3>Solution</h3>
The equation for volume of a sphere is
... V = (4/3)π·r³
Differentiating gives
... dV = 4π·r²·dr
Filling in the given numbers gives
... change in volume ≈ 4π·(15 cm)²·(0.4 cm)
... = 360π cm³ ≈ 1130.97 cm³ . . . . . . volume of layer 4mm thick
<h2>11.</h2><h3>Given</h3>
- an x by x by 2x cuboid with surface area 129.6 cm²
- rate of change of x is 0.01 cm/s
<h3>Find</h3>
<h3>Solution</h3>
The area is that of two cubes of dimension x joined together. The area of each such cube is 6x², but the two joined faces don't count in the external surface area. Thus the area of the cuboid is 10x².
The volume of the cuboid is that of two cubes joined, so is 2x³. Then the rate of change of volume is
... dV/dt = (d/dt)(2x³) = 6x²·dx/dt
We know x² = A/10, where A is the area of the cuboid, so the rate of change of volume is ...
... dV/dt = (6/10)A·dx/dt = 0.6·(129.6 cm²)(0.01 cm/s)
... dV/dt = 0.7776 cm³/s
