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Mathematics 22 Online
OpenStudy (anonymous):

please help!!! A high-altitude spherical weather balloon expands as it rises due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.02 inches per second and that r = 36 inches at time t = 0. Determine the equation that models the volume V of the balloon at time t and find the volume when t = 360 seconds. a. V(t) = 4π(0.02t)2; 651.44 in.3 b. V(t) = 4π(36 + 0.02t)2; 1,694,397.14 in.3 c. V(t) =4pi(0.02t)^3/3 ; 4690.37 in.3 d. V(t) = 4pi(36+0.02t)^3/3; 337,706.83 in.3

OpenStudy (anonymous):

given: \(\Large{dr\over dt}=0.02\\r(t=0)=36\) Find V(t) for sphere: \(V = {4\over3}\pi r^3\)

OpenStudy (anonymous):

differentiate V with respect to "t"

OpenStudy (anonymous):

would it be answer b @electrokid ?

OpenStudy (anonymous):

did you differentiate?

OpenStudy (anonymous):

I believe so

OpenStudy (anonymous):

ok, lets see it

OpenStudy (anonymous):

I am sorry I meant choice d but I think I am doing it wrong, can you plaease explain it more @electrokid

OpenStudy (anonymous):

yes. enter the derivative that you got here using the equation button.

OpenStudy (anonymous):

\[V=\frac{ 4 }{ 3 }\pi(r+\Delta rt)^{3}\]

OpenStudy (anonymous):

@electrokid

OpenStudy (anonymous):

no.

OpenStudy (anonymous):

well, i may have confused you. It is a little bit easier than that

OpenStudy (anonymous):

what is r(t) = radius as a function of time? \[r(t)=\int dr=\int(0.02)dt=0.02t+C\] we know that when t=0, r=36\[36=0+C\] so, we have \[r(t)=0.02t+36\]

OpenStudy (anonymous):

Okay what is the next step?

OpenStudy (anonymous):

now, \[V(t)={4\over3}\pi r^3(t)\\ \boxed{V(t)={4\over3}\pi(0.02t+36)^3}\]

OpenStudy (anonymous):

so is the answer D?

OpenStudy (anonymous):

D and B have the same functoin for V(t) did you plug-in the value of "t" and find "V"?

OpenStudy (anonymous):

yeah is it 1928838.459728091213476?

OpenStudy (anonymous):

good. then thats your answer.

OpenStudy (anonymous):

I did not calculate. plug in "t=360" in the function V(t)

OpenStudy (anonymous):

so 1928838.459728091213476(360)? or \[\frac{ 4 }{ 3 }\Pi(0.02(360)+36)^3\]?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

which one?

OpenStudy (anonymous):

I do not know. solve it. :)

OpenStudy (anonymous):

well the answers I got do not match wat the answer should be...

OpenStudy (anonymous):

694381845.5 and 3333032.9

OpenStudy (anonymous):

nope. it comes out as 3.3771*10^5

OpenStudy (anonymous):

Oh okay how did you get that?

OpenStudy (anonymous):

What did I do wrong?

OpenStudy (anonymous):

putting the numbers in the calculator, i guess!!

OpenStudy (anonymous):

\[ {4\over3}\pi[0.02(360)+36]^3 \]

OpenStudy (anonymous):

Ohhhh okay lol Thank you so much for all of your help!

OpenStudy (anonymous):

yw.

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