It has been observed that the following formula accurately models the relationship between the size of a tumor and the amount of time it has been growing.

V(t)= 1100[1023e^(-0.02415t)+1]^4

Where t is months and V(t) is in cubic centimeters.

A. Find the volume at 240 months
B. if a tumor of size 0.5 cm is detected, according to the formula, how long has it been growing? What does that imply?
C. Find lim V(t) as t reaches infinity
D. Calculate the rate of change of tumor volume at 240 months

Question

It has been observed that the following formula accurately models the relationship between the size of a tumor and the amount of time it has been growing.

V(t)= 1100[1023e^(-0.02415t)+1]^4

Where t is months and V(t) is in cubic centimeters.

A. Find the volume at 240 months
B. if a tumor of size 0.5 cm is detected, according to the formula, how long has it been growing?  What does that imply?
C.  Find lim V(t) as t reaches infinity
D. Calculate the rate of change of tumor volume at 240 months

anonymous · Answer

I got A, which is 3.85cm i believe

anonymous · Answer

cubic cm that is

mertsj · Answer

Is the time supposed to be in months?

anonymous · Answer

yes

anonymous · Answer

I think i worked through it- and got 214 months for B.  I think the limit is 1100.  The last one im not sure, but i got .28

mertsj · Answer

I can't verify your answer to A.  I get a much bigger answer.

anonymous · Answer

hm, ok ill try it again- but i cant imagine a tumor bigger than that

anonymous · Answer

It may be the formula, i can retype it if you like

mertsj · Answer

$$V(t)=1100[1023e ^{-.024215t}+1]^4$$

mertsj · Answer

Is that right?

mertsj · Answer

Gotta go

anonymous · Answer

$$V(t)= 1100[1023e ^{-.02415t}+1]^{-4}$$

anonymous · Answer

Ah, i see what I did...it should be raised to negative four, not four...sorry about that