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About Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience
Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook


 
   

You are here:  Experts > Teens > Homework/Study Tips > Calculus > Related Rates

Topic: Calculus



Expert: Abe Mantell
Date: 7/8/2008
Subject: Related Rates

Question
Hi, I am having a problem solving a related rates problem.  
The height of a rectangular box is 10 inches.  Its length increases at a rate of 2 in/sec and its width decreases at the rate of 4 in/sec.  When the length is 8inches and the width is 6 inches, the volume of the box is changing atthe rate of?  
do i want to use the relation V=lwh?
i set up:
dV/dt to be found
dw/dt = -4
dl/dt = 2
is there a dh/dt and how do i find that if i am already looking for dV/dt?
I'm lost...any help is appreciated!  Thankyou.

Answer
Hello Stephanie,

Since h is constant, dh/dt=0 and no need to treat h as a variable.
So, V=L*W*10 or V=10LW...now differentiate w.r.t. "t"...
==> dV/dt= 10(L*dW/dt + dL/dt*W), product rule since both L & W are
functions of time (t).  Now substitute L=8, W=6, dL/dt=2, dW/dt=-4
==> dV/dt=10[8*(-4) + 2*6]=-200
So the volume is decreasing at the rate of 200 in^3/min.

OK?

Abe


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