Question

A tumor is a collection of cancerous cells. Suppose each cancerous cell has a radius of 5 times 10^{-3} cm, and that the doubling time for a population of cancerous cells is 50 hours.


If the tumor has an initial volume of 0.1 cm^3 (For simplicity, assume that the tumor is entirely made of cells, that there is no space between them.) Round your answer up to the next whole number.


Find a formula for V(t), the total volume of the cancerous cells, after t hours. V(t)=


How many hours will it take before the volume of the tumor reaches 1 cm^3?

Answer 1

you are given your initial volume of 0.1 cm^3, so that is your 'a'
you are given that at t = 20 hours the volume doubles to 0.2 cm^3. use this and the initial volume to find b. Then put it into a*b^t form. don't round, use the exact answer you found b to be. Meaning when you solve for it using algebra, don't plug anything into your calculator. As soon as you use your calculator you are rounding.

Answer 2

This is the reply from my instructor but I don't follow

Answer 3

if total volume/ volume of a single cell = # of cells.
Then total volume = (# of cells)* volume of a single cell
now the # of cells is an exponential equation and the volume of a single cell never changes. So, you will need to use your initial # of cells to find your 'a' and then the point of doubling to find your 'b' and then you have
total volume = (a*b^t)* volume of a single cell.

Answer 4

Can you write the equation for the number of tumor cells as a function of \(t\)?

Answer 5

note well that we've gotten two different doubling times specified here, 20 hours and 50 hours!

Answer 6

This is as far as I got on it.
4/3*pi*r^3=5.23598*10^-7
1.91*10^5=191,000

Answer 7

If the tumor has an initial volume of 0.1 cm^3, how many cancerous cells does it contain? (For simplicity, assume that the tumor is entirely made of cells, that there is no space between them.) Round your answer up to the next whole number.

The answer above is for this first part of the problem

Answer 8

I am kind of lost on how to get V(t) given the initial size and find b with a

Answer 9

190,986 cells is what I get, but what's 14 malignant cells between friends? :-)

Answer 10

So let's call that \(P_0\) for the initial population of tumor cells. Now we want to develop an equation that will give us \(P(t)\) or the population of tumor cells at time \(t\).

Answer 11

did you decide if you're going to use 20 or 50 hours as the doubling time?

Answer 12

20

Answer 13

okay, so \(t_{dbl} = 20\)

We can write our population equation as \[P(t) = P_0 (2)^{t/t_{dbl}}\]

At \(t=0\), it reduces to \[P(0) = P_0(2)^{0} = P_0\]

At \(t = t_{dbl}\), it becomes \[P(t_{dbl}) = P_0(2)^{t_{dbl}/t_{dbl}} = P_0(2)^1 = 2P_0\]

Do you agree those both square with your understanding of the scenario

Answer 14

Is double basically 2 and I would replace dbl with 2?

Answer 15

Well, replace it with 20 as our doubling time

Answer 16

\(t_{dbl}\) is just the time it takes for the population to double

Answer 17

you replace it with 20 hours if you're going with 20 hours as the doubling time

Answer 18

Ok, so we have the population equation and I have substituted 20 into the original equation for dbl

Answer 19

uh, what is the "original equation"?

Answer 20

Oh, into our population equation

Answer 21

Sorry, there is no original

Answer 22

so, the volume equation is just the population equation times the volume of a single cell, right?

Answer 23

yes, that is right

Answer 24

I have to come up with the formula to satisfy Find a formula for V(t), the total volume of the cancerous cells, after t hours. V(t)=

Answer 25

Okay, so write out that equation and evaluate it at \(t = 0\). Then set \(V(t) = 2V(0)\) and solve for \(t\)

Answer 26

That is where I am having the problem... don't know what equation to write out and what values to use

Answer 27

Whoops, reread the problem, never mind that last post!

You want to set \(V(t) = 1 \text{cm}^3\) and solve for \(t\)

Answer 28

Look, didn't we just agree that the volume equation is the size of a single cell * P(t)?
What more do you need?

Answer 29

Yes, so in order to find t, I would set V(20) = 1cm^3^t/20 (confused at this part)

Answer 30

the only place you get to use 20 is where I have written \(t_{dbl}\). It is a constant in the equation.

Answer 31

please, write out the volume equation as you understand it for me.

Answer 32

ok, V=Po(2)^t/20

Answer 33

you need some extra parentheses in there to make it say what you mean if you aren't going to format it...

V(t) = P0 (2)^(t/20)

so, V(t) at the moment of interest is 1 cm^3

1 cm^3 = P0 (2)^(t/20)

\[1 = P_0 (2)^{t/20}\]\(P_0\) is a constant you can put in once you've rearranged to solve for \(t\); do you know how to solve that equation for \(t\

Answer 34

equation for \(t\)?

