Question

3.4 derivatives as rates of change - da1: problem 14
(1 point)
results for this submission
1 of the answers is not correct.
(a) sodium chlorate crystals are easy to grow in the shape of cubes by allowing the solution of water and sodium chlorate to evaporate slowly. if v is the volume of such a cube with side length x, calculate dv/dx when x = 3 mm. explain what this means.
(b) find the rate of change of the volume of the cube with respect to its edge length when the surface area of the cube is 16 square mm.
(a)v(3) = 27 cubic mm/ mm
(b)v(?) = 1 cubic mm/mm
note: you can earn partial credit on this problem.

Explanation:

Step1: Find the volume formula of a cube

The volume \(V\) of a cube with side length \(x\) is \(V = x^{3}\).

Step2: Differentiate the volume formula

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we differentiate \(V\) with respect to \(x\). So \(V^\prime=\frac{dV}{dx}=3x^{2}\).

Step3: Relate surface area to side - length

The surface area \(S\) of a cube is \(S = 6x^{2}\). Given \(S = 16\), then \(6x^{2}=16\), so \(x^{2}=\frac{16}{6}=\frac{8}{3}\), and \(x=\sqrt{\frac{8}{3}}\).

Step4: Calculate the derivative at the relevant \(x\) - value

Substitute \(x=\sqrt{\frac{8}{3}}\) into \(V^\prime = 3x^{2}\). We get \(V^\prime=3\times\frac{8}{3}=8\).

Answer:

\(8\) cubic mm/mm