Question

Calculus
The following formula accurately models the relationship between the size of a certain type of a tumor and the amount of time that it has been growing:
V(t)=400(1-e^ -.0024t)^3

where t is in months and V(t) is measured in cubic centimeters. Calculate the rate of change of tumor volume at 100 months.

Answer 1

have you differentiated yet?

Answer 2

@mthompson440 are you there?

Answer 3

hi

Answer 4

well, I have tried to differentiate but I must not be doing it right.
V'(100)= 3* (-.0024)* (1-e^ -.24)^2 but that is not correct

Answer 5

\[V=400(1-e^{-0.0024t})^3\]
this right for V?

Answer 6

yes

Answer 7

I'm going to start on the most outside
I do notice your 400 constant multiple is missing right off the bat
and you have changed -.0024 to -.24
but let's go through it slowly

Answer 8

oh you plug in a 100 :p

Answer 9

yes )

Answer 10

\[V'=400 \cdot 3(1-e^{-0.0024 t})^2 \cdot (1-e^{-0.0024t})'\]
so here I brought down constant multiply of 400
and applied chain rule to the ( )^3 part

Answer 11

now we nee to also different (1-exp(-0.0024t)

Answer 12

\[\frac{d}{dt}(1-e^{-0.0024 t})=\frac{d}{dt}(1)-\frac{d}{dt}e^{-0.0024t}=0-\frac{d}{dt}e^{-0.0024t } \\ -\frac{d}{dt}e^{-0.0024t}=?\]
for this part you need chain rule also

Answer 13

I thought that the 400 being a constant = 0 ?

Answer 14

400 is a constant
but it is a constant multiple
it does not stand alone

Answer 15

\[\frac{d}{dx}400=0 \\ \text{ but } \frac{d}{dx}400x=400\]

Answer 16

= e^.0024t
or e^.0024(100)
or e^.24

Answer 17

can you differentiated what I asked you to differentiate above?

Answer 18

\[\frac{d}{dx}e^{ax} =(ax)'e^{ax}=ae^{ax}\]

Answer 19

so
\[-\frac{d}{dt} e^{-0.0024 t}=?\]

Answer 20

the two negative become positive
- (-e^0.0024t)

Answer 21

yes but you are still forgetting something

Answer 22

\[-\frac{d}{dt} e^{-0.0024t}=-(-0.0024t)'e^{-0.0024t}=-(-0.0024)e^{-0.0024t}=0.0024e^{-0.0024t}\]

Answer 23

-
.0024e^-.0024t

Answer 24

\[V'=400 \cdot 3(1-e^{-0.0024 t})^2 \cdot (1-e^{-0.0024t})' \\ V'=1200(1-e^{-0.0024t})^2 \cdot 0.0024 e^{-0.0024 t}\]

Answer 25

now we can replace t with 100

Answer 26

we can do one more thing before do that

Answer 27

\[V'=2.88(1-e^{-0.0024t})^2 e^{-0.0024 t}\]

Answer 28

any questions?

Answer 29

Thank you so much. I will have to sit here and digest this. But it all looks good. I would not have figured this out. Thank you so much

Answer 30

do you understand about the constant multiple part?

Answer 31

\[\frac{d}{dx}(c \cdot f(x)) = c \cdot \frac{d}{dx} f(x) \text{ this constant multiple rule }\]

Answer 32

you could also use product rule to come up with that

Answer 33

\[\frac{d}{dx} (c \cdot f(x)) \\ =f (x) \cdot \frac{d}{dx} c+c \cdot \frac{d}{dx} f(x) \\ =f(x) \cdot 0+ c \cdot \frac{d}{dx}f(x) \\ =0+c \cdot \frac{d}{dx}f(x) \\ =c \cdot \frac{d}{dx}f(x)\]

Answer 34

400 being the c here

Answer 35

of course we could use the rule again in this same function because we have a bit of chain rule going on