1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Related rates, clarification sought

  1. Apr 28, 2005 #1
    Given, "if a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10 cm."

    [tex]\frac{dD}{dt} = \frac{dD}{dA} \frac{dA}{dt}[/tex] were D is my diameter, A is (surface) area and t is time.

    I relate D to A by, [tex] A = 4pi(\frac{D}{2})^2 = piD^2; A' = 2piD[/tex] and then solve,

    [tex]\frac{dD}{dt} = 2piD * -1 cm^2/min =[/tex]
    [tex]2pi(10cm) * -1cm^2/min = -20 cm^3/min[/tex]

    I know I am wrong because I got volume, not length. When I solve for variables, like A, and differentiate, do i always place result where ever A shows up in the differentiation equation? In this case, dD/dA, so I should write 1/(2piD)?

    The book give the answer as [tex]\frac{1}{20 cm/min}[/tex]
  2. jcsd
  3. Apr 28, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You solved for dA/dD

    You want dD/dA
  4. Apr 28, 2005 #3
    yeah I was suspecting that. thanks.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Related rates, clarification sought
  1. Related Rates (Replies: 3)

  2. Related Rates (Replies: 2)

  3. Related Rates tank drain (Replies: 16)