Question

A spherical balloon is being inflated so that its volume is increasing at the rate of 5m^3/min. At what rate is the Diameter increasing when the diameter is 12m?

Answer 1

i know this:
dv/dt=5m^3/min
d=12 and radius= 6m
Formula of sphere: V=4/3pi(r)^3

Answer 2

then I got this:
dr/dt= 5/144pi

Answer 3

I'm not sure how to get the rate of the diameter... please help...

Answer 4

\(d\color{red}{r}/dt\) u have found. but u need to find \(d\color{red}{d}/dt\)

as the rate of diameter is asked not radius

Answer 5

i know, I'm sure how to get dd/dt

Answer 6

\(\large\color{black}{\begin{align}
V&=\dfrac{4}{3}\pi(r)^3\hspace{.33em}\\~\\
V&=\dfrac{4}{3}\pi(\dfrac{d}{2})^3\hspace{.33em}\\~\\
V&=\dfrac{4}{3}\pi\dfrac{d^3}{8}\hspace{.33em}\\~\\
V&=\dfrac{1}{6}\pi\cdot d^3\hspace{.33em}\\~\\
\dfrac{dV}{dt}&=\dfrac{1}{6}\pi\cdot \dfrac{d'd^3}{dt}\hspace{.33em}\\~\\
5&=\dfrac{1}{6}\pi\cdot \dfrac{dd^3 }{d'd'}\dfrac{d'd'}{dt}\hspace{.33em}\\~\\
5&=\dfrac{1}{6}\pi\cdot3d^2\dfrac{d'd'}{dt}\hspace{.33em}\\~\\
5&=\dfrac{1}{2}\pi\cdot12^2\dfrac{d'd'}{dt}\hspace{.33em}\\~\\
5&=72\pi\cdot\dfrac{d'd'}{dt}\hspace{.33em}\\~\\
\dfrac{d'd'}{dt} =\dfrac{5}{72\pi}\quad m/min\hspace{.33em}\\~\\
\end{align} }\)

Answer 7

i have just used \('\quad '\) to seperate the two \(d\)

Answer 8

okay! i got it. Thank you!

Answer 9

u got \(dr/dt= 5/144\pi\)

by just multiplying it by \(2\)

u could have reached

\(2dr/dt= 5/72\pi\\
dd/dt= 5/72\pi\\
\)

Answer 10

what do you mean by multiplying?

Answer 11

@mathmath333