Calculating Related Rates of a Metal Cube

[max0005]

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.

Homework Equations

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? :(

 

[Mark44]

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.

Homework Equations

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?

 

[max0005]

Dear Mark44, thanks for your answer.

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]

Dear Mark44, thanks for your answer.

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.

 

[max0005]

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]

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

 

[max0005]

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

 

[Mark44]

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.

 

[max0005]

Ok, getting that too! :D

Thank you very much! :D

 

[max0005]

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) [tex]\frac{dF}{dt} = 0.05km/hr... Is this correct?

 

[Mark44]

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:

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

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?