# Related Rates

• Mar 25th 2007, 08:53 PM
camherokid
Related 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 PM
Jhevon
Quote:

Originally Posted by camherokid
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
• Mar 26th 2007, 12:23 AM
camherokid
Quote:

Originally Posted by Jhevon
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

They are looking for db/dt and the answer is -1.6cm/min
• Mar 26th 2007, 12:30 AM
Jhevon
Quote:

Originally Posted by camherokid
They are looking for db/dt and the answer is -1.6cm/min

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