Answer 35

no, I got part of the volume equation right, but not sure how to solve for t

Answer 36

don't remember your properties of logarithms, eh?

\[\frac{1}{P_0} = 2^{t/20}\]
Take the log of both sides:

\[\log\frac{1}{P_0} = \frac{t}{20}\log 2\]because \(\log u^a = a\log u\)

\[\frac{\log \frac{1}{P_0}}{{\log 2}} = \frac{t}{20}\]

Answer 37

it's funny that you should say that... I failed my last exam on logarithms (it's one of the areas I am not so good at)

Answer 38

Actually, we're going from a volume of 0.1 cm^3 to 1.0 cm^3 which is a factor of 10. Because the volume varies directly with the population, we could just find the time for the population to grow by a factor of 10.

\[P(t) = 10 P_0 = P_0 2^{t/20}\]\[10 = 2^{t/20}\]\[\log 10 = \frac{t}{20}\log 2\]\[\frac{1}{0.30103}*20 = t\]

Answer 39

thanks for staying up late to help me with this problem... sadly I have been working on it off and on for a couple hours

Answer 40

my equation above I forgot to multiply by the volume of a single cell!

the second works much better :-)

Answer 41

so t = 66.438561

Answer 42

should have been
\[1 = V_{cell} P_0 (2)^{t/20}\]where \(V_{cell}\) is the value you worked out for a single cell's volume


Yes, that's the correct value for \(t\)

Answer 43

or just 66.44

Answer 44

I never know if I should round up to 2 or 4 decimal places

Answer 45

Here's a plot, showing each doubling (1st 3 vertical lines) and then the final arrival at V=1 cm^3 at t = 66.44 hours

Answer 46

you are probably in a higher level math such as calculus or beyond

Answer 47

uh, well, I was the last time I took a math class, but that was many years ago :-)

Answer 48

I have to take precalc for my b.s. in software development

Answer 49

well, you'd probably get more use out of a discrete math class than precalc, but depending on what you'll be developing, no lasting harm should come from exposure to this material :-)

Answer 50

so now that I got t, I plug that in to get the formula for V(t)

Answer 51

I am still on that part of the question :)

Answer 52

no, you already have the formula for V(t). It is the volume of a single cell * P(t)

Answer 53

if you want a nice general formula,

\[V(t) = \frac{4}{3}\pi r^3 P_0 (2)^{t/t_{dbl}}\]where \(r\) is radius of a single cell, \(P_0\) is the initial population of cells, \(t_{dbl}\) is the doubling time

Answer 54

that is going back to this formula: \[V(t)=2^(t/20)\
]

Answer 55

oh, I was sort of on the right track

Answer 56

this problem is difficult because I am trying to wrap my head around volume of population of cancer cells... an abstract idea

Answer 57

Actually, it could just as easily be the weight of the tumor — the only change would be that we'd find the weight of an individual cell and multiply by that instead of the volume of an individual cell.

Answer 58

yes, it could be anything... so given the general formula, it is still does not accept that answer although I am not sure if I put in the equation correctly

Answer 59

Find a formula for V(t), the total volume of the cancerous cells, after t hours. V(t)=

Answer 60

I put in 4/3*pi*r^3Po(2)^t/20, maybe I need to simplify this when I put in my answer

Answer 61

you need to substitute your values in...

you're going to have \[V(t) = <number>2^{t/20}\]

another possibility is that you are supposed to be using 50 hours as the doubling time (remember, that's what it says in the initial problem statement)

Answer 62

oh, so that is what the instructor is talking about where I was given the values for initial value and t=20

Answer 63

yes

Answer 64

ok, 4/3*pi*r^3 is just the formula for volume in general

Answer 65

yes

Answer 66

but it doesn't look like I used that formula in the problem at all

Answer 67

except to find the first value

Answer 68

you used it to find the volume of a single tumor cell.

Answer 69

the volume of a single tumor cell is part of the equation

Answer 70

it is also a constant for this problem

Answer 71

ok so probably still missing something but for the formula I have V(t)=0.1^(t/20)

Answer 72

yes, you're missing the 2!

Answer 73

and the initial population

Answer 74

so the formula we are working with is Ae^kt

Answer 75

I remember some of this

Answer 76

ok, second part is correct
V(t)=0.1(2)^t/20

Answer 77

that is what they asked for in V(t)=

Answer 78

one last part to all of this
How many hours will it take before the volume of the tumor reaches 1 cm^3? Round your answer up to the next whole number.

Answer 79

so hours = t

Answer 80

We solved that already! Remember the stuff with the logarithms?

Answer 81

t is in units of hours, yes

Answer 82

log10 = t/20 log2

Answer 83

1/0.30103 * 20 = t

Answer 84

yes, go on

Answer 85

t = 66.44

Answer 86

close enough for government work...

Answer 87

so we already solved that at the same time as finding V(t)=

Answer 88

that makes things easier

Answer 89

66.44 is not going through

Answer 90

Round to the nearest whole number, t=66.438561

Answer 91

67 = t

Answer 92

no!

Answer 93

nearest whole number is not 67

Answer 94

66.438561-67 = -0.561439
66.438561-66 = 0.438561

which number is closer, 66, or 67?

Answer 95

66

Answer 96

ok, I put in 66 and still not what they are looking for... going back to the value of t

Answer 97

that answer is going to depend on what you used for the doubling time, of course.