Question

Suppose you have a cube of length s. The volume of that cube is V = s^3. Now let's suppose the dimensions of that cube (and hence its volume) depend on time. We are wondering about the relationship between the growth of the length versus the growth of the volume. Suppose
s(t) = t.

Then

s'(t) = t______________ and V'(t) = t ___________

Next suppose V(t) = t.

s'(t) = t______________ and V'(t) = t____________

Answer 1

Answer:

Part 1

s'(t)=1

$V'(t)=3t^2$

Part 2

$s'(t)=\frac{1}{3\sqrt[3]{t^2}}$

V'(t)=1

Step-by-step explanation:

As stated in the question, if s is the length of a cube, its volume is

$V(s)=s^3$

When s changes in time, the volume will change too. The challenge here is to find the growth of the volume in terms of those changes in length

Part 1

$s(t)=t$

The change of s with respect to time is found by differentiating the relation to get

$s'=1$

The volume will also change, and its derivative is

$V'(s)=3s^2.s'$

Since s'=1

$V'(s)=3s^2$

How s=t

$V'(t)=3t^2$

Part 2

$V(t)=t$

Replacing this into the formula for V(s)

$V(s)=s^3=t$

So we have

$s=\sqrt[3]{t}=t^{\frac{1}{3}}$

Computing the derivatives:

V(t)=t =>

V'(t)=1

$s'(t)=\frac{1}{3}t^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{t^2}}$