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Homework Help: Related Rates (rectangular prism)

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A rectangular prism has its length increasing by 12 cm/min, its width increasing by 4 cm/min and its height increasing by 2 cm/min. How fast is it's volume changing when the dimensions are 200 cm in length, 50 cm in width and 30 cm in height?


    2. Relevant equations



    3. The attempt at a solution

    I am reworking a few things for an online calculus course that I am about to finish. This questions seems to have me baffled though. I have worked out the equations to arrive at dv/dt, however when I try to complete ( by attempting to arrive at (t) or use the quadratic equations the values I arrive at are always negative?

    Here we go,

    dL/dt= 12cm/min dW/dt=4cm/min dH/dt=3cm/min

    How fast is the change occuring when∶ L=200 W=50 H=30

    V=LWH

    V=(200+12t)(50+4t)(30+3t)

    dv/dt=10,000+800t+600t+48t^2 (30+3t)

    =300000+20000t+24000t+1600t+18000t+1200t^2+1440t^2+96t^3

    = 96t^3+ 2648t^2+63600t+300000

    =2〖(96t)〗^2+2(2648t)+63600

    =192t^2+5296t+63600

    ∴at t=2cm/min

    This is where I get lost, I have tried using v(t)=x(t)y(t)z(t) and keep running into (-) values.Hopefully someone can shed a little light on this one for me.
     
  2. jcsd
  3. Aug 7, 2008 #2
    Aren't you trying to find the volume when t = 0? So wouldn't that mean the volume is changing at a rate of 63600 cm^3/min? (Assuming your calculations are correct)
     
  4. Aug 7, 2008 #3
    Relevant Equations:

    V (prism) = Bh = LWH

    How to Tackle The Problem:

    Whenever starting any related rates problem, always 1. )STATE your givens.

    You already did it--but here's to summarize:

    dL/dt = 12 cm/min
    dw/dt = 4 cm/min
    dh/dt = 2 cm/min

    W = 50
    L = 200
    H = 30

    Next 2.) Find out what it's ASKING.


    "How fast is it's volume changing"

    Ding ding ding! Basically, in English, it's asking you to solve for dV/dt.

    Since V = lwh

    Find dV/dt ... don't do any substitutions yet. After you've solved for dV/dt THEN 3.) substitute your givens in the equation.
    I got an answer of 62,000 cm^3 / min . Not too far from moemoney's answer.
     
  5. Aug 7, 2008 #4
    carlodelmundo's is correct and is far simpler.

    You just have to remember the multiplication identity for taking derivatives of the product of 3 variables. This will save you time rather than having to expand everything out.

    Y = abc ==> Y' = a'bc + ab'c + abc' if I remember correctly.
     
  6. Aug 7, 2008 #5
    Yep thats correct, just an extension of the product rule.
     
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