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
Answer:
9(m+H) = a
Step-by-step explanation:
m=a/9-H
The first step is to add H to each side.
m+H = a/9 -H+H
m+H = a/9
Now multiply both side by 9
9(m+H) = a/9*9
9(m+H) = a
Answer:
d
Step-by-step explanation:
if u substitute -4 in the equation it will give u a math error
Start at 60
hike down 137 below so new is 60-137=-77m
then go down again 57
-77-57=-134
new altitude is -134m
21 because you have to have at least 1 side as 3m so the other two should be 9m to have the greatest possible length
