Question

How do I do question 9 & 11? Please help thank you!

Answer 1

9.

Given

• sphere with radius 15 cm

Find

• linear approximation to the volume when the radius increases 0.4 cm

Solution

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

11.

Given

• an x by x by 2x cuboid with surface area 129.6 cm²
• rate of change of x is 0.01 cm/s

Find

• rate of change of volume

Solution

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