subaruturbo_18

If x = 2t − ln(2t) and y = t2 − ln(t2) where t > 0, find the value of
t at the point on the curve at which the gradient is 2.


What the F.

corvid

Differenciate x and y. Divide dx/dt by dy/dt to get dy/dx.
dy/dx = 2.
Solve for t.

scud8

iTrader: (1)

I don't think the question is transcribed right. What is t2 in "y = t2 - ln(t2)"? If it is a different variable to t (ie. t subscript 2) then you can't solve the question as there is not way of relating t and t2. If it should be 2*t then x is always equal to y so there will be no point where the gradient is 2.

SJ_Skyline

or t squared?

It's over 20 years since I had to do any differentiation. I don't intend to start again now!

alcazar

iTrader: (2)

Nor me

stilover

If your too stupid not to know this simple equation, then I fear for your future.

In fact, it's so easy I'm not even going to bother telling you the answer.














































Probably because I don't have a ******* clue.

scud8

iTrader: (1)

Assuming t2 is t^2, the gradient is 2 when t=0 and t=2, so the answer is t=2. I'm not going to show my workings! Just follow corvid's instructions.