Question

77. the volume of a cube is increasing at a rate of
`10 \mathrm{~cm}^(3) / \mathrm{min}` . how fast is the surface area increasing when the length of an edge is
`30 \mathrm{~cm} ?`

Answer 1

`\displaystyle (4)/(3)\text{cm}^2/\text{min}`

Explanation:

Given

`\displaystyle (dV)/(dt)=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\(d(SA))/(dt)=?}\:;s=30\text{cm}`

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

`\displaystyle V=s^3\\\\(dV)/(dt)=3s^2(ds)/(dt)\\ \\10=3(30)^2(ds)/(dt)\\ \\10=3(900)(ds)/(dt)\\\\10=2700(ds)/(dt)\\\\(10)/(2700)=(ds)/(dt)\\\\(ds)/(dt)=(1)/(270)\text{cm}/\text{min}`

(2) Find the rate of the cube's surface area with respect to time at s=30:

`\displaystyle SA=6s^2\\\\(d(SA))/(dt)=12s(ds)/(dt)\\ \\(d(SA))/(dt)=12(30)\biggr((1)/(270)\biggr)\\\\(d(SA))/(dt)=(360)/(270)\biggr\\\\(d(SA))/(dt)=(4)/(3)\text{cm}^2/\text{min}`

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of
`\displaystyle (4)/(3)\text{cm}^2/\text{min}`.