3√10^4x is equivalent to -√10 3/4^x and -4√10^3x they are all equivalent
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
Substitute 
, so that 
. Then the ODE is equivalent to
which is separable as
Split the left side into partial fractions,
so that integrating both sides is trivial and we get
Given the initial condition 
, we find
so that the ODE has the particular solution,
