Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Air is pumped into a spherical balloon at a rate of 2 cm^3/s (two cubic centimetres per second). Denote by V the volume of the balloon, r the radius of the balloon, and S the surface area of the balloon. (a) Find dr/dt when V = 30 cm^3. (b) Find dS/dt when V = 30 cm^3. You may need these formulas: V = 4/3πr^3, S = 4πr^2.

OpenStudy (anonymous):

V=4/3pi r^3 Find dV/dr from here

OpenStudy (anonymous):

can you?

OpenStudy (anonymous):

but its not dV/dt that i need to find, its dr/dt. Thats what i dont understand :(

OpenStudy (campbell_st):

ok... to start, you need to find r, when V = 30 cm^3 when you know that, use the relationship \[\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\] you know DV/dt... and can find DV/dr when V = 30 once you have them, you can find dr/dt

OpenStudy (anonymous):

ummm ok. So dies dV/dt = 2 then?

OpenStudy (campbell_st):

try to keep r as an exact value... makes it easier when substituting

OpenStudy (campbell_st):

thats correct... rate of change in volume with respect to time is 2 cm^3/sec

OpenStudy (anonymous):

so where does the V=30cm^3 come in then?

OpenStudy (campbell_st):

you need to find the radius when the volume is 30cm... then you substitute that into dV/dr

OpenStudy (anonymous):

but i need to keep the answer in terms of dr/dt

OpenStudy (campbell_st):

yep.. and you need dr/dt when V = 30.... so you need to find the value of r when V = 30... so solve \[30 = \frac{4}{3} \pi r^3\] for r... this is the 1st step

OpenStudy (anonymous):

so r would equal 1.9276? would i put it as a decimal?

OpenStudy (campbell_st):

ok... now find dV/dr when \[V = \frac{4}{3} \pi r^3\]

OpenStudy (campbell_st):

I would have left r as \[r = \sqrt[3]{\frac{90}{4\pi}}\]

OpenStudy (campbell_st):

but a decimal will work fine

OpenStudy (anonymous):

how do i find DV/dr?

OpenStudy (anonymous):

do i subtitute the r value i found into the V formula?

OpenStudy (campbell_st):

differentiate the volume equation with respect to the radius...

OpenStudy (campbell_st):

do the derivative 1st... then substitute

OpenStudy (anonymous):

so V'=4(pi)r^2 and therefore dV/dr = 46.692 once i substitute r in

OpenStudy (campbell_st):

great... so then using the equation \[\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \] so \[2 = 46.692 \times \frac{dr}{dt} \] just solve for dr/dt

OpenStudy (anonymous):

awesome thank you so much. so would i take the same steps for part b?

OpenStudy (campbell_st):

ok... so for the 2nd part... you know dr/dt you need to find dS/dr.... using the surface area formula above....S = 4pi r^2 substitute your value of r.... you ahve calculated above then its a little easier \[\frac{dS}{dt} = \frac{dS}{dr} \times \frac{dr}{dt}\] hope this make sense

OpenStudy (anonymous):

you are a legend, thank you so much. Heaps of help!!!!!

OpenStudy (campbell_st):

lol... hope its all correct... and good luck

OpenStudy (anonymous):

do you mind checking part b. i got ds/dr=8(pi)(1.9276( which equals 48.446 so ds/dt = 48.446 x 0.0429 = 2.0783

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!