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How quickly is the mass of the tree increasing?

  • Thread starter mvgio124
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  • #1
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A tree trunk is approximated by a circular cylinder of height 50 meters and diameter 4 meters. The tree is growing taller at a rate of 4 meters per year and the diameter is increasing at a rate of 5 cm per year. The density of the wood is 4000 Kg per cubic meter. How quickly is the mass of the tree increasing?
 

Answers and Replies

  • #2
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Mass = Density*Volume. [tex]\frac{d(Mass(t))}{dt}[/tex] = Density*[tex]\frac{d(Volume(t))}{dt}[/tex]

Does that help?
 
  • #3
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yes it does but my webwork homework site keeps saying it got it wrong. wouldnt d/dt volume be 4(4pi)+50pi(5/200)^(2)? i dont know where im going wrong in the calculation
 
  • #4
631
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yes it does but my webwork homework site keeps saying it got it wrong. wouldnt d/dt volume be 4(4pi)+50pi(5/200)^(2)? i dont know where im going wrong in the calculation
Yes the second term there looks wrong. Remember, Volume is [tex]\pi[/tex]*r2*h.
 
  • #5
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well im completely stumped now. i have no clue what numbers to input. wouldnt d/dt of volume be (d/dheight)(pi*r^(2))+height(d/dr radius)pi? how do u input those numbers correctly based on the info?
 
  • #6
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Take a breather :smile:

[tex]\frac{d(r*r)}{dt}[/tex] = 2*r*[tex]\frac{dr}{dt}[/tex]. So that (5/200)2 you plugged in, is wrong.
 
Last edited:
  • #7
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well diameters rate of change is 5 cm or 5/100 half of that is the radius' rate of change 5/200 or 1/40. so b/c of the formula u just gave d/dt r^(2) would be 2*(1/40)
 
  • #8
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Yes. (There was a typo in my post(corrected now)). So it would really be 2*2*5/200 instead of (5/200)2
 
  • #9
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ok well i do the entire d/dt volume and get 4(4pi)+50pi(2/40)and which come to 16pi+100/40pi. that comes to d/dt volume is 37pi/2 or 58.119. i then multiplied that by density 4000 and got 232476. but when i submit that answer it keeps saying that it is incorrect
 
  • #10
631
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It's not 2/40, it's 4/40. You should be more careful with your calculations..
 

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