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Related Rates of change (pebble/ripple problem)

  1. Jul 15, 2011 #1

    Femme_physics

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    I'm not sure whether that's considered physics, algebra or calculus, but it might include all three that I just thought to post it here.


    1. The problem statement, all variables and given/known data


    http://img824.imageshack.us/img824/924/ripples.jpg [Broken]

    A radius of a ripple is increasing at a rate of 6 inches per second

    Find how the area of the circle is changing when time = 2 sec


    3. The attempt at a solution


    http://img820.imageshack.us/img820/5668/pebble.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 15, 2011 #2

    SteamKing

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    I think the problem is asking you to calculate dA/dt when t = 2 sec. From the diagram, you also know that r = 6 inches when r = 1 s, and r = 12 inches when t = 2 s.
     
  4. Jul 15, 2011 #3

    HallsofIvy

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    The problem asked for the rate of change of the area of the circle. I see nowhere in there a formula for area.
     
  5. Jul 15, 2011 #4

    I like Serena

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    Hey Fp! :smile:

    Your formulas are for circumferences and not for areas of circles.
    Although the proper formula does have a circumference-like component in it.

    I think you need to treat this as a calculus question.
    Let's see if I can "step" you through this problem (if you want).

    You need the formula [itex]A = \pi R^2[/itex] for the area of a circle.
    And take the derivative as ST suggested.
    That is [itex]\frac {dA} {dt} = 2 \pi R \frac {dR} {dt}[/itex]
    (As you can see this formula does contain a circumference-like component in it. :wink:)

    Now you only need to fill in a value for R corresponding to t = 2 seconds, and you need to fill in the rate that the radius increases.
     
  6. Jul 15, 2011 #5

    Femme_physics

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    Yes *smacks forehead*..used the wrong formula for area! Was confusing it with perimeter.

    So this should make slightly more sense... although when u take the derivative of area u get perimeter! Hmmm!

    http://img190.imageshack.us/img190/9265/bestshot.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Jul 15, 2011 #6

    I like Serena

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    Yeah, well, taking the derivative is slightly more complex here.
    But I'm not sure you've heard of the chain-rule yet....
    Did you?

    And are you familiar with the notation [itex]\frac {dA} {dt}[/itex]?
    This denotes the derivative of the area to the time.

    The point is that you do not take the derivative to R, but you take the derivative to t.
    This means that the resulting formula has to be multiplied with the derivative of R to t, that is, the rate that R increases.
     
  8. Jul 15, 2011 #7

    Femme_physics

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    Damn, I'm more terrible in calculus than I thought! I just thought to practice it a bit as it would help me in more advanced dynamics!


    I think it goes something like "the first term times the derivative of the second minus the second term times the derivative of the first".


    Oh yes, Leibovitz notation if I'm not mistaken!

    Right! And we should take the derivative WITH RESPECT TO TIME because we're looking at how things change over time!

    That means I should have an equation with R = (something)t ?


    Hmm...I'll mull over that.
     
  9. Jul 15, 2011 #8

    I like Serena

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    Yes it would! :smile:

    Very true! Neat isn't it?


    Uhh, no. I think that is the first part of the quotient rule.

    The chain rule goes something like "the derivative of the surrounding function, multiplied by the derivative of the enclosed function".

    In symbols: (u(v))' = u'(v) . v'



    Yes!
    (That is, Leibnitz notation. :wink:)


    Exactly! I'm getting all excited here!


    Yes, I think that "(something)" is given as part of your problem....


    You do that.
     
    Last edited: Jul 15, 2011
  10. Jul 15, 2011 #9

    Mark44

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    Um, that would be Leibniz, as in Gottfried Wilhelm (von) Leibniz. The "niz" syllable of his last name is pronounced as if it were "nitz."
     
  11. Jul 15, 2011 #10
    You are given [itex]\frac{dr}{dt}[/itex]. You want [itex]\frac{dA}{dt}[/itex] so that you can solve it for your given time (radius.)

    You are right in that the derivative of [itex]\pi r^2[/itex] is [itex]2\pi r[/itex], but we need to implicitly differentiate with respect to time. So, multiply this derivative by the rate of change of radius with respect to time.

    [itex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/itex]

    If time is 2, radius is 12. The rate of area change is above, solving it with the value of the radius when t = 2 explains how area is changing when t = 2.
     
  12. Jul 15, 2011 #11

    I like Serena

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    Lol. I looked it up just before I wrote it down and got a hit on Leibnitz.
    It turns out I was redirected to Leibniz without me noticing!
     
  13. Jul 16, 2011 #12

    Femme_physics

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    Noted :)

    To my understand "implicitly differentiate" means using the chain rule. But why should we use the chain rule if there is one term (pi x R^2 ) to differentiate with respect to time?

    That's the part I'm struggling with atm.
     
  14. Jul 16, 2011 #13

    chiro

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    As other people have stated above, you are finding out rates of change with respect to another quantity: namely t.

    Think of a composition of functions: A is in terms of R and R is in terms of t. In terms of a "chain" this is basically A = f(R) and R = g(t). If g(t) was trivial (example g(t) = t), then dR/dt would be just 1 which means you could ignore dR/dt.

    If you do multivariable calculus later (if you haven't already done it), this will help you understand what is going on. Its the same sort of idea in calculating derivatives of multivariable systems, but you deal with a matrix style approach.
     
  15. Jul 16, 2011 #14

    Femme_physics

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    Aha, so I got TWO functions here to deal with here.

    A = pi R^2
    And
    6R = t

    Yes?
     
  16. Jul 16, 2011 #15

    chiro

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    Basically dR/dt = 6 so R = 6t (initial condition is t = 0, R = 0).

    This example is nice because R(t) is a simple analytic function, but you will find with later physics that your rates of change will be more complex and may not have nice analytic solutions.

    But yeah you're right with A(R) = pi * R^2, and R(t) = 6t (t is in seconds, R is in inches).
     
  17. Jul 16, 2011 #16
    Because you aren't just finding the derivative, you are finding the derivative with respect to time. Area isn't an equation that involves time, this creates a new function, "with respect to time," essentially. I'll show you why it's a special case chain rule below:

    The chain rule involves little work in this example, check it out:

    [itex]A = \pi r ^2[/itex]

    Lets differentiate both sides, with respect to time. Set it up:

    [itex]\frac{d}{dt}[A] =\frac{d}{dt}[ \pi r ^2 ][/itex]

    What is [itex]\frac{d}{dt} [A][/itex] ?

    We apply the chain rule.

    [itex]\frac{dA}{dt}[/itex] times the derivative of A, which is one.

    Remember, derivative of outer function, evaluated to the inner function, times the derivative of the inner function.

    So, we now have

    [itex]\frac{dA}{dt} =\frac{d}{dt}[ \pi r ^2 ][/itex]

    And yes, you can actually skip the chain rule and just move the A, it's just the "reason" why it works. You will have to actually apply the chain rule often, though, so remember it!

    Now we have to differentiate the right side with respect to time. To do this, find the derivative of [itex][ \pi r ^2 ][/itex] as you've successfully done, and multiply by the derivative of radius with respect to time.


    [itex]\frac{dA}{dt} =\frac{d}{dt}[ \pi r ^2 ][/itex]
    equals

    [itex]\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}[/itex]

    It's just about ready to solve. You know the radius you want to solve for, it's 12.

    Look at
    [itex]2 \pi r \frac{dr}{dt}[/itex]

    What is [itex]\frac{dr}{dt}[/itex]?

    The derivative, or rate of change of radius, with respect to time, which you are given. :)

    You want the rate of change of area, with respect to time, which is sitting by itself on the left side.

    You know that when time is 2, radius is 12. You know [itex]\frac{dr}{dt}[/itex], is just 6. You want [itex]\frac{dA}{dt}[/itex].

    SPOILER ALERT::










    [itex]\frac{dA}{dt} = 2 \pi (12) (6)[/itex][itex][/itex]
     
    Last edited: Jul 16, 2011
  18. Jul 16, 2011 #17

    Femme_physics

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    Great! Does it mean I can just plug in the R form the second function ( R(t) = 6t) to the area function, which would turn out [ A(R) = pi x (6t)^2 ]

    Would that be okay?
     
  19. Jul 16, 2011 #18
    Yes, you can do that. Albeit it's slightly unnecessary since the picture reveals that radius is 12 when time is 2. Mathematically, it's correct though.

    Now derive both sides, with respect to time, solve for t = 2, and you have your answer.
     
  20. Jul 16, 2011 #19

    Femme_physics

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    I am a bit confused when you do this notation

    [itex]\frac{d}{dt}[A] =\frac{d}{dt}[ \pi r ^2 ][/itex]

    A is just the function name. How can you differentiate it? I mean, I understand how to different the function itself, but why write dA/dt at the "A" side of things?


    Oh, can I just do...

    http://img204.imageshack.us/img204/2202/dadtf.jpg [Broken]

    Hmm but I end up with the same answer
     
    Last edited by a moderator: May 5, 2017
  21. Jul 16, 2011 #20
    No, you cannot. It is purely coincidental. We will only arrive at the same answer when t = 2. What if they asked for area behavior when t = 3? Our formulas yield the same answer because I am multiplying 2, 6, and 12, while you are squaring 12. (12)(x) = x^2 ONLY when x is 12.

    Not to get too analytical, but look at it this way, your equation shows area increasing at a rate of [itex]\pi r^2[/itex]. That IS area. If your rate of area change equals the area formula, that would mean your area was doubling every second. A circle can't double in area every second when the radius is increasing at a constant speed! Does that make sense?

    The reason why we got the same answer when t = 2, is because area in fact is doubling when t = 2, but ONLY when t = 2.
     
    Last edited: Jul 16, 2011
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