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

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- Mar 25th 2007, 08:53 PMcamherokidRelated Rates
The altitude of a triangle is increasing at a rate of 1cm/min while the area of the triangle is in creasing at a rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100cm^2.

- Mar 25th 2007, 09:25 PMJhevon
Let A be the area, b be the base and h be the height (or altitude)

Let dh/dt be the rate at which the height is changing

Let dhA/dt be the rate at which the area is changing

Let db/dt be the rate at which the base is changing

A = (1/2)bh

when A = 100, h = 10

100 = (1/2)(10)b

=> b = 20

now, A = (1/2)bh

=> dA/dt = (1/2)h db/dt + (1/2)b dh/dt

=> dh/dt = [2dA/dt - h db/dt]/b

now plug in all the values we know

=> dh/dt = [2(2) - 10(1)]/20

=> dh/dt = -3/10

so the base is decreasing at a rate of -3/10 cm/min - Mar 26th 2007, 12:23 AMcamherokid
- Mar 26th 2007, 12:30 AMJhevon
sorry about that:

Let A be the area, b be the base and h be the height (or altitude)

Let dh/dt be the rate at which the height is changing

Let dhA/dt be the rate at which the area is changing

Let db/dt be the rate at which the base is changing

A = (1/2)bh

when A = 100, h = 10

100 = (1/2)(10)b

=> b = 20

now, A = (1/2)bh

=> dA/dt = (1/2)h db/dt + (1/2)b dh/dt

=> db/dt = [2dA/dt - b dh/dt]/h

now plug in all the values we know

=> db/dt = [2(2) - 20(1)]/10

=> db/dt = -16/10 = -8/5

=> db/dt = -1.6 cm/min

so the base is decreasing at a rate of -1.6 cm/min