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Mathematics 8 Online
OpenStudy (anonymous):

The radius of a spherical balloon is growing at the rate of 50 cm/s. How quickly is the volume increasing when the radius is 6 cm?

OpenStudy (campbell_st):

well you know these things \[\frac{dr}{dt} = 50\] you also know \[V = \frac{4}{3} \pi r^3\] so you then need to find the 1st derivative of volume with respect to the radius \[\frac{dV}{dr}\] to find the rate of change in volume with respect to time its \[\frac{dV}{dt}= \frac{dV}{dr} \times \frac{dr}{dt}\] the last task is to substitute r = 6 into dV/dt to find the answer. Hope it helps

OpenStudy (anonymous):

?

OpenStudy (campbell_st):

ok... so you haven't studied related rates..?

OpenStudy (campbell_st):

if you have studied calculus can you find the 1st derivative of the volume, with respect to r. which means finding dV/dr

OpenStudy (anonymous):

i dont get it.

OpenStudy (campbell_st):

so you don't get calculus... or related rates..?

OpenStudy (amistre64):

you dont get, anything that campbell is saying? what course is this in?

OpenStudy (amistre64):

we want how V chainges with respect to time, so i just do dV/dt implicitly and fill in the rest

OpenStudy (amistre64):

i spose %50 is the same regardless of how you approach it :)

OpenStudy (amistre64):

*50 ... my fingers hate me

OpenStudy (anonymous):

will my answer be 245 pi?

OpenStudy (campbell_st):

no it's not

OpenStudy (campbell_st):

can I ask, do you know how to differentiate..?

OpenStudy (anonymous):

yes,just confused with the porblem

OpenStudy (campbell_st):

ok... so make it smaller problems... if \[V = \frac{4}{3} \pi r^3\] can you find \[\frac{dV}{dr}\] or the 1st derivative of the volume with respect to r... ?

OpenStudy (campbell_st):

so the value \[\frac{4}{3} \pi\] is a coefficient like 5 in 5x so you are really just need to find the derivative of V = r^3

OpenStudy (anonymous):

216?

OpenStudy (campbell_st):

no, that is just substituting r = 6.... the substitution is the last step so I need you to do \[If~~V = r^3~~~what~~is ~~~\frac{dV}{dr} = \]

OpenStudy (radar):

Where did everybody go?

OpenStudy (anonymous):

idk

OpenStudy (campbell_st):

so any thoughts on what the derivative would be..?

OpenStudy (anonymous):

i was out sick for the class-so no not really

OpenStudy (campbell_st):

here is some information that may help you understand related rates http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx https://www.khanacademy.org/math/differential-calculus the notes are worth reading, and the Khan academy stuff contains notes and hopefully some videos... I think you need to understand differentiation before doing this question on related rates... Its about have sound basics... to achieve success

OpenStudy (anonymous):

what about this maths problem?

OpenStudy (campbell_st):

So I'll leave... I'm sure someone will provide you with an answer if the question is left open... But I think you need to do some revision 1st

OpenStudy (anonymous):

alright.

OpenStudy (radar):

Well it appears that knowing how to differentiate is prerequisite for related rates.

OpenStudy (radar):

If I differentiated for dV/dr do you think you could do the rest of the math?

OpenStudy (radar):

I just checked out those links provided by campbell_st, they are really good, especially the second one. I believe after watching several videos you will have no problem solving this problem.

OpenStudy (anonymous):

i will check it out.

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