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
given: \(\Large{dr\over dt}=0.02\\r(t=0)=36\) Find V(t) for sphere: \(V = {4\over3}\pi r^3\)
differentiate V with respect to "t"
would it be answer b @electrokid ?
did you differentiate?
I believe so
ok, lets see it
I am sorry I meant choice d but I think I am doing it wrong, can you plaease explain it more @electrokid
yes. enter the derivative that you got here using the equation button.
\[V=\frac{ 4 }{ 3 }\pi(r+\Delta rt)^{3}\]
@electrokid
no.
well, i may have confused you. It is a little bit easier than that
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 \[r(t)=0.02t+36\]
Okay what is the next step?
now, \[V(t)={4\over3}\pi r^3(t)\\ \boxed{V(t)={4\over3}\pi(0.02t+36)^3}\]
so is the answer D?
D and B have the same functoin for V(t) did you plug-in the value of "t" and find "V"?
yeah is it 1928838.459728091213476?
good. then thats your answer.
I did not calculate. plug in "t=360" in the function V(t)
so 1928838.459728091213476(360)? or \[\frac{ 4 }{ 3 }\Pi(0.02(360)+36)^3\]?
yep
which one?
I do not know. solve it. :)
well the answers I got do not match wat the answer should be...
694381845.5 and 3333032.9
nope. it comes out as 3.3771*10^5
Oh okay how did you get that?
What did I do wrong?
putting the numbers in the calculator, i guess!!
\[ {4\over3}\pi[0.02(360)+36]^3 \]
Ohhhh okay lol Thank you so much for all of your help!
yw.
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