Hard Maths? plz help...?

Question

I hav a few questions on differentiation...

I'm doing GCSE's soon, and need to be able to do them, so a good explanation would be helpful plz =]

The lead, "L" meters, of a runner in the last 75minutes of a marathon is given by the formula

L = 1000 + 6t - t^2/4, where t is time in minutes.

a) work out dL/dt

b) calculate the time at which the runner has the greatest lead

c) At what rate is the runer's lead being cut when t = 60minutes?

thanks a lot! =]

Answer 1

A. telling you tu differentiate tu the dL/dt.

so L= 6-2/4t^-1/2 (ask me if you didnt knw how i goh it )

B. calculate the time so make dL/dt = 0 and find the value of t.

...

now in the equation sub in 60 insted of the T then you work out the rate

Answer 2

L = 1000 + 6t - t^2/4

L = 1000 + 6t - (t^2)/4

dL/dt = 6 - t/2

0 = 6 -t/2

t = 12 minutes is the answer to b. This is a relative max. note the second derivative is negative.

dL/dt = 6 -30

dL/dt = -24 meter/min is the rate the lead is cut at t=60min.

Answer 3

To differentiate an equation like this, you need to know a set of rules of differentiation.  The only one that applies here is that

d t^n

------ = n*t^(n-1)

d t

You can split up the sections of the equation and differentiate them seperately, so:

d(1000).dt = 0,

d(6t)/dt = 6,

d(t^2/4)/dt = 2t/4 = t/2

So, dL/dt = 0+6-t/2 = 6-t/2.

when L is a maximum (or a minimum) the rate of change of L is 0, i.e.

dL/dt = 0 = 6-t/2,

so t/2 = 6, t=12 minutes.

When t=60, dL/dt = 6-60/2 = -24metres per minute (the minus sign means that his lead is being cut).