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

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

Answer 2

The volume of a cube is dependent on its dimensions. The rate of volume and dimension with respect to time for given cases are:

• When $s(t) = t$:

$s'(t) = 1\\v'(t) = 3t^2$

• When $V(t) = t$:

$s'(t) = \dfrac{1}{3 \times \: ^3\sqrt{t^2}}\\\\v'(t) = 1$

How to calculate the instantaneous rate of growth of a function?

Suppose that a function is defined as;

$y = f(x)$

Then, suppose that we want to know the instantaneous rate of the growth of the function with respect to the change in x, then its instantaneous rate is given as:

$\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}$

For the given case, we have:

$V = s^3$

Getting their rate with respect to time:

• Case 1: s(t) = t

Then, we  get

$s'(t) = 1\\v(t) = s^3 = t^3\\v'(t) = \dfrac{dt^3}{dt} = 3t^2$

• Case 2: When V(t) = t

Then, we get:

$V = s^3\\s = V^{1/3}\\\\V'(t) = 1\\\\s'(t) = \dfrac{d(s(t))}{dt} = \dfrac{d(V^{1/3})}{dt} = \dfrac{d(t^{1/3})}{dt} = \dfrac{1}{3}t^{1/3 - 1} = \dfrac{1}{3} t^{-2/3} = \dfrac{1}{3 \times\: ^3\sqrt{t^2}}$

The rate of volume and dimension with respect to time for given cases are:

• When $s(t) = t$:

$s'(t) = 1\\v'(t) = 3t^2$

• When $V(t) = t$:

$s'(t) = \dfrac{1}{3 \times \: ^3\sqrt{t^2}}\\\\v'(t) = 1$

Learn more about rate of function here:

https://brainly.com/question/15125607