# Related Rates of a metal cube

## Homework Statement

Dear All,

I am having problems understanding how to deal with related rates. The problem is the following:

A solid 400gm metal cube of size length 10cm expands uniformly when heated. If the length of its side expand at 0.5cm(hr), find the rate at which, after 5 hours, its volume will increase.

## The Attempt at a Solution

What I need to know is how the Volume will increase after five hours. Hence:

$$\frac{dV}{dt}$$

As

$$V=S^3$$

(V = Volume, S = Side, T = Time)

Then

$$\frac{dV}{dt} = 3L^2\frac{dS}{dt}$$

Which is equal to

$$3L^2*0.5$$

As to find the length of the sides:

$$f(S) = 10+0.5(T)$$

Where T is time in hours. However, by substituting values I don't obtain the correct answer. Any help? :(

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Mark44
Mentor

## Homework Statement

Dear All,

I am having problems understanding how to deal with related rates. The problem is the following:

A solid 400gm metal cube of size length 10cm expands uniformly when heated. If the length of its side expand at 0.5cm(hr), find the rate at which, after 5 hours, its volume will increase.
Kind of a nit, but the expansion rate is 0.5 cm per hour, or 0.5 cm/hr.

## The Attempt at a Solution

What I need to know is how the Volume will increase after five hours. Hence:

$$\frac{dV}{dt}$$

As

$$V=S^3$$

(V = Volume, S = Side, T = Time)

Then

$$\frac{dV}{dt} = 3L^2\frac{dS}{dt}$$

Which is equal to

$$3L^2*0.5$$

As to find the length of the sides:

$$f(S) = 10+0.5(T)$$

Where T is time in hours. However, by substituting values I don't obtain the correct answer. Any help? :(
You need to evaluate dV/dt at time t = 5 hours.

This equation is not very helpful:
$$f(S) = 10+0.5(T)$$
You have f as a function of S, but S doesn't appear in the formula. The length really is a function of time, so L(t) = 10 + 0.5t.

What is the length of each side at t = 5 hours?

I'm sorry, I always mistake terminology and syntax. What I meant is that the length of the sides at time T is 10 (original size) + 0.5 (expansion/hour) * T (number of hours): 10+0.5(T)... Is this correct?

Mark44
Mentor

I'm sorry, I always mistake terminology and syntax. What I meant is that the length of the sides at time T is 10 (original size) + 0.5 (expansion/hour) * T (number of hours): 10+0.5(T)... Is this correct?
Yes.

So L(t) = 10 + 0.5t gives the length of a side at time t hours.

Note that t (lower case) is usually used for time, while T (upper case) is usually used for temperature.

So L(5) is 12.5. Therefore I should now substitute values in the equation for $$\frac{dV}{dt}$$, but doing so returns a wrong answer according to my textbook...

Mark44
Mentor
What do you get for dV/dt at t = 5, and what does the textbook give as the answer?

I get about 234, while the book gives 37.5 ..

Mark44
Mentor
That's what I get (i.e., your value). Are you sure you have copied the problem correctly? Are you sure that the expansion rate figure is correct? It seems pretty high to me.

Ok, getting that too! :D

Thank you very much! :D

I'd have another question, always on the same type of problem.

A girl approaches a tower 75m high at 5km/hr. At what rate is the distance from the top of the tower changing when she is 50m from the foot of the tower?

What is given is:

$$\frac{dX}{dt} = 5km/hr$$ )

Where X is the horizontal displacement of the girl. What I need to find is $$\frac{dF}{dt}$$, where F is the distance from the girl to the top of the tower.

Hence:

$$\frac{dF}{dt} = \frac{dX}{dt} * \frac{dF}{dX}$$

What remains to be found is $$\frac{dF}{dX}$$:

$$\frac{dF}{dX} = \frac{d}{dX}(x^2+75^)$$ (Pythagora)

Returning

$$\frac{dF}{dX} = 2x$$

Hence

$$\frac{dF}{dt} = 5km/hr * 2x$$ and, for x=50 (or 0.05km) $$\frac{dF}{dt} = 0.05km/hr... Is this correct? Mark44 Mentor I'd have another question, always on the same type of problem. A girl approaches a tower 75m high at 5km/hr. At what rate is the distance from the top of the tower changing when she is 50m from the foot of the tower? What is given is: [tex]\frac{dX}{dt} = 5km/hr$$
To keep the books straight, dx/dt = - 5 km/hr. She's getting closer to the tower, so her horizontal distance is decreasing, which means that its derivative is negative.
Where X is the horizontal displacement of the girl. What I need to find is $$\frac{dF}{dt}$$, where F is the distance from the girl to the top of the tower.

Hence:

$$\frac{dF}{dt} = \frac{dX}{dt} * \frac{dF}{dX}$$

What remains to be found is $$\frac{dF}{dX}$$:

$$\frac{dF}{dX} = \frac{d}{dX}(x^2+75^)$$ (Pythagora)
There's a typo above - I know you intended for there to be an exponent on 75, but you forgot to include it.
$$\frac{dF}{dX} = \frac{d}{dX}(x^2+75^2)$$ (Pythagoras)

Technically, F as you are using it doesn't represent the distance between her and the top of the tower. It's the square of the distance, so F = D2.

The relationship between the variables is x2 + 752 = D2, so the relationship between the rates is 2x*dx/dt = 2D * dD/dt. You can solve for dD/dt in this equation.
Returning

$$\frac{dF}{dX} = 2x$$
Try to be consistent in your use of upper and lower case letters. Here you are using X and x. In some problems, both cases are used, and if you aren't careful to distinguish between them, you'll definitely get confused.
Hence

$$\frac{dF}{dt} = 5km/hr * 2x$$ and, for x=50 (or 0.05km) [tex]\frac{dF}{dt} = 0.05km/hr... Is this correct?