Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (goformit100):

Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm.

OpenStudy (goformit100):

@agent0smith @Noemi95 @Shannon20150 @91004775

OpenStudy (anonymous):

take the formula for the area of a circle you are trying to find dA/dr

OpenStudy (anonymous):

\[A=\pi r^2\]\[\frac{ dA }{ dt }=2 \pi r \frac{ dr }{ dt }\]

OpenStudy (anonymous):

Evaluate the derivative at r = 5. To find the the dA/dt however, we also need dr/dt, the rate at which the radius is changing.

OpenStudy (agent0smith):

Did they give a dr/dt...?

OpenStudy (anonymous):

with respect to radius, not with respect to time

OpenStudy (goformit100):

ok then

OpenStudy (agent0smith):

@Peter14 "Find the rate of change of the area of a circle per second with respect to its radius" The wording is kinda confusing, really...

OpenStudy (agent0smith):

Unless we just have to leave it in terms of dr/dt.

OpenStudy (anonymous):

OOps lol. Sorry I thought it was with respect to time. So it actually becomes:\[\frac{ dA }{ dr }=2 \pi r\] So at r = 5, it becomes:\[\frac{ dA }{ dr }=10 \pi\]

OpenStudy (anonymous):

And you need to include the units.

OpenStudy (agent0smith):

@genius12 that's what I meant about the misleading wording: "Find the rate of change of the area of a circle per second with respect to its radius"

OpenStudy (goformit100):

Thanks Sir

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!