The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters. T = 2pi*sqrt(l/9.8) (a) How fast does the period increase? need some help please, these rate problems give me trouble...
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Originally Posted by mathlete The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters. T = 2pi*sqrt(l/9.8) (a) How fast does the period increase? need some help please, these rate problems give me trouble... With respect to what? If you double the length 2l then So doubling the length increases the period by
no sorry, it wants us to find dT/dl, so we're supposed to differentiate with respect to l, sorry...
Originally Posted by mathlete The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters. T = 2pi*sqrt(l/9.8) (a) How fast does the period increase? need some help please, these rate problems give me trouble... Now taking the derivative we get
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