# Math Help - Related Rates

1. ## Related Rates

I am having a problem understanding how the chain rule was used on $\frac{d(x^2)}{dt}$ to arrive at $2x \cdot \frac{dx}{dt}$ in the question below (sections highlighted in red)

Question: The edge of an expanding square is changing at the rate of 2 cm/s. Determine the rate of change of its area at the instant when its edge is 6cm long.

Here is the working:

$A = x^2$ (state the problem mathematically)

$\frac{dA}{dt}=\frac{d(x^2)}{dt}$ (differentiate with respect to time)

$\frac{dA}{dt}=2x \cdot \frac{dx}{dt}$ (chain rule - I don't understand how this was arrived at $2x \cdot \frac{dx}{dt}$ )

Use x = 6cm and $\frac{dx}{dt}$ =2cm/s (and I don't understand why dx/dt =2. In other words, from reading the question above, how would I know that dx/dt=2?)

$\frac{dA}{dt}=2(6 cm) \cdot (2cm/s)$

$\frac{dA}{dt}=24cm^2/2$

Answer: The area of the square is changing at the rate of $24cm/s^2$ at the instant when its side is 6 cm long

2. x represents the side length of the square ... the problem states that the edge (side) is expanding at a rate of 2 cm/sec ... x is changing over time and $\dfrac{dx}{dt}$ represents that rate of change ... it's symbology can be read as "the rate that x changes w/r to time".

since x is an implicit function of time, then $\dfrac{d}{dt}(x) = \dfrac{dx}{dt}$, and because of the chain rule $\dfrac{d}{dt}(x^2) = 2(x)^1 \cdot \dfrac{dx}{dt}$

I recommend you review your previous lessons in finding implicit derivatives. The mechanics of taking derivatives are the same in this situation.

3. Thank you Skeeter

4. $A = x^2$

$\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dt}$ Using the chain rule. Can you see why?

And, we can work out that $\frac{dA}{dx}=2x$

So $\frac{dA}{dt}=2x\times\frac{dx}{dt}$

Edit: This post was moved from a duplicate thread by a moderator.