Expert: Paul Klarreich - 2/29/2008

Question
Slag left over from the manufacture of cement is being piled outside a cement plant. The resulting slag heap is a cone. The side of the cone make an angle of 40degrees with the horizontal. Environmentalists measure the circumference of the heap one day, finding it to be 3000ft, and increasing at 10 ft/day. About how fast is the cement plant generating slag?

Answer
Questioner:   cameron
Category:  Calculus
Private:  No
 
Subject:  calc
Question:  Sand is being piled onto a cone. The sides of the cone makes an angle of 40 degrees with the horizontal. The circumference one day is 3000ft, and increasing at 10 ft/day. About how fast is the volume of sand increasing?
...............................
Hi, Cameron,

You will note I took the liberty of correcting some of the wording and eliminating excess verbosity, but the problem is exactly the same.  You should try that some time.

[Be sure to Browse the Past Answers, looking for Related Rates as the subject line. Yuo can see I fixed yours, too.  There are probably 50-100 by now.]

Your variables:

r = radius of cone.
h = height of cone.
V = Volume .......
c = circumference of base of cone.

Rates:
dr/dt not known.
dh/dt not known.
dV/dt TO BE FOUND.
dc/dt = 10 ft/day

Relations:

V = (1/3) pi r^2 h

c = 2 pi r

h = r tan 40   (see diagram)

    /|\  
   / | \  
  /  |  \  
 /  h|   \  
/    |    \  
/40   |     \  
------+------
  r

We want to express V in terms of c alone, since dc/dt is given.

r = c/2pi

h = (c tan 40)/2pi

V = (1/3) pi (c/2pi)^2 (c tan 40)/2pi

V = (1/3) pi (c^3 tan 40)/8pi^3

V = (c^3 tan 40)/24pi^2

OK, now differentiate:

dV/dt = 3c^2 tan 40/24pi^2 dc/dt

dV/dt = c^2 dc/dt tan 40/8pi^2

Now substitute values:

c = 3000,  dc/dt = 10:

dV/dt = (3000)^2 (10) tan 40/8pi^2

Just use your calculator.  [It's a lot of sand, or whatever.  No wonder the planet is a mess.]