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Thread: Related Rates

  1. #1
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    Related Rates

    I am having a problem understanding how the chain rule was used on $\displaystyle \frac{d(x^2)}{dt}$ to arrive at $\displaystyle 2x \cdot \frac{dx}{dt}$ 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:

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

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

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

    Use x = 6cm and $\displaystyle \frac{dx}{dt}$ =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?)

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

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

    Answer: The area of the square is changing at the rate of $\displaystyle 24cm/s^2$ at the instant when its side is 6 cm long
    Last edited by sparky; Feb 26th 2011 at 05:23 AM.
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  2. #2
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    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 $\displaystyle \dfrac{dx}{dt}$ 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 $\displaystyle \dfrac{d}{dt}(x) = \dfrac{dx}{dt}$, and because of the chain rule $\displaystyle \dfrac{d}{dt}(x^2) = 2(x)^1 \cdot \dfrac{dx}{dt}$

    I recommend you review your previous lessons in finding implicit derivatives. The mechanics of taking derivatives are the same in this situation.
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  3. #3
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    Thank you Skeeter
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  4. #4
    Super Member Quacky's Avatar
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    $\displaystyle A = x^2$

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

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

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

    Edit: This post was moved from a duplicate thread by a moderator.
    Last edited by Quacky; Feb 26th 2011 at 01:50 PM. Reason: Moved from thread created by duplicate post.
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