Here's the actual question and my working so far:

Find the rate of change of I when t=3,

where I = 16 - 16(0.5)^t

Answer = ln 4

My working so far:

I = 16(1-0.5^t)

1 - \frac {I}{16}=0.5^t

ln 1 - \frac {I}{16} = t ln 0.5

When I differentiate the RHS wrt t I get ln 05.

But how do I differentiate the LHS wrt t?

Thanks.

Reply 5

vc94

You didn't ln the LHS correctly.
Any how, leave it as I = 16 - 16(0.5)^t and work out dI/dt.

a^t becomes (a^t)ln(a)

Reply 6

H3rrW4rum

Original post by vc94

You didn't ln the LHS correctly.
Any how, leave it as I = 16 - 16(0.5)^t and work out dI/dt.

a^t becomes (a^t)ln(a)

Why can you not just

ln (1 - (I/16)) ?

And how can you differentiate 16 - 16(0.5)^t straight away?

Reply 7

foolsihboy

\frac{d}{dt}f(I) = (\frac{df}{dI})(\frac{dI}{dt})

Reply 8

StephenP91

Original post by H3rrW4rum

Here's the actual question and my working so far:

Find the rate of change of I when t=3,

where I = 16 - 16(0.5)^t

Answer = ln 4

My working so far:

I = 16(1-0.5^t)

1 - \frac {I}{16}=0.5^t

ln 1 - \frac {I}{16} = t ln 0.5

When I differentiate the RHS wrt t I get ln 05.

But how do I differentiate the LHS wrt t?

Thanks.

Implicit Differentiation.

(edited 13 years ago)

Reply 9

H3rrW4rum

I know if

y = a^t

\frac {dy}{dx} = a^t ln a

So in this case

I = 16 - 16(0.5^t)

\frac {dI}{dt} = -16(0.5^t) ln 0.5

I just can't see how you get there.

EDIT: I can see it now, -16 is a constant you can just take out right? Then proceed as normal, put the -16 back in and you get the answer.