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Homework Help: Use chain rule to find rate of area increase

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data
    32) A stone dropped into a pond at time t=0 seconds causes a circular ripple that ravels out from the point of impact at 5 metres per second. At what rate (in square metres per second) is the area within the circle increasing when t=10?

    2. Relevant equations
    I need to use the chain rule dy/dx = dy/du x du/dx

    3. The attempt at a solution
    The area of a circle is A=∏r2
    Differentiating this formula will tell me the rate at which the area increases for a specific radius

    The rate at which the radius increases is 5 metres per second, so I think that would be expressed as

    I need to find the rate of change of the area of the circle at 10 seconds. Using this chain rule, I think this would be found with
    dA/dt=dA/dr x dr/dt
    where dA/dt is the derivative of area as a function of time. So multiplying these derivatives gives me
    dA/dt=dA/dr x dr/dt
    The radius at 10 seconds by multiplying 5m/s x 10s =50m. So I substitute 50m in for r.

    I'm struggling a bit in my calculus class and I'm a bit unsure about all of this. Am I going in the right direction with my solution attempt above?

    Also I noticed on these forums that people are entering equations and formulas in "fancy text", if I may call it that. What I mean is making math stuff look like what one would see in a textbook, as opposed to using unicode characters like I did above. How can I do that?
  2. jcsd
  3. Mar 28, 2013 #2
    ## A = \pi r^2 ## and ## r = vt ##, so $$ \frac {dA} {dt} = \frac {dA} {dr} \frac {dr} {dt} = 2 \pi r v = 2 \pi v^2 t $$

    You did well till this point. But you made a mistake in arithmetic then.

    To see how to write in the "fancy text" (which is properly called LaTeX), hit the Quote button on my message.
  4. Mar 28, 2013 #3
    How did you get ##r=vt##?
  5. Mar 28, 2013 #4
    I got that exactly the same way you got "The radius at 10 seconds by multiplying 5m/s x 10s =50m. So I substitute 50m in for r." I just used letters instead of numbers.
  6. Mar 28, 2013 #5
    Ah of course. ##v## for velocity and ##t## for time.

    I'm still a bit confused about what you did after though. What does ##dr / dt## equal?
  7. Mar 28, 2013 #6
    $$ \frac {dr} {dt} = \frac {d(vt)} {dt} = ? $$
  8. Mar 29, 2013 #7
    I am so confused by all of this T_T

    Okay so...if I am understanding this right ##dr/dt## is the rate at which the radius changes with time. If the radius is changing at ##5m/s##, does that mean that ##dr/dt=r/t##?
  9. Mar 29, 2013 #8
    Wait I think I get it.
    ##\frac {dr}{dt}= \frac {5m}{s}= \frac {r}{t}= \frac {vt}{t}= v##
  10. Mar 29, 2013 #9
    Why are my fractions so tiny?
  11. Mar 29, 2013 #10
    If you agree that ## r = vt ##, then you obtain the derivative by just following the rules.

    If you do not know what ## r ## really is, you need to make some assumption. The assumption in this problem is that the wave propagation speed is constant, which brings you back to ## r = vt ##.
  12. Mar 29, 2013 #11
    ## foo ## makes inline expressions (small). $$ bar $$ makes "display" expressions (big).
  13. Mar 29, 2013 #12
    $$\frac {dA}{dr}=2πr$$
    $$\frac {dr}{dt}=v$$
    $$\frac {dA}{dt}=2πr×v$$
    $$=2π \frac {5m}{s}×10s× \frac {5m}{s}$$
    $$=2π×50m× \frac {5m}{s}$$
    $$= \frac{1570.796m^2}{s}$$

    Is this correct?
    Oh I messed something up there with my formatting.
    Ok fixed my formatting.
    Last edited: Mar 29, 2013
  14. Mar 29, 2013 #13
    ##π∏## Just testing different pi symbols
  15. Mar 29, 2013 #14
    Your answer is correct. Good job!
  16. Mar 29, 2013 #15
    Thank you so much for helping me Voko :)
  17. Mar 29, 2013 #16
    One more thing. You can have multiline math text in the display mode:
    line \ 1
    line \ 2
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