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.
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
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