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Related rates question conical reservoir

  • Thread starter shanshan
  • Start date
  • #1
24
0

Homework Statement


A conical reservoir is 50m deep and 200m across the top. Water is being pumped in at a rate of 7000m^3/min, and at the same time, water is being drained out at a rate of 9000 m^3/min. What is the change in diameter when the height of the water in the reservoir is 20m?


Homework Equations


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


The Attempt at a Solution


h=(.5r)
v=(1/3)(pi)(r^2)(.5r)
=(1/6)(pi)(r^3)
v'=(.5)(pi)(r^2)(r')

7000=.5pi(40^2)(r')
= 2.785
-9000= .5pi(40^2)(r')
=-3.581

2.785-3.581 = -0.796 m/min

my answer, however, is supposed to be -1.6m/min
 

Answers and Replies

  • #2
cronxeh
Gold Member
961
10

Homework Statement


A conical reservoir is 50m deep and 200m across the top. Water is being pumped in at a rate of 7000m^3/min, and at the same time, water is being drained out at a rate of 9000 m^3/min. What is the change in diameter when the height of the water in the reservoir is 20m?


Homework Equations


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


The Attempt at a Solution


h=(.5r)
v=(1/3)(pi)(r^2)(.5r)
=(1/6)(pi)(r^3)
v'=(.5)(pi)(r^2)(r')

7000=.5pi(40^2)(r')
= 2.785
-9000= .5pi(40^2)(r')
=-3.581

2.785-3.581 = -0.796 m/min

my answer, however, is supposed to be -1.6m/min
What is (40^2) ?
 
  • #3
24
0
I got the 40 from h=.5r and r=20
 
  • #4
cronxeh
Gold Member
961
10
I got the 40 from h=.5r and r=20
Well then, very well I suppose.

Your question is to find change in diameter, hence multiply it by 2 and you get your -1.592 ~ -1.6
 
  • #5
24
0
AHHH silly mistake. thankyou!!
 

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