How quickly is the mass of the tree increasing?

[mvgio124]

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?

 

[Sourabh N]

Mass = Density*Volume. $$\frac{d(Mass(t))}{dt}$$ = Density*$$\frac{d(Volume(t))}{dt}$$

Does that help?

 

[mvgio124]

yes it does but my webwork homework site keeps saying it got it wrong. wouldn't d/dt volume be 4(4pi)+50pi(5/200)^(2)? i don't know where I am going wrong in the calculation

 

[Sourabh N]

yes it does but my webwork homework site keeps saying it got it wrong. wouldn't d/dt volume be 4(4pi)+50pi(5/200)^(2)? i don't know where I am going wrong in the calculation

Yes the second term there looks wrong. Remember, Volume is $$\pi$$*r²*h.

 

[mvgio124]

well I am completely stumped now. i have no clue what numbers to input. wouldn't 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?

 

[Sourabh N]

Take a breather

$$\frac{d(r*r)}{dt}$$ = 2*r*$$\frac{dr}{dt}$$. So that (5/200)² you plugged in, is wrong.

 

[mvgio124]

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)

 

[Sourabh N]

Yes. (There was a typo in my post(corrected now)). So it would really be 2*2*5/200 instead of (5/200)²

 

[mvgio124]

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

 

[Sourabh N]

It's not 2/40, it's 4/40. You should be more careful with your calculations..