simple differentiation??

The population x, in thousands, of red squirrels is modelled by the equation
x = a/1+kt
where t is the time in years, and a and k are constants. When ,

Show that
dx/dt = -kx^2/a

can anyone do this step by step cos i'm clearly missing something simple, I'm using the quotient rule

Reply 1

CalculusMan

10

londongirl

The population x, in thousands, of red squirrels is modelled by the equation
x = a/1+kt
where t is the time in years, and a and k are constants. When ,

Is that 'When ,' meant to continue...?

Reply 2

gyrase

3

londongirl

The population x, in thousands, of red squirrels is modelled by the equation
x = a/1+kt
where t is the time in years, and a and k are constants. When ,

Show that
dx/dt = -kx^2/a

can anyone do this step by step cos i'm clearly missing something simple, I'm using the quotient rule

x = a/(1+kt)
.. = a(1+kt)^(-1)
dx/dt = -ak(1+kt)^(-2) = -ak/(1+kt)^2

dx/dt = -ak/(1+kt)^2

but OMG! (1+kt) = a/x, so dx/dt = -ak/(a/x)^2 = -k/(a/x^2) = -kx^2/a

Reply 3

yusufu

16

MrTrig

Is that 'When ,' meant to continue...?

Not necessary.
x = a/1+kt = a(1+kt)^-1
using chain rule (u = 1+kt):
dx/dt = -ak (1+kt)^-2
Now x/a = 1/(1+kt)
so dx/dt = -ak(x/a)^2 = (-kx^2)/a
QED

Reply 4

Rabite

8

a(1+kt)^-1
Goes to
-ak(1+kt)^-2
From the first thing:
1+kt=a/x
-ak(a/x)^-2

And that rearranges to your thingy. I think xD

[edit] Okay okay. You win this time.
But I still take the bronze. >>;;

simple differentiation??

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