The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm2?

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The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm2?

Vaing1990 · Accepted Answer

Let h=altitude b=base given: dhdt=1 cmmin, dAdt=2 cm2min Find the rate of change of the base, dbdt, at the moment when h=10 cm, A=100 m3 Find the value of b at the same moment A=12bh 100=12b(10) b=20 cm Differentiate the triangle area equation with respect to t (in minutes) A=12bh dAdt=12(hdbdt+bdhdt) Product rule Now plug in all the know values 2=12(10dbdt+20(1)) 2=5dbdt+10 −5dbdt=8 dbdt=−1.6 cmmin It is negative, so it is decreasing. Result: −1.6 cmmin

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of

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