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 cm2 / min. at what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2?

Ya, calculus and related rates, such fun!

everything is changing with respect to t

altitude rate will be dh/dt and that is 1cm/min

dh/dt=1cm/min

area will be da/dt which is increasing at 2cm²/min

da/dt=2cm²/min

base=db/dt

alright

area=1/2bh

take dervitivie of both sides

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

solve for db/dt

distribute

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

move

da/dt-1/2 (dh/dt) (b) = 1/2 (db/dt) (h)

times 2 both sides

2da/dt - (dh/dt) (b) = (db/dt) (h)

divide by h

(2da/dt - (dh/dt) (b)) / h=db/dt

ok

we know

height=10

area=100

so

a=1/2bh

100=1/2b10

100=5b

20=b

so

h=10

b=20

da/dt=2cm²/min

dh/dt=1cm/min

therefor

(2 (2cm²/min) - (1cm/min) (20cm)) / 10cm=db/dt

(4cm²/min-20cm²/min) / 10cm=db/dt

(-16cm²/min) / 10cm=db/dt

-1.6cm/min=db/dt

the base is decreasing at 1.6cm/